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3.1: Formula Mass and the Mole Concept
What is the total mass (amu) of carbon in each of the following molecules?
- (a) CH 4
- (b) CHCl 3
- (c) C 12 H 10 O 6
- (d) CH 3 CH 2 CH 2 CH 2 CH 3
(a) 12.01 amu; (b) 12.01 amu; (c) 144.12 amu; (d) 60.05 amu
What is the total mass of hydrogen in each of the molecules?
- (a) CH 4
- (b) CHCl 3
- (c) C 12 H 10 O 6
- (d) CH 3 CH 2 CH 2 CH 2 CH 3
Calculate the molecular or formula mass of each of the following:
(a) P 4
(b) H 2 O
(c) Ca(NO 3 ) 2
(d) CH 3 CO 2 H (acetic acid)
(e) C 12 H 22 O 11 (sucrose, cane sugar).
(a) 123.896 amu; (b) 18.015 amu; (c) 164.086 amu; (d) 60.052 amu; (e) 342.297 amu
Determine the molecular mass of the following compounds:
(a)
(b)
(c)
(d)
Determine the molecular mass of the following compounds:
(a)
(b)
(c)
(d)
- (a) 56.107 amu;
- (b) 54.091 amu;
- (c) 199.9976 amu;
- (d) 97.9950 amu
Which molecule has a molecular mass of 28.05 amu?
(a)
(b)
(c)
Write a sentence that describes how to determine the number of moles of a compound in a known mass of the compound if we know its molecular formula.
Use the molecular formula to find the molar mass; to obtain the number of moles, divide the mass of compound by the molar mass of the compound expressed in grams.

- (a) Which has the largest number of molecules? Explain why.
- (b) Which has the greatest mass? Explain why.
Which contains the greatest mass of oxygen: 0.75 mol of ethanol (C 2 H 5 OH), 0.60 mol of formic acid (HCO 2 H), or 1.0 mol of water (H 2 O)? Explain why.
Formic acid. Its formula has twice as many oxygen atoms as the other two compounds (one each). Therefore, 0.60 mol of formic acid would be equivalent to 1.20 mol of a compound containing a single oxygen atom.
Which contains the greatest number of moles of oxygen atoms: 1 mol of ethanol (C 2 H 5 OH), 1 mol of formic acid (HCO 2 H), or 1 mol of water (H 2 O)? Explain why.
How are the molecular mass and the molar mass of a compound similar and how are they different?
The two masses have the same numerical value, but the units are different: The molecular mass is the mass of 1 molecule while the molar mass is the mass of 6.022 × 10 23 molecules.
Calculate the molar mass of each of the following compounds:
- (a) hydrogen fluoride, HF
- (b) ammonia, NH 3
- (c) nitric acid, HNO 3
- (d) silver sulfate, Ag 2 SO 4
- (e) boric acid, B(OH) 3
Calculate the molar mass of each of the following:
- (a) S 8
- (b) C 5 H 12
- (c) Sc 2 (SO 4 ) 3
- (d) CH 3 COCH 3 (acetone)
- (e) C 6 H 12 O 6 (glucose)
(a) 256.528 g/mol; (b) 72.150 g mol −1 ; (c) 378.103 g mol −1 ; (d) 58.080 g mol −1 ; (e) 180.158 g mol −1
Calculate the empirical or molecular formula mass and the molar mass of each of the following minerals:
- (a) limestone, CaCO 3
- (b) halite, NaCl
- (c) beryl, Be 3 Al 2 Si 6 O 18
- (d) malachite, Cu 2 (OH) 2 CO 3
- (e) turquoise, CuAl 6 (PO 4 ) 4 (OH) 8 (H 2 O) 4
Calculate the molar mass of each of the following:
- (a) the anesthetic halothane, C 2 HBrClF 3
- (b) the herbicide paraquat, C 12 H 14 N 2 Cl 2
- (c) caffeine, C 8 H 10 N 4 O 2
- (d) urea, CO(NH 2 ) 2
- (e) a typical soap, C 17 H 35 CO 2 Na
(a) 197.382 g mol −1 ; (b) 257.163 g mol −1 ; (c) 194.193 g mol −1 ; (d) 60.056 g mol −1 ; (e) 306.464 g mol −1
Determine the number of moles of compound and the number of moles of each type of atom in each of the following:
- (a) 25.0 g of propylene, C 3 H 6
- (b) 3.06 × 10 −3 g of the amino acid glycine, C 2 H 5 NO 2
- (c) 25 lb of the herbicide Treflan, C 13 H 16 N 2 O 4 F (1 lb = 454 g)
- (d) 0.125 kg of the insecticide Paris Green, Cu 4 (AsO 3 ) 2 (CH 3 CO 2 ) 2
- (e) 325 mg of aspirin, C 6 H 4 (CO 2 H)(CO 2 CH 3 )
Determine the mass of each of the following:
- (a) 0.0146 mol KOH
- (b) 10.2 mol ethane, C 2 H 6
- (c) 1.6 × 10 −3 mol Na 2 SO 4
- (d) 6.854 × 10 3 mol glucose, C 6 H 12 O 6
- (e) 2.86 mol Co(NH 3 ) 6 Cl 3
- (a) 0.819 g;
- (b) 307 g;
- (c) 0.23 g;
- (d) 1.235 × 10 6 g (1235 kg);
- (e) 765 g
Determine the number of moles of the compound and determine the number of moles of each type of atom in each of the following:
- (a) 2.12 g of potassium bromide, KBr
- (b) 0.1488 g of phosphoric acid, H 3 PO 4
- (c) 23 kg of calcium carbonate, CaCO 3
- (d) 78.452 g of aluminum sulfate, Al 2 (SO 4 ) 3
- (e) 0.1250 mg of caffeine, C 8 H 10 N 4 O 2
Determine the mass of each of the following:
- (a) 2.345 mol LiCl
- (b) 0.0872 mol acetylene, C 2 H 2
- (c) 3.3 × 10 −2 mol Na 2 CO 3
- (d) 1.23 × 10 3 mol fructose, C 6 H 12 O 6
- (e) 0.5758 mol FeSO 4 (H 2 O) 7
- (a) 99.41 g;
- (b) 2.27 g;
- (c) 3.5 g;
- (d) 222 kg;
- (e) 160.1 g
The approximate minimum daily dietary requirement of the amino acid leucine, C 6 H 13 NO 2 , is 1.1 g. What is this requirement in moles?
Determine the mass in grams of each of the following:
- (a) 0.600 mol of oxygen atoms
- (b) 0.600 mol of oxygen molecules, O 2
- (c) 0.600 mol of ozone molecules, O 3
(a) 9.60 g; (b) 19.2 g; (c) 28.8 g
A 55-kg woman has 7.5 × 10 −3 mol of hemoglobin (molar mass = 64,456 g/mol) in her blood. How many hemoglobin molecules is this? What is this quantity in grams?
Determine the number of atoms and the mass of zirconium, silicon, and oxygen found in 0.3384 mol of zircon, ZrSiO 4 , a semiprecious stone.
zirconium: 2.038 × 10 23 atoms; 30.87 g; silicon: 2.038 × 10 23 atoms; 9.504 g; oxygen: 8.151 × 10 23 atoms; 21.66 g
Determine which of the following contains the greatest mass of hydrogen: 1 mol of CH 4 , 0.6 mol of C 6 H 6 , or 0.4 mol of C 3 H 8 .
Determine which of the following contains the greatest mass of aluminum: 122 g of AlPO 4 , 266 g of Al 2 Cl 6 , or 225 g of Al 2 S 3 .
AlPO 4 : 1.000 mol
Al 2 Cl 6 : 1.994 mol
Al 2 S 3 : 3.00 mol
Diamond is one form of elemental carbon. An engagement ring contains a diamond weighing 1.25 carats (1 carat = 200 mg). How many atoms are present in the diamond?
The Cullinan diamond was the largest natural diamond ever found (January 25, 1905). It weighed 3104 carats (1 carat = 200 mg). How many carbon atoms were present in the stone?
3.113 × 10 25 C atoms
One 55-gram serving of a particular cereal supplies 270 mg of sodium, 11% of the recommended daily allowance. How many moles and atoms of sodium are in the recommended daily allowance?
A certain nut crunch cereal contains 11.0 grams of sugar (sucrose, C 12 H 22 O 11 ) per serving size of 60.0 grams. How many servings of this cereal must be eaten to consume 0.0278 moles of sugar?
0.865 servings, or about 1 serving.
A tube of toothpaste contains 0.76 g of sodium monofluorophosphate (Na 2 PO 3 F) in 100 mL.
- What mass of fluorine atoms in mg was present?
- How many fluorine atoms were present?
Which of the following represents the least number of molecules?
- 20.0 g of H 2 O (18.02 g/mol)
- 77.0 g of CH 4 (16.06 g/mol)
- 68.0 g of CaH 2 (42.09 g/mol)
- 100.0 g of N 2 O (44.02 g/mol)
- 84.0 g of HF (20.01 g/mol)
20.0 g H 2 O represents the least number of molecules since it has the least number of moles.
3.2: Determining Empirical and Molecular Formulas
What information do we need to determine the molecular formula of a compound from the empirical formula?
Calculate the following to four significant figures:
- (a) the percent composition of ammonia, NH 3
- (b) the percent composition of photographic “hypo,” Na 2 S 2 O 3
- (c) the percent of calcium ion in Ca 3 (PO 4 ) 2
(a) % N = 82.24%
% H = 17.76%;
(b) % Na = 29.08%
% S = 40.56%
% O = 30.36%;
(c) % Ca 2+ = 38.76%
Determine the following to four significant figures:
- the percent composition of hydrazoic acid, HN 3
- the percent composition of TNT, C 6 H 2 (CH 3 )(NO 2 ) 3
- the percent of SO 4 2– in Al 2 (SO 4 ) 3
Determine the percent ammonia, NH 3 , in Co(NH 3 ) 6 Cl 3 , to three significant figures.
% NH 3 = 38.2%
Determine the percent water in CuSO 4 ∙5H 2 O to three significant figures.
Determine the empirical formulas for compounds with the following percent compositions:
(a) 15.8% carbon and 84.2% sulfur
(b) 40.0% carbon, 6.7% hydrogen, and 53.3% oxygen
(a) CS 2
(b) CH 2 O
Determine the empirical formulas for compounds with the following percent compositions:
(a) 43.6% phosphorus and 56.4% oxygen
(b) 28.7% K, 1.5% H, 22.8% P, and 47.0% O
A compound of carbon and hydrogen contains 92.3% C and has a molar mass of 78.1 g/mol. What is its molecular formula?
C 6 H 6
Dichloroethane, a compound that is often used for dry cleaning, contains carbon, hydrogen, and chlorine. It has a molar mass of 99 g/mol. Analysis of a sample shows that it contains 24.3% carbon and 4.1% hydrogen. What is its molecular formula?
Determine the empirical and molecular formula for chrysotile asbestos. Chrysotile has the following percent composition: 28.03% Mg, 21.60% Si, 1.16% H, and 49.21% O. The molar mass for chrysotile is 520.8 g/mol.
Mg 3 Si 2 H 3 O 8 (empirical formula), Mg 6 Si 4 H 6 O 16 (molecular formula)
Polymers are large molecules composed of simple units repeated many times. Thus, they often have relatively simple empirical formulas. Calculate the empirical formulas of the following polymers:
- Lucite (Plexiglas); 59.9% C, 8.06% H, 32.0% O
- Saran; 24.8% C, 2.0% H, 73.1% Cl
- polyethylene; 86% C, 14% H
- polystyrene; 92.3% C, 7.7% H
- Orlon; 67.9% C, 5.70% H, 26.4% N
A major textile dye manufacturer developed a new yellow dye. The dye has a percent composition of 75.95% C, 17.72% N, and 6.33% H by mass with a molar mass of about 240 g/mol. Determine the molecular formula of the dye.
C 15 H 15 N 3
3.3: Molarity
Questions
Explain what changes and what stays the same when 1.00 L of a solution of NaCl is diluted to 1.80 L.
What information do we need to calculate the molarity of a sulfuric acid solution?
We need to know the number of moles of sulfuric acid dissolved in the solution and the volume of the solution.
What does it mean when we say that a 200-mL sample and a 400-mL sample of a solution of salt have the same molarity? In what ways are the two samples identical? In what ways are these two samples different?
Determine the molarity for each of the following solutions:
- 0.444 mol of CoCl 2 in 0.654 L of solution
- 98.0 g of phosphoric acid, H 3 PO 4 , in 1.00 L of solution
- 0.2074 g of calcium hydroxide, Ca(OH) 2 , in 40.00 mL of solution
- 10.5 kg of Na 2 SO 4 ·10H 2 O in 18.60 L of solution
- 7.0 × 10 −3 mol of I 2 in 100.0 mL of solution
- 1.8 × 10 4 mg of HCl in 0.075 L of solution
- (a) 0.679 M ;
- (b) 1.00 M ;
- (c) 0.06998 M ;
- (d) 1.75 M ;
- (e) 0.070 M ;
- (f) 6.6 M
Determine the molarity of each of the following solutions:
- 1.457 mol KCl in 1.500 L of solution
- 0.515 g of H 2 SO 4 in 1.00 L of solution
- 20.54 g of Al(NO 3 ) 3 in 1575 mL of solution
- 2.76 kg of CuSO 4 ·5H 2 O in 1.45 L of solution
- 0.005653 mol of Br 2 in 10.00 mL of solution
- 0.000889 g of glycine, C 2 H 5 NO 2 , in 1.05 mL of solution
Answers:
a.) 0.9713 M
b.) 5.25x10 -3 M
c.) 6.122x10 -2 M
d.) 7.62 M
e.) 0.5653 M
f.) 1.13x10 -2 M
Consider this question: What is the mass of the solute in 0.500 L of 0.30 M glucose, C 6 H 12 O 6 , used for intravenous injection?
(a) Outline the steps necessary to answer the question.
(b) Answer the question.
(a) determine the number of moles of glucose in 0.500 L of solution; determine the molar mass of glucose; determine the mass of glucose from the number of moles and its molar mass; (b) 27 g
Consider this question: What is the mass of solute in 200.0 L of a 1.556- M solution of KBr?
- (a) Outline the steps necessary to answer the question.
- (b) Answer the question.
Answer:
(a)
- Calculate to moles of KBr by multiplying the Molarity by the amount of solution (200.0 L)
- Find the Molar Mass of KBr and convert moles of solute to grams
(b)
\(\dfrac{1.556\:moles\:\ce{KBr}}{1\:\cancel{L}}\times 200.0\:\cancel{L}=311.2\:moles\:\ce{KBr}\)
\(311.2\:\cancel{moles}\:\ce{KBr}\times\dfrac{119.0\:g\:\ce{KBr}}{1\:\cancel{mole}\:\ce{KBr}}=37,030\:g\)
37,030g; 37.03 kg
Calculate the number of moles and the mass of the solute in each of the following solutions:
- (a) 2.00 L of 18.5 M H 2 SO 4 , concentrated sulfuric acid
- (b) 100.0 mL of 3.8 × 10 −5 M NaCN, the minimum lethal concentration of sodium cyanide in blood serum
- (c) 5.50 L of 13.3 M H 2 CO, the formaldehyde used to “fix” tissue samples
- (d) 325 mL of 1.8 × 10 −6 M FeSO 4 , the minimum concentration of iron sulfate detectable by taste in drinking water
(a) 37.0 mol H 2 SO 4 ;
3.63 × 10 3 g H 2 SO 4 ;
(b) 3.8 × 10 −6 mol NaCN;
1.9 × 10 −4 g NaCN;
(c) 73.2 mol H 2 CO;
2.20 kg H 2 CO;
(d) 5.9 × 10 −7 mol FeSO 4 ;
8.9 × 10 −5 g FeSO 4
Calculate the number of moles and the mass of the solute in each of the following solutions:
- 325 mL of 8.23 × 10 −5 M KI, a source of iodine in the diet
- 75.0 mL of 2.2 × 10 −5 M H 2 SO 4 , a sample of acid rain
- 0.2500 L of 0.1135 M K 2 CrO 4 , an analytical reagent used in iron assays
- 10.5 L of 3.716 M (NH 4 ) 2 SO 4 , a liquid fertilizer
Answers:
a. 2.67x10 -5 moles KI; 4.44x10 -3 g KI
b. 1.7x10 -6 moles H 2 SO 4 ; 1.6x10 -4 g H 2 SO 4
c. 2.838x10 -2 moles K 2 CrO 4 ; 5.510g K 2 CrO 4
d. 39.0 moles (NH 4 ) 2 SO 4 ; 5,160 g (NH 4 ) 2 SO 4
Consider this question: What is the molarity of KMnO 4 in a solution of 0.0908 g of KMnO 4 in 0.500 L of solution?
- (a) Outline the steps necessary to answer the question.
- (b) Answer the question.
Consider this question: What is the molarity of HCl if 35.23 mL of a solution of HCl contain 0.3366 g of HCl?
- (a) Outline the steps necessary to answer the question.
- (b) Answer the question.
Answer:
(a)
- Convert g of HCl to moles of HCl and convert mL of solution to L of solution
- Divide moles of HCl by L of solution
(b)
\(0.3366\:\cancel{g}\:\ce{HCl}\times\dfrac{1\:mole\:\ce{HCl}}{36.46\:\cancel{g}\:\ce{HCl}}=9.232\times10^{-3}\:moles\:\ce{HCl}\)
\(35.23\:mL = 0.03523\:L\)
\(\dfrac{9.232\times10^{-3}\:moles\:\ce{HCl}}{0.03523\:L}=0.2621\:M\:\ce{HCl}\)
0.2621 M ;
Calculate the molarity of each of the following solutions:
(a) 0.195 g of cholesterol, C 27 H 46 O, in 0.100 L of serum, the average concentration of cholesterol in human serum
(b) 4.25 g of NH 3 in 0.500 L of solution, the concentration of NH 3 in household ammonia
(c) 1.49 kg of isopropyl alcohol, C 3 H 7 OH, in 2.50 L of solution, the concentration of isopropyl alcohol in rubbing alcohol
(d) 0.029 g of I 2 in 0.100 L of solution, the solubility of I 2 in water at 20 °C
(a) 5.04 × 10 −3 M ;
(b) 0.499 M ;
(c) 9.92 M ;
(d) 1.1 × 10 −3 M
Calculate the molarity of each of the following solutions:
- 293 g HCl in 666 mL of solution, a concentrated HCl solution
- 2.026 g FeCl 3 in 0.1250 L of a solution used as an unknown in general chemistry laboratories
- 0.001 mg Cd 2+ in 0.100 L, the maximum permissible concentration of cadmium in drinking water
- 0.0079 g C 7 H 5 SNO 3 in one ounce (29.6 mL), the concentration of saccharin in a diet soft drink.
There is about 1.0 g of calcium, as Ca 2+ , in 1.0 L of milk. What is the molarity of Ca 2+ in milk?
0.025 M
What volume of a 1.00- M Fe(NO 3 ) 3 solution can be diluted to prepare 1.00 L of a solution with a concentration of 0.250 M ?
If 0.1718 L of a 0.3556- M C 3 H 7 OH solution is diluted to a concentration of 0.1222 M , what is the volume of the resulting solution?
0.5000 L
If 4.12 L of a 0.850 M -H 3 PO 4 solution is be diluted to a volume of 10.00 L, what is the concentration the resulting solution?
What volume of a 0.33- M C 12 H 22 O 11 solution can be diluted to prepare 25 mL of a solution with a concentration of 0.025 M ?
1.9 mL
What is the concentration of the NaCl solution that results when 0.150 L of a 0.556- M solution is allowed to evaporate until the volume is reduced to 0.105 L?
What is the molarity of the diluted solution when each of the following solutions is diluted to the given final volume?
- (a) 1.00 L of a 0.250- M solution of Fe(NO 3 ) 3 is diluted to a final volume of 2.00 L
- (b) 0.5000 L of a 0.1222- M solution of C 3 H 7 OH is diluted to a final volume of 1.250 L
- (c) 2.35 L of a 0.350- M solution of H 3 PO 4 is diluted to a final volume of 4.00 L
- (d) 22.50 mL of a 0.025- M solution of C 12 H 22 O 11 is diluted to 100.0 mL
- (a) 0.125 M ;
- (b) 0.04888 M ;
- (c) 0.206 M ;
- (e) 0.0056 M
What is the final concentration of the solution produced when 225.5 mL of a 0.09988- M solution of Na 2 CO 3 is allowed to evaporate until the solution volume is reduced to 45.00 mL?
A 2.00-L bottle of a solution of concentrated HCl was purchased for the general chemistry laboratory. The solution contained 868.8 g of HCl. What is the molarity of the solution?
11.9 M
An experiment in a general chemistry laboratory calls for a 2.00- M solution of HCl. How many mL of 11.9 M HCl would be required to make 250 mL of 2.00 M HCl?
What volume of a 0.20- M K 2 SO 4 solution contains 57 g of K 2 SO 4 ?
1.6 L
The US Environmental Protection Agency (EPA) places limits on the quantities of toxic substances that may be discharged into the sewer system. Limits have been established for a variety of substances, including hexavalent chromium, which is limited to 0.50 mg/L. If an industry is discharging hexavalent chromium as potassium dichromate (K 2 Cr 2 O 7 ), what is the maximum permissible molarity of that substance?
3.4: Other Units for Solution Concentrations
Questions
-
Consider this question: What mass of a concentrated solution of nitric acid (68.0% HNO3 by mass) is needed to prepare 400.0 g of a 10.0% solution of HNO3 by mass?
- Outline the steps necessary to answer the question.
- Answer the question.
- What mass of a 4.00% NaOH solution by mass contains 15.0 g of NaOH?
- What mass of solid NaOH (97.0% NaOH by mass) is required to prepare 1.00 L of a 10.0% solution of NaOH by mass? The density of the 10.0% solution is 1.109 g/mL.
- What mass of HCl is contained in 45.0 mL of an aqueous HCl solution that has a density of 1.19 g cm–3 and contains 37.21% HCl by mass?
- The hardness of water (hardness count) is usually expressed in parts per million (by mass) of \(\ce{CaCO_3}\), which is equivalent to milligrams of \(\ce{CaCO_3}\) per liter of water. What is the molar concentration of Ca 2+ ions in a water sample with a hardness count of 175 mg CaCO 3 /L?
- The level of mercury in a stream was suspected to be above the minimum considered safe (1 part per billion by weight). An analysis indicated that the concentration was 0.68 parts per billion. Assume a density of 1.0 g/mL and calculate the molarity of mercury in the stream.
- In Canada and the United Kingdom, devices that measure blood glucose levels provide a reading in millimoles per liter. If a measurement of 5.3 mM is observed, what is the concentration of glucose (C 6 H 12 O 6 ) in mg/dL?
- A throat spray is 1.40% by mass phenol, \(\ce{C_6H_5OH}\), in water. If the solution has a density of 0.9956 g/mL, calculate the molarity of the solution.
- Copper(I) iodide (CuI) is often added to table salt as a dietary source of iodine. How many moles of CuI are contained in 1.00 lb (454 g) of table salt containing 0.0100% CuI by mass?
- A cough syrup contains 5.0% ethyl alcohol, C 2 H 5 OH, by mass. If the density of the solution is 0.9928 g/mL, determine the molarity of the alcohol in the cough syrup.
- D5W is a solution used as an intravenous fluid. It is a 5.0% by mass solution of dextrose (\(\ce{C_6H_{12}O_6}\)) in water. If the density of D5W is 1.029 g/mL, calculate the molarity of dextrose in the solution.
- Find the molarity of a 40.0% by mass aqueous solution of sulfuric acid, \(\ce{H_2SO_4}\), for which the density is 1.3057 g/mL.
Solutions
1
- (a) The dilution equation can be used, appropriately modified to accommodate mass-based concentration units: \[\mathrm{\%\,mass_1 \times mass_1=\%\;mass_2 \times mass_2}\] This equation can be rearranged to isolate \(\mathrm{mass_1}\) and the given quantities substituted into this equation.
- (b) 58.8 g


11.12: Discussion- Rethinking Motivation
Post 1: Initial Post
Instructions
This posting should be a minimum of one short paragraph and a maximum of two paragraphs. Word totals for this post should be in the 100–200-word range. Whether you agree or disagree, explain why with supporting evidence and concepts from the readings or a related experience. Include a reference, link, or citation when appropriate.
Preparation
This discussion is based on a TED Talk by Dan Pink, as well as drawing on readings from the Motivation module. Pink, a former speechwriter, career analyst, and author of the book Drive: The Surprising Truth About What Motivates Us , reveals some remarkable facts about motivation that have been confirmed in many social science experiments, but aren’t widely recognized. Pink argues that the way we’re motivating people is all wrong, and he proposes a new model to replace traditional incentives.
Watch Pink’s TED Talk, below:
You can view the transcript for “The Puzzle of Motivation” (opens in new window) or the text alternative for “The Puzzle of Motivation | Dan Pink” (opens in new window).
For Discussion
Please respond to both of the following questions in your initial post.
- Did you find Pink’s talk surprising? Why or why not?
-
Pink concludes:
There is a mismatch between what science knows and what business does. Here is what science knows. One: Those 20th century rewards, those motivators we think are a natural part of business, do work, but only in a surprisingly narrow band of circumstances. Two: Those if-then rewards often destroy creativity. Three: The secret to high performance isn’t rewards and punishments, but that unseen intrinsic drive–the drive to do things for their own sake. The drive to do things cause they matter.
How do you think managers can use this knowledge to motivate their employees? How will you use this knowledge in your own pursuits?
Posts 2 and 3: Respond to Classmates’ Posts
Instructions
After you have created your initial post, look over the discussion posts of your classmates and give at least two thoughtful responses to two different classmates (one per classmate).
A response posting should be a minimum of one short paragraph. Word totals for these posts should be in the 75–100-word range. Whether you agree or disagree, explain why with supporting evidence and concepts from the readings or a related experience. Include a reference, link, or citation when appropriate. The goal of your response posts is to extend discussions already taking place or pose new possibilities or opinions not previously voiced. Your goal should be to motivate the group discussion and present a creative approach to the topic.
Contributors and Attributions
- Discussion: Rethinking Motivation. Authored by : Lumen Learning. License : CC BY: Attribution
- Dan Pink: The Puzzle of Motivation. Provided by : TED Talks. Located at : https://www.ted.com/talks/dan_pink_on_motivation . License : CC BY-NC-ND: Attribution-NonCommercial-NoDerivatives
- General Discussion Instructions. Authored by : Linda Williams and Lumen Learning. Located at : https://courses.lumenlearning.com/wmopen-introbusiness/chapter/discussion-external-factors-internal-pressure/ . License : CC BY: Attribution


7.10: Types of Love
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Sternberg’s Triangle of Love: Three Components
Sternberg (1988) suggests that there are three main components of love: passion, intimacy, and commitment. Love relationships vary depending on the presence or absence of each of these components. Passion refers to the intense, physical attraction partners feel toward one another. Intimacy involves the ability the share feelings, personal thoughts and psychological closeness with the other. Commitment is the conscious decision to stay together. Passion can be found in the early stages of a relationship, but intimacy takes time to develop because it is based on knowledge of the partner. Once intimacy has been established, partners may resolve to stay in the relationship. Although many would agree that all three components are important to a relationship, many love relationships do not consist of all three. Let’s look at other possibilities.
Image shows the 3 properties of love and how it creates the 7 different loves according to theory. (Public Domain; AnonMoos ).
Liking : In this relationship, intimacy or knowledge of the other and a sense of closeness is present. Passion and commitment, however, are not. Partners feel free to be themselves and disclose personal information. They may feel that the other person knows them well and can be honest with them and let them know if they think the person is wrong. These partners are friends. However, being told that your partner ‘thinks of you as a friend’ can be a devastating blow if you are attracted to them and seek a romantic involvement.
Infatuation : Perhaps, this is Sternberg’s version of “love at first sight”. Infatuation consists of an immediate, intense physical attraction to someone. A person who is infatuated finds it hard to think of anything but the other person. Brief encounters are played over and over in one’s head; it may be difficult to eat and there may be a rather constant state of arousal. Infatuation is rather short-lived, however, lasting perhaps only a matter of months or as long as a year or so. It tends to be based on chemical attraction and an image of what one thinks the other is all about.
Fatuous Love : However, some people who have a strong physical attraction push for commitment early in the relationship. Passion and commitment are aspects of fatuous love. There is no intimacy and the commitment is premature. Partners rarely talk seriously or share their ideas. They focus on their intense physical attraction and yet one, or both, is also talking of making a lasting commitment. Sometimes this is out of a sense of insecurity and a desire to make sure the partner is locked into the relationship.
Empty Love : This type of love may be found later in a relationship or in a relationship that was formed to meet needs other than intimacy or passion (money, childrearing, status). Here the partners are committed to staying in the relationship (for the children, because of a religious conviction, or because there are no alternatives perhaps), but do not share ideas or feelings with each other and have no physical attraction for one another.
Romantic Love : Intimacy and passion are components of romantic love, but there is no commitment. The partners spend much time with one another and enjoy their closeness but have not made plans to continue ‘no matter what’. This may be true because they are not in a position to make such commitments or because they are looking for passion and closeness and are afraid it will die out if they commit to one another and start to focus on other kinds of obligations.
Companionate Love : Intimacy and commitment are the hallmarks of companionate love. Partners love and respect one another and they are committed to staying together. But their physical attraction may have never been strong or may have just died out. This may be interpreted as ‘just the way things are’ after so much time together or there may be a sense of regret and loss. Nevertheless, partners are good friends committed to one another.
Consummate Love : Intimacy, passion, and commitment are present in consummate love. This is often the ideal type of love. The couple shares passion; the spark has not died, and the closeness is there. They feel like best friends as well as lovers and they are committed to staying together.
Types of Lovers
Lee (1973) offers a theory of love styles or types of lovers derived from an analysis of writings about love through the centuries. As you read these, think about how these styles might become part of the types of love described above.
- Pragma is a style of love that emphasizes the practical aspects of love. The pragmatic lover considers compatibility and the sensibility of their choice of partners. This lover will be concerned with goals in life, status, family reputation, attitudes about parenting, career issues and other practical concerns.
- Mania is a style of love characterized by volatility, insecurity, and possessiveness. This lover gets highly upset during arguments or breakups, may have trouble sleeping when in love, and feels emotions very intensely.
- Agape is an altruistic, selfless love. These partners give of themselves without expecting anything in return. Such a lover places the partner’s happiness above their own and is self-sacrificing to benefit the partner.
- Eros is an erotic style of loving in which the person feels consumed. Physical chemistry and emotional involvement are important to this type of lover.
- Ludus refers to a style of loving that emphasizes the game of seduction and fun. Such a lover stays away from commitment and often has several love interests at the same time. This lover does not self-disclose and in fact may prefer to keep the other guessing. This lover can end a relationship easily.
- Storge is a style of love that develops slowly over time. It often begins as a friendship and becomes sexual much later. These partners are likely to remain friends even after the breakup.
Frames of Relationships
| A | H | M |
Another useful way to consider relationships is to consider the amount of dependency in the relationship. Davidson (1991) suggests three models. The A-frame relationship is one in which the partners lean on one another and are highly dependent on the other for survival. If one partner changes, the other is at risk of ‘falling over’. This type of relationship cannot easily accommodate change and the partners are vulnerable should change occur. A breakup could be devastating.
The H-frame relationship is one in which the partners live parallel lives. They rarely spend time with one another and tend to have separate lives. What time they do share is usually spent meeting obligations rather than sharing intimacies. This independent type of relationship can end without suffering emotionally.
The M-frame relationship is interdependent. Partners have a strong sense of connection but also are able to stand alone without suffering devastation. If this relationship ends, partners will be hurt and saddened, but will still be able to stand alone. This ability comes from a strong sense of self-love. Partners can love each other without losing a sense of self. And each individual has self-respect and confidence that enriches the relationship as well as strengthens the self.
We have been looking at love in the context of many kinds of relationships. In our next lesson, we will focus more specifically on marital relationships. But before we do, we examine the dynamics of falling in and out of love.
The Process of Love and Breaking Up
Reiss (1960) provides a theory of love as process. Based on the wheel theory of love, love relationships begin with the establishment of rapport.
Rapport involves sharing likes, preferences, establishing some common interests. The next step is to begin to disclose more personal information through self-revelation. When one person begins to open up, the social expectation is that the other will follow and also share more personal information so that each has made some risk and trust is built. Sexual intimacy may also become part of the relationship. Gradually, partners begin to disclose even more about themselves and are met with support and acceptance as they build mutual dependency. With time, partners come to rely on each other for need fulfillment. The wheel must continue in order for love to last. It becomes important for partners to continue to establish rapport by discussing the day’s events, communicating about their goals and desires, and showing signs of trust. Partners must continue to rely on one another to have certain needs fulfilled. If the wheel turns backward, partners talk less and less, rely less on one another and are less likely to disclose.
Process of Disaffection: Breaking Up
When relationships are new, partners tend to give one another the benefit of the doubt and focus on what they like about one another. Flaws and imperfections do not go unnoticed; rather, they are described as endearing qualities. So, for example, the partner who has a very large nose is described as ‘distinguished’ or as having a ‘striking feature.’ This is very exhilarating because features that someone may have previously felt self-conscious about are now accepted or even appreciated. However, once partners begin the process of breaking up, these views are abandoned and questionable qualities are once again flaws and imperfections.
Kersten (1990) provides a look at the dynamics of breaking up. Although this work is primarily about divorce, the dynamics of dissolving any long-term relationship are similar. The beginning phase of breaking up involves seeing imperfections in the relationship but remaining hopeful that things will improve. This improvement will require the partner’s cooperation because they are primarily at fault. So, as long as the offending partner makes the necessary changes, and of course the offended partner will provide the advice, support, and guidance required, the relationship will continue. (If you are thinking that this is not going to work-you are right. Attempts to change one’s partner are usually doomed to failure. Would you want your partner to try to change you?)
Once it becomes clear that efforts to change are futile, the middle phase is entered. This phase is marked by disappointment. Partners talk less and less, make little eye contact, and grow further apart. One may still try to make contact, but the other is clearly disengaged and is considering the benefits and costs of leaving the relationship.
In the end phase, the decision to leave has been made. The specific details are being worked out. Turning a relationship around is very difficult at this point. Trust has diminished, and thoughts have turned elsewhere. This stage is one of hopelessness.
We will explore marriage, divorce, and cohabitation more fully in our next lesson.


5: Circuit Components and their Schematic Symbols
All electrical components are rated with different current, voltage, and other values, depending on their use. Replacing an electrical component with one of a different value could present a serious safety hazard to the person using the equipment. Should you have to replace an electrical component at any time, always be sure that the current, voltage, and other electrical ratings of the replacement match those of the original.
In addition, to ensure public safety from electrical and fire hazards, the Canadian Standards Association (CSA) approves electrical components. Each component is tested before it can be sold on the Canadian market. Always use only CSA-approved equipment. The CSA approval should be clearly visible on the component or the package.
You may have to assemble devices in a circuit from plans or drawings in which symbols are used to represent basic electrical devices and components. Although some plans use pictures of the devices instead of symbols, you should be familiar with the symbols so that you can identify each individual device. Not all electrical symbols are standard, but most symbols can be easily recognized.


10.10: Controlling Your Search
Now that you have learned about tools and some of the issues surrounding searching online, you need to be able to control your search. Just like with library databases, there are strategies you can employ with online searches. Recall that in our chapter scenario , you needed to find information about antiracism efforts of government and private organizations. When forming your search for this scenario, you might use the following keywords: antiracism, government, organizations, policies, and efforts. But you might want to take advantage of some more advanced search commands.
Limiting by Domain
One search command to make note of is site:. This command allows you to search for, or within, specific websites based on domain name. For our scenario, we need government and organizations, so it would be helpful to limit our search to those types of websites. You can set up your search to look something like this:
antiracism policies site:gov
This search example will look for the keywords antiracism and policies within government websites only. The site: command will limit results to those with gov in the website address (also known as the URL). The domain gov is reserved for U.S. government websites.
Other common domains that you could use include .com, .org, and .edu. The domain .org stands for “organization,” however, anyone can use the .org domain, not just non-profit organizations. It is very important to use these domain reservations as guidelines rather than definitive markers of credibility. For example, an .edu website (reserved for educational institutions) could host papers written by students, which might not be appropriate sources for a college-level research paper. Likewise, an .org website could be hosted by a commercial enterprise or even a hate group.
Limiting by File Type
The scenario also asks for online documents. The meaning of this word, “documents,” can be vague and unclear, but its use in the scenario indicates that we’re looking for something you can download, such as a PDF or maybe a Word document. You can also limit your results to file types when searching by using the filetype command. Your search would look something like this:
antiracism policies site:gov filetype:pdf
With the search example above, you will be limiting your search results to government websites that have PDFs related to antiracism policies.
Other Filters
In addition to those commands, there are easy-to-find tools built into search engines. You can search for images, news, maps and more when you search online. You can limit your results to the past year, the past hour, or other time periods. All of these additional limiters can be very powerful and helpful when conducting research, depending on your topic. Explore your search engine to see how much control you can have over your online searching.


5.E: Acid-Base Equilibria (Exercises)
Brønsted-Lowry Acids and Bases
1. Write equations that show NH 3 as both a conjugate acid and a conjugate base.
- Answer
- One example for NH 3 as a conjugate acid: \(\ce{NH2- + H+ ⟶ NH3}\); as a conjugate base: \(\ce{NH4+}(aq)+\ce{OH-}(aq)⟶\ce{NH3}(aq)+\ce{H2O}(l)\)
2. Write equations that show \(\ce{H2PO4-}\) acting both as an acid and as a base.
3. Show by suitable net ionic equations that each of the following species can act as a Brønsted-Lowry acid:
- \(\ce{H3O+}\)
- HCl
- NH 3
- CH 3 CO 2 H
- \(\ce{NH4+}\)
- \(\ce{HSO4-}\)
- Answer
- a. \(\ce{H3O+}(aq)⟶\ce{H+}(aq)+ \ce{H_2O}_{(l)}\) ;
- b. \(\ce{HCl}(l)⟶\ce{H+}(aq)+\ce{Cl-}(aq)\);
- c. \(\ce{NH3}(aq)⟶\ce{H+}(aq)+\ce{NH2-}(aq)\);
- d. \(\ce{CH3CO2H}(aq)⟶\ce{H+}(aq)+\ce{CH3CO2-}(aq)\);
- e. \(\ce{NH4+}(aq)⟶\ce{H+}(aq)+\ce{NH3}(aq)\);
- f. \(\ce{HSO4-}(aq)⟶\ce{H+}(aq)+\ce{SO4^2-}(aq)\)
4. Show by suitable net ionic equations that each of the following species can act as a Brønsted-Lowry acid:
- HNO 3
- \(\ce{PH4+}\)
- H 2 S
- CH 3 CH 2 COOH
- \(\ce{H2PO4-}\)
- HS −
5. Show by suitable net ionic equations that each of the following species can act as a Brønsted-Lowry base:
- H 2 O
- \(\ce{OH-}\)
- NH 3
- CN −
- S 2−
- \(\ce{H2PO4-}\)
- Answer
- a. \(\ce{H_2O}_{(l)} + \ce{H^+} (aq)⟶\ce{H3O+}(aq)\)
- b. \(\ce{OH-} (aq) + \ce{H^+} (aq)⟶ \ce{H_2O}_{(l)}\)
- c. \(\ce{NH3}(aq) + \ce{H^+} (aq)⟶\ce{NH4+}(aq)\);
- d. \(\ce{CN-}(aq) + \ce{H^+} (aq)⟶\ce{HCN}(aq)\)
- e. \(\ce{S^2-}(aq) + \ce{H^+} (aq)⟶\ce{HS-}(aq)\)
- f. \(\ce{H2PO4-}(aq) + \ce{H^+} (aq)⟶\ce{H3PO4}(aq)\)
6. Show by suitable net ionic equations that each of the following species can act as a Brønsted-Lowry base:
- HS −
- \(\ce{PO4^3-}\)
- \(\ce{NH2-}\)
- C 2 H 5 OH
- O 2−
- \(\ce{H2PO4-}\)
7. What is the conjugate acid of each of the following? What is the conjugate base of each?
- \(\ce{OH-}\)
- H 2 O
- \(\ce{HCO3-}\)
- NH 3
- \(\ce{HSO4-}\)
- H 2 O 2
- HS −
- \(\ce{H5N2+}\)
- Answer
- H 2 O, O 2− ; H 3 O + , \(\ce{OH^-}\) ; H 2 CO 3 , \(\ce{CO3^2-}\); \(\ce{NH4+}\), \(\ce{NH2-}\); H 2 SO 4 , \(\ce{SO4^2-}\); \(\ce{H3O2+}\), \(\ce{HO2-}\); H 2 S; S 2− ; \(\ce{H6N2^2+}\), H 4 N 2
8. What is the conjugate acid of each of the following? What is the conjugate base of each?
- H 2 S
- \(\ce{H2PO4-}\)
- PH 3
- HS −
- \(\ce{HSO3-}\)
- \(\ce{H3O2+}\)
- H 4 N 2
- CH 3 OH
9. Identify and label the Brønsted-Lowry acid, its conjugate base, the Brønsted-Lowry base, and its conjugate acid in each of the following equations:
- \(\ce{HNO3 + H2O ⟶ H3O+ + NO3-}\)
- \(\ce{CN- + H2O ⟶ HCN + OH-}\)
- \(\ce{H2SO4 + Cl- ⟶ HCl + HSO4-}\)
- \(\ce{HSO4- + OH- ⟶ SO4^2- + H2O}\)
- \(\ce{O^2- + H2O ⟶ 2OH-}\)
- \(\ce{[Cu(H2O)3(OH)]+ + [Al(H2O)6]^3+ ⟶ [Cu(H2O)4]^2+ + [Al(H2O)5(OH)]^2+}\)
- \(\ce{H2S + NH2- ⟶ HS- + NH3}\)
- Answer
- The labels are Brønsted-Lowry acid = BA; its conjugate base = CB; Brønsted-Lowry base = BB; its conjugate acid = CA. HNO 3 (BA), H 2 O(BB), H 3 O + (CA), \(\ce{NO3- (CB)}\); CN − (BB), H 2 O(BA), HCN(CA), \(\ce{OH^-}\) (CB); H 2 SO 4 (BA), Cl − (BB), HCl(CA), \(\ce{HSO4- (CB)}\); \(\ce{HSO4- (BA)}\), OH - (BB), \(\ce{SO4^2- (CB)}\), H 2 O(CA); O 2− (BB), H2O(BA) \(\ce{OH^-}\) (CB and CA); [Cu(H 2 O) 3 (OH)] + (BB), [Al(H 2 O) 6 ] 3+ (BA), [Cu(H 2 O) 4 ] 2+ (CA), [Al(H 2 O) 5 (OH)] 2+ (CB); H 2 S(BA), \(\ce{NH2- (BB)}\), HS − (CB), NH 3 (CA)
10. Identify and label the Brønsted-Lowry acid, its conjugate base, the Brønsted-Lowry base, and its conjugate acid in each of the following equations:
- \(\ce{NO2- + H2O ⟶ HNO2 + OH-}\)
- \(\ce{HBr + H2O ⟶ H3O+ + Br-}\)
- \(\ce{HS- + H2O ⟶ H2S + OH-}\)
- \(\ce{H2PO4- + OH- ⟶HPO4^2- + H2O}\)
- \(\ce{H2PO4- + HCl ⟶ H3PO4 + Cl-}\)
- \(\ce{[Fe(H2O)5(OH)]^2+ + [Al(H2O)6]^3+ ⟶ [Fe(H2O)6]^3+ + [Al(H2O)5(OH)]^2+}\)
- \(\ce{CH3OH + H- ⟶ CH3O- + H2}\)
11. What are amphiprotic species? Illustrate with suitable equations.
- Answer
- Amphiprotic species may either gain or lose a proton in a chemical reaction, thus acting as a base or an acid. An example is H 2 O.
- As an acid: \(\ce{H2O}(aq) + \ce{NH3}(aq) \rightleftharpoons \ce{NH4+}(aq) + \ce{OH-}(aq)\).
- As a base: \(\ce{H2O}(aq) + \ce{HCl}(aq) \rightleftharpoons \ce{H3O+}(aq) + \ce{Cl-}(aq)\)
12. State which of the following species are amphiprotic and write chemical equations illustrating the amphiprotic character of these species:
- H 2 O
- \(\ce{H2PO4-}\)
- S 2−
- \(\ce{CO3^2-}\)
- \(\ce{HSO4-}\)
13. State which of the following species are amphiprotic and write chemical equations illustrating the amphiprotic character of these species.
- NH 3
- \(\ce{HPO4-}\)
- Br −
- \(\ce{NH4+}\)
- \(\ce{ASO4^3-}\)
- Answer
- amphiprotic: \(\ce{NH3 + H3O+ ⟶ NH4OH + H2O}\), \(\ce{NH3 + OCH3- ⟶ NH2- + CH3OH}\); \(\ce{HPO4^2- + OH- ⟶ PO4^3- + H2O}\), \(\ce{HPO4^2- + HClO4 ⟶ H2PO4- + ClO4-}\); not amphiprotic: Br − ; \(\ce{NH4+}\); \(\ce{AsO4^3-}\)
14. Is the self ionization of water endothermic or exothermic? The ionization constant for water ( K w ) is \(2.9 \times 10^{-14}\) at 40 °C and \(9.6 \times 10^{-14}\) at 60 °C.
pH and pOH
15. Explain why a sample of pure water at 40 °C is neutral even though [H 3 O + ] = 1.7 × 10 −7 M . K w is 2.9 × 10 −14 at 40 °C.
- Answer
- In a neutral solution [H 3 O + ] = [OH − ]. At 40 °C,
- [H 3 O + ] = [OH − ] = (2.910 −14 ) 1/2 = 1.7 × 10 −7 .
16. The ionization constant for water ( K w ) is 2.9 × 10 −14 at 40 °C. Calculate [H 3 O + ], [OH − ], pH, and pOH for pure water at 40 °C.
17. The ionization constant for water ( K w ) is 9.614 × 10 −14 at 60 °C. Calculate [H 3 O + ], [OH − ], pH, and pOH for pure water at 60 °C.
- Answer
- x = 3.101 × 10 −7 M = [H 3 O + ] = [OH − ]
- pH = -log 3.101 × 10 −7 = −(−6.5085) = 6.5085
- pOH = pH = 6.5085
18. Calculate the pH and the pOH of each of the following solutions at 25 °C for which the substances ionize completely:
- 0.200 M HCl
- 0.0143 M NaOH
- 3.0 M HNO 3
- 0.0031 M Ca(OH) 2
19. Calculate the pH and the pOH of each of the following solutions at 25 °C for which the substances ionize completely:
- 0.000259 M HClO 4
- 0.21 M NaOH
- 0.000071 M Ba(OH) 2
- 2.5 M KOH
- Answer
- pH = 3.587; pOH = 10.413; pH = 0.68; pOH = 13.32; pOH = 3.85; pH = 10.15; pH = −0.40; pOH = 14.4
20. What are the pH and pOH of a solution of 2.0 M HCl, which ionizes completely?
21. What are the hydronium and hydroxide ion concentrations in a solution whose pH is 6.52?
- Answer
- [H 3 O + ] = 3.0 × 10 −7 M ; [OH − ] = 3.3 × 10 −8 M
22. Calculate the hydrogen ion concentration and the hydroxide ion concentration in wine from its pH. See below Figure for useful information.
23. Calculate the hydronium ion concentration and the hydroxide ion concentration in lime juice from its pH. See Figure for useful information.
- Answer
- [H 3 O + ] = 1 × 10 −2 M ; [OH − ] = 1 × 10 −12 M
24. The hydronium ion concentration in a sample of rainwater is found to be 1.7 × 10 −6 M at 25 °C. What is the concentration of hydroxide ions in the rainwater?
25. The hydroxide ion concentration in household ammonia is 3.2 × 10 −3 M at 25 °C. What is the concentration of hydronium ions in the solution?
- Answer
- [OH − ] = 3.1 × 10 −12 M
Relative Strengths of Acids and Bases
Q14.3.1
Explain why the neutralization reaction of a strong acid and a weak base gives a weakly acidic solution.
Q14.3.2
Explain why the neutralization reaction of a weak acid and a strong base gives a weakly basic solution.
The salt ionizes in solution, but the anion slightly reacts with water to form the weak acid. This reaction also forms OH − , which causes the solution to be basic.
Q14.3.3
Use this list of important industrial compounds (and Figure ) to answer the following questions regarding: CaO, Ca(OH) 2 , CH 3 CO 2 H, CO 2, HCl, H 2 CO 3 , HF, HNO 2 , HNO 3 , H 3 PO 4 , H 2 SO 4 , NH 3 , NaOH, Na 2 CO 3 .
- Identify the strong Brønsted-Lowry acids and strong Brønsted-Lowry bases.
- List those compounds in that can behave as Brønsted-Lowry acids with strengths lying between those of H 3 O + and H 2 O.
- List those compounds in that can behave as Brønsted-Lowry bases with strengths lying between those of H 2 O and OH − .
Q14.3.4
The odor of vinegar is due to the presence of acetic acid, CH 3 CO 2 H, a weak acid. List, in order of descending concentration, all of the ionic and molecular species present in a 1- M aqueous solution of this acid.
S14.3.4
[H 2 O] > [CH 3 CO 2 H] > \(\ce{[H3O+]}\) ≈ \(\ce{[CH3CO2- ]}\) > [OH − ]
Q14.3.5
Household ammonia is a solution of the weak base NH 3 in water. List, in order of descending concentration, all of the ionic and molecular species present in a 1- M aqueous solution of this base.
Q14.3.4
Explain why the ionization constant, K a , for H 2 SO 4 is larger than the ionization constant for H 2 SO 3 .
S14.3.4
The oxidation state of the sulfur in H 2 SO 4 is greater than the oxidation state of the sulfur in H 2 SO 3 .
Q14.3.7
Explain why the ionization constant, K a , for HI is larger than the ionization constant for HF.
Q14.3.8
Gastric juice, the digestive fluid produced in the stomach, contains hydrochloric acid, HCl. Milk of Magnesia, a suspension of solid Mg(OH) 2 in an aqueous medium, is sometimes used to neutralize excess stomach acid. Write a complete balanced equation for the neutralization reaction, and identify the conjugate acid-base pairs.
S14.3.8
\(\underset{\large\ce{BB}}{\ce{Mg(OH)2}(s)}+\underset{\large\ce{BA}}{\ce{HCl}(aq)}⟶\underset{\large\ce{CB}}{\ce{Mg^2+}(aq)}+\underset{\large\ce{CA}}{\ce{2Cl-}(aq)}+\underset{\:}{\ce{2H2O}(l)}\)
Q14.3.9
Nitric acid reacts with insoluble copper(II) oxide to form soluble copper(II) nitrate, Cu(NO 3 ) 2 , a compound that has been used to prevent the growth of algae in swimming pools. Write the balanced chemical equation for the reaction of an aqueous solution of HNO 3 with CuO.
Q14.3.10
What is the ionization constant at 25 °C for the weak acid \(\ce{CH3NH3+}\), the conjugate acid of the weak base CH 3 NH 2 , K b = 4.4 × 10 −4 .
S14.3.10
\(K_\ce{a}=2.3×10^{−11}\)
Q14.3.11
What is the ionization constant at 25 °C for the weak acid \(\ce{(CH3)2NH2+}\), the conjugate acid of the weak base (CH 3 ) 2 NH, K b = 7.4 × 10 −4 ?
Q14.3.12
Which base, CH 3 NH 2 or (CH 3 ) 2 NH, is the strongest base? Which conjugate acid, \(\ce{(CH3)2NH2+}\) or (CH 3 ) 2 NH, is the strongest acid?
S14.3.12
The strongest base or strongest acid is the one with the larger K b or K a , respectively. In these two examples, they are (CH 3 ) 2 NH and \(\ce{CH3NH3+}\).
Q14.3.3
Which is the stronger acid, \(\ce{NH4+}\) or HBrO?
Q14.3.14
Which is the stronger base, (CH 3 ) 3 N or \(\ce{H2BO3-}\)?
S14.3.14
triethylamine.
Q14.3.15
Predict which acid in each of the following pairs is the stronger and explain your reasoning for each.
- H 2 O or HF
- B(OH) 3 or Al(OH) 3
- \(\ce{HSO3-}\) or \(\ce{HSO4-}\)
- NH 3 or H 2 S
- H 2 O or H 2 Te
Q14.3.16
Predict which compound in each of the following pairs of compounds is more acidic and explain your reasoning for each.
- \(\ce{HSO4-}\) or \(\ce{HSeO4-}\)
- NH 3 or H 2 O
- PH 3 or HI
- NH 3 or PH 3
- H 2 S or HBr
S14.3.16
- \(\ce{HSO4-}\); higher electronegativity of the central ion. H 2 O;
- NH 3 is a base and water is neutral, or decide on the basis of K a values. HI;
- PH 3 is weaker than HCl; HCl is weaker than HI. Thus, PH 3 is weaker than HI.
- PH 3 ; in binary compounds of hydrogen with nonmetals, the acidity increases for the element lower in a group.
- HBr; in a period, the acidity increases from left to right; in a group, it increases from top to bottom. Br is to the left and below S, so HBr is the stronger acid.
Q14.3.17
Rank the compounds in each of the following groups in order of increasing acidity or basicity, as indicated, and explain the order you assign.
- acidity: HCl, HBr, HI
- basicity: H 2 O, OH − , H − , Cl −
- basicity: Mg(OH) 2 , Si(OH) 4 , ClO 3 (OH) (Hint: Formula could also be written as HClO 4 ).
- acidity: HF, H 2 O, NH 3 , CH 4
Q14.3.18
Rank the compounds in each of the following groups in order of increasing acidity or basicity, as indicated, and explain the order you assign.
- acidity: NaHSO 3 , NaHSeO 3 , NaHSO 4
- basicity: \(\ce{BrO2-}\), \(\ce{ClO2-}\), \(\ce{IO2-}\)
- acidity: HOCl, HOBr, HOI
- acidity: HOCl, HOClO, HOClO 2 , HOClO 3
- basicity: \(\ce{NH2-}\), HS − , HTe − , \(\ce{PH2-}\)
- basicity: BrO − , \(\ce{BrO2-}\), \(\ce{BrO3-}\), \(\ce{BrO4-}\)
S14.3.18
- NaHSeO 3 < NaHSO 3 < NaHSO 4 ; in polyoxy acids, the more electronegative central element—S, in this case—forms the stronger acid. The larger number of oxygen atoms on the central atom (giving it a higher oxidation state) also creates a greater release of hydrogen atoms, resulting in a stronger acid. As a salt, the acidity increases in the same manner.
- \(\ce{ClO2- < BrO2- < IO2-}\); the basicity of the anions in a series of acids will be the opposite of the acidity in their oxyacids. The acidity increases as the electronegativity of the central atom increases. Cl is more electronegative than Br, and I is the least electronegative of the three.
- HOI < HOBr < HOCl; in a series of the same form of oxyacids, the acidity increases as the electronegativity of the central atom increases. Cl is more electronegative than Br, and I is the least electronegative of the three.
- HOCl < HOClO < HOClO 2 < HOClO 3 ; in a series of oxyacids of the same central element, the acidity increases as the number of oxygen atoms increases (or as the oxidation state of the central atom increases).
- \(\ce{HTe- < HS- << PH2- < NH2-}\); \(\ce{PH2-}\) and \(\ce{NH2-}\) are anions of weak bases, so they act as strong bases toward H + . \(\ce{HTe-}\) and HS − are anions of weak acids, so they have less basic character. In a periodic group, the more electronegative element has the more basic anion.
- \(\ce{BrO4- < BrO3- < BrO2- < BrO-}\); with a larger number of oxygen atoms (that is, as the oxidation state of the central ion increases), the corresponding acid becomes more acidic and the anion consequently less basic.
Q14.3.19
Both HF and HCN ionize in water to a limited extent. Which of the conjugate bases, F − or CN − , is the stronger base? See Table .
Q14.3.20
The active ingredient formed by aspirin in the body is salicylic acid, C 6 H 4 OH(CO 2 H). The carboxyl group (−CO 2 H) acts as a weak acid. The phenol group (an OH group bonded to an aromatic ring) also acts as an acid but a much weaker acid. List, in order of descending concentration, all of the ionic and molecular species present in a 0.001- M aqueous solution of C 6 H 4 OH(CO 2 H).
\(\ce{[H2O] > [C6H4OH(CO2H)] > [H+]0 > [C6H4OH(CO2)- ] ≫ [C6H4O(CO2H)- ] > [OH- ]}\)
What do we represent when we write:
\[\ce{CH3CO2H}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{CH3CO2-}(aq)?\]
Q14.3.21
Explain why equilibrium calculations are not necessary to determine ionic concentrations in solutions of certain strong electrolytes such as NaOH and HCl. Under what conditions are equilibrium calculations necessary as part of the determination of the concentrations of all ions of some other strong electrolytes in solution?
S14.3.21
Strong electrolytes are 100% ionized, and, as long as the component ions are neither weak acids nor weak bases, the ionic species present result from the dissociation of the strong electrolyte. Equilibrium calculations are necessary when one (or more) of the ions is a weak acid or a weak base.
Q14.3.22
Are the concentrations of hydronium ion and hydroxide ion in a solution of an acid or a base in water directly proportional or inversely proportional? Explain your answer.
Q14.3.23
What two common assumptions can simplify calculation of equilibrium concentrations in a solution of a weak acid?
S14.3.23
- Assume that the change in initial concentration of the acid as the equilibrium is established can be neglected, so this concentration can be assumed constant and equal to the initial value of the total acid concentration.
- Assume we can neglect the contribution of water to the equilibrium concentration of H 3 O + .
Q14.3.24
What two common assumptions can simplify calculation of equilibrium concentrations in a solution of a weak base?
Q14.3.25
Which of the following will increase the percent of NH 3 that is converted to the ammonium ion in water (Hint: Use LeChâtelier’s principle.)?
- addition of NaOH
- addition of HCl
- addition of NH 4 Cl
S14.3.25
The addition of HCl
Q14.3.26
Which of the following will increase the percent of HF that is converted to the fluoride ion in water?
- addition of NaOH
- addition of HCl
- addition of NaF
Q14.3.27
What is the effect on the concentrations of \(\ce{NO2-}\), HNO 2 , and OH − when the following are added to a solution of KNO 2 in water:
- HCl
- HNO 2
- NaOH
- NaCl
- KNO
The equation for the equilibrium is:
\[\ce{NO2-}(aq)+\ce{H2O}(l)⇌\ce{HNO2}(aq)+\ce{OH-}(aq)\]S14.3.27
- Adding HCl will add H 3 O + ions, which will then react with the OH − ions, lowering their concentration. The equilibrium will shift to the right, increasing the concentration of HNO 2 , and decreasing the concentration of \(\ce{NO2-}\) ions.
- Adding HNO 2 increases the concentration of HNO 2 and shifts the equilibrium to the left, increasing the concentration of \(\ce{NO2-}\) ions and decreasing the concentration of OH − ions.
- Adding NaOH adds OH − ions, which shifts the equilibrium to the left, increasing the concentration of \(\ce{NO2-}\) ions and decreasing the concentrations of HNO 2 .
- Adding NaCl has no effect on the concentrations of the ions.
- Adding KNO 2 adds \(\ce{NO2-}\) ions and shifts the equilibrium to the right, increasing the HNO 2 and OH − ion concentrations.
Q14.3.28
What is the effect on the concentration of hydrofluoric acid, hydronium ion, and fluoride ion when the following are added to separate solutions of hydrofluoric acid?
- HCl
- KF
- NaCl
- KOH
- HF
The equation for the equilibrium is:
\[\ce{HF}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{F-}(aq)\]Q14.3.29
Why is the hydronium ion concentration in a solution that is 0.10 M in HCl and 0.10 M in HCOOH determined by the concentration of HCl?
S14.3.29
This is a case in which the solution contains a mixture of acids of different ionization strengths. In solution, the HCO 2 H exists primarily as HCO 2 H molecules because the ionization of the weak acid is suppressed by the strong acid. Therefore, the HCO 2 H contributes a negligible amount of hydronium ions to the solution. The stronger acid, HCl, is the dominant producer of hydronium ions because it is completely ionized. In such a solution, the stronger acid determines the concentration of hydronium ions, and the ionization of the weaker acid is fixed by the [H 3 O + ] produced by the stronger acid.
Q14.3.30
From the equilibrium concentrations given, calculate K a for each of the weak acids and K b for each of the weak bases.
CH 3 CO 2 H: \(\ce{[H3O+]}\) = 1.34 × 10 −3 M ;
\(\ce{[CH3CO2- ]}\) = 1.34 × 10 −3 M ;[CH 3 CO 2 H] = 9.866 × 10 −2 M ;
ClO − : [OH − ] = 4.0 × 10 −4 M ;
[HClO] = 2.38 × 10 −5 M ;
[ClO − ] = 0.273 M ;
HCO 2 H: [HCO 2 H] = 0.524 M ;
\(\ce{[H3O+]}\) = 9.8 × 10 −3 M ; \(\ce{[HCO2- ]}\) = 9.8 × 10 −3 M ;\(\ce{C6H5NH3+ : [C6H5NH3+]}\) = 0.233 M ;
[C 6 H 5 NH 2 ] = 2.3 × 10 −3 M ;
\(\ce{[H3O+]}\) = 2.3 × 10 −3 MFrom the equilibrium concentrations given, calculate K a for each of the weak acids and K b for each of the weak bases.
NH 3 : [OH − ] = 3.1 × 10 −3 M ;
\(\ce{[NH4+]}\) = 3.1 × 10 −3 M ;[NH 3 ] = 0.533 M ;
HNO 2 : \(\ce{[H3O+]}\) = 0.011 M ;
\(\ce{[NO2- ]}\) = 0.0438 M ;[HNO 2 ] = 1.07 M ;
(CH 3 ) 3 N: [(CH 3 ) 3 N] = 0.25 M ;
[(CH 3 ) 3 NH + ] = 4.3 × 10 −3 M ;[OH − ] = 4.3 × 10 −3 M ;
\(\ce{NH4+ : [NH4+]}\) = 0.100 M ;
[NH 3 ] = 7.5 × 10 −6 M ;
[H 3 O + ] = 7.5 × 10 −6 M- \(K_\ce{b}=1.8×10^{−5};\)
- \(K_\ce{a}=4.5×10^{−4};\)
- \(K_\ce{b}=7.4×10^{−5};\)
- \(K_\ce{a}=5.6×10^{−10}\)
Q14.3.31
Determine K b for the nitrite ion, \(\ce{NO2-}\). In a 0.10- M solution this base is 0.0015% ionized.
Q14.3.32
Determine K a for hydrogen sulfate ion, \(\ce{HSO4-}\). In a 0.10- M solution the acid is 29% ionized.
S14.3.32
\(K_\ce{a}=1.2×10^{−2}\)
Q14.3.33
Calculate the ionization constant for each of the following acids or bases from the ionization constant of its conjugate base or conjugate acid:
- F −
- \(\ce{NH4+}\)
- \(\ce{AsO4^3-}\)
- \(\ce{(CH3)2NH2+}\)
- \(\ce{NO2-}\)
- \(\ce{HC2O4-}\) (as a base)
Q14.3.52
Calculate the ionization constant for each of the following acids or bases from the ionization constant of its conjugate base or conjugate acid:
- HTe − (as a base)
- \(\ce{(CH3)3NH+}\)
- \(\ce{HAsO4^3-}\) (as a base)
- \(\ce{HO2-}\) (as a base)
- \(\ce{C6H5NH3+}\)
- \(\ce{HSO3-}\) (as a base)
S14.3.52
- \(K_\ce{b}=4.3×10^{−12};\)
- \(K_\ce{a}=1.4×10^{−10};\)
- \(K_\ce{b}=1×10^{−7};\)
- \(K_\ce{b}=4.2×10^{−3};\)
- \(K_\ce{b}=4.2×10^{−3};\)
- \(K_\ce{b}=8.3×10^{−13}\)
Q14.3.53
For which of the following solutions must we consider the ionization of water when calculating the pH or pOH?
- 3 × 10 −8 M HNO 3
- 0.10 g HCl in 1.0 L of solution
- 0.00080 g NaOH in 0.50 L of solution
- 1 × 10 −7 M Ca(OH) 2
- 0.0245 M KNO 3
Q14.3.54
Even though both NH 3 and C 6 H 5 NH 2 are weak bases, NH 3 is a much stronger acid than C 6 H 5 NH 2 . Which of the following is correct at equilibrium for a solution that is initially 0.10 M in NH 3 and 0.10 M in C 6 H 5 NH 2 ?
- \(\ce{[OH- ]}=\ce{[NH4+]}\)
- \(\ce{[NH4+]}=\ce{[C6H5NH3+]}\)
- \(\ce{[OH- ]}=\ce{[C6H5NH3+]}\)
- [NH 3 ] = [C 6 H 5 NH 2 ]
- both a and b are correct
is the correct statement.
Q14.3.55
Calculate the equilibrium concentration of the nonionized acids and all ions in a solution that is 0.25 M in HCO 2 H and 0.10 M in HClO.
Q14.3.56
Calculate the equilibrium concentration of the nonionized acids and all ions in a solution that is 0.134 M in HNO 2 and 0.120 M in HBrO.
S14.3.56
[H 3 O + ] = 7.5 × 10 −3 M
[HNO 2 ] = 0.126 [OH − ] = 1.3 × 10 −12 M [BrO − ] = 3.2 × 10 −8 M [HBrO] = 0.120 MQ14.3.57
Calculate the equilibrium concentration of the nonionized bases and all ions in a solution that is 0.25 M in CH 3 NH 2 and 0.10 M in C 5 H 5 N ( K b = 1.7 × 10 −9 ).
Q14.3.58
Calculate the equilibrium concentration of the nonionized bases and all ions in a solution that is 0.115 M in NH 3 and 0.100 M in C 6 H 5 NH 2 .
S14.3.58
[OH − ] = \(\ce{[NO4+]}\) = 0.0014 M
[NH 3 ] = 0.144 M [H 3 O + ] = 6.9 × 10 −12 M \(\ce{[C6H5NH3+]}\) = 3.9 × 10 −8 M [C 6 H 5 NH 2 ] = 0.100 MQ14.3.59
Using the K a values in Appendix H , place \(\ce{Al(H2O)6^3+}\) in the correct location in Figure .
Q14.3.60
Calculate the concentration of all solute species in each of the following solutions of acids or bases. Assume that the ionization of water can be neglected, and show that the change in the initial concentrations can be neglected. Ionization constants can be found in Appendix H and Appendix I .
- 0.0092 M HClO, a weak acid
- 0.0784 M C 6 H 5 NH 2 , a weak base
- 0.0810 M HCN, a weak acid
- 0.11 M (CH 3 ) 3 N, a weak base
- 0.120 M \(\ce{Fe(H2O)6^2+}\) a weak acid, K a = 1.6 × 10 −7
S14.3.60
\(\ce{\dfrac{[H3O+][ClO- ]}{[HClO]}}=\dfrac{(x)(x)}{(0.0092−x)}≈\dfrac{(x)(x)}{0.0092}=3.5×10^{−8}\)
Solving for x gives 1.79 × 10 −5 M . This value is less than 5% of 0.0092, so the assumption that it can be neglected is valid. Thus, the concentrations of solute species at equilibrium are:
[H 3 O + ] = [ClO] = 1.8 × 10 −5 M [HClO] = 0.00092 M [OH − ] = 5.6 × 10 −10 M ;
\(\ce{\dfrac{[C6H5NH3+][OH- ]}{[C6H5NH2]}}=\dfrac{(x)(x)}{(0.0784−x)}≈\dfrac{(x)(x)}{0.0784}=4.6×10^{−10}\)
Solving for x gives 6.01 × 10 −6 M .
This value is less than 5% of 0.0784, so the assumption that it can be neglected is valid. Thus, the concentrations of solute species at equilibrium are: \(\ce{[CH3CO2- ]}\) = [OH − ] = 6.0 × 10 −6 M [C 6 H 5 NH 2 ] = 0.00784 [H 3 O + ] = 1.7× 10 −9 M ; \(\ce{\dfrac{[H3O+][CN- ]}{[HCN]}}=\dfrac{(x)(x)}{(0.0810−x)}≈\dfrac{(x)(x)}{0.0810}=4×10^{−10}\) Solving for x gives 5.69 × 10 −6 M . This value is less than 5% of 0.0810, so the assumption that it can be neglected is valid. Thus, the concentrations of solute species at equilibrium are: [H 3 O + ] = [CN − ] = 5.7 × 10 −6 M [HCN] = 0.0810 M [OH − ] = 1.8 × 10 −9 M ; \(\ce{\dfrac{[(CH3)3NH+][OH- ]}{[(CH3)3N]}}=\dfrac{(x)(x)}{(0.11−x)}≈\dfrac{(x)(x)}{0.11}=7.4×10^{−5}\) Solving for x gives 2.85 × 10 −3 M . This value is less than 5% of 0.11, so the assumption that it can be neglected is valid. Thus, the concentrations of solute species at equilibrium are: [(CH 3 ) 3 NH + ] = [OH − ] = 2.9 × 10 −3 M [(CH 3 ) 3 N] = 0.11 M [H 3 O + ] = 3.5 × 10 −12 M ; \(\ce{\dfrac{[Fe(H2O)5(OH)+][H3O+]}{[Fe(H2O)6^2+]}}=\dfrac{(x)(x)}{(0.120−x)}≈\dfrac{(x)(x)}{0.120}=1.6×10^{−7}\) Solving for x gives 1.39 × 10 −4 M . This value is less than 5% of 0.120, so the assumption that it can be neglected is valid. Thus, the concentrations of solute species at equilibrium are: [Fe(H 2 O) 5 (OH) + ] = [H 3 O + ] = 1.4 × 10 −4 M \(\ce{[Fe(H2O)6^2+]}\) = 0.120 M [OH − ] = 7.2 × 10 −11 M
Q14.3.61
Propionic acid, C 2 H 5 CO 2 H ( K a = 1.34 × 10 −5 ), is used in the manufacture of calcium propionate, a food preservative. What is the hydronium ion concentration in a 0.698- M solution of C 2 H 5 CO 2 H?
Q14.3.62
White vinegar is a 5.0% by mass solution of acetic acid in water. If the density of white vinegar is 1.007 g/cm 3 , what is the pH?
S14.3.62
pH = 2.41
Q14.3.63
The ionization constant of lactic acid, CH 3 CH(OH)CO 2 H, an acid found in the blood after strenuous exercise, is 1.36 × 10 −4 . If 20.0 g of lactic acid is used to make a solution with a volume of 1.00 L, what is the concentration of hydronium ion in the solution?
Q14.3.64
Nicotine, C 10 H 14 N 2 , is a base that will accept two protons ( K 1 = 7 × 10 −7 , K 2 = 1.4 × 10 −11 ). What is the concentration of each species present in a 0.050- M solution of nicotine?
S14.3.64
[C 10 H 14 N 2 ] = 0.049 M
[C 10 H 14 N 2 H + ] = 1.9 × 10 −4 M \(\ce{[C10H14N2H2^2+]}\) = 1.4 × 10 −11 M [OH − ] = 1.9 × 10 −4 M [H 3 O + ] = 5.3 × 10 −11 MQ14.3.65
The pH of a 0.20- M solution of HF is 1.92. Determine K a for HF from these data.
Q14.3.66
The pH of a 0.15- M solution of \(\ce{HSO4-}\) is 1.43. Determine K a for \(\ce{HSO4-}\) from these data.
S14.3.66
\(K_\ce{a}=1.2×10^{−2}\)
Q14.3.67
The pH of a 0.10- M solution of caffeine is 11.16. Determine K b for caffeine from these data:
\(\ce{C8H10N4O2}(aq)+\ce{H2O}(l)⇌\ce{C8H10N4O2H+}(aq)+\ce{OH-}(aq)\)Q14.3.68
The pH of a solution of household ammonia, a 0.950 M solution of NH 3 , is 11.612. Determine K b for NH 3 from these data.
S14.3.68
\(K_\ce{b}=1.77×10^{−5}\)
Hydrolysis of Salt Solutions
Q14.4.1
Determine whether aqueous solutions of the following salts are acidic, basic, or neutral:
- Al(NO 3 ) 3
- RbI
- KHCO 2
- CH 3 NH 3 Br
Q14.4.2
Determine whether aqueous solutions of the following salts are acidic, basic, or neutral:
- FeCl 3
- K 2 CO 3
- NH 4 Br
- KClO 4
S14.4.2
acidic; basic; acidic; neutral
Q14.4.3
Novocaine, C 13 H 21 O 2 N 2 Cl, is the salt of the base procaine and hydrochloric acid. The ionization constant for procaine is 7 × 10 −6 . Is a solution of novocaine acidic or basic? What are [H 3 O + ], [OH − ], and pH of a 2.0% solution by mass of novocaine, assuming that the density of the solution is 1.0 g/mL.


2: Sediment Creation and Transport
Sediment is formed by the weathering of existing rocks and minerals. This sediment can be eroded (transported from one location to another) in a variety of ways by moving fluids or under the direct influence of gravity. Sediment will eventually be deposited and might eventually be transformed into sedimentary rock.
- Explain the difference between physical and chemical weathering and provide examples of each.
- Define the most important properties of fluids and sedimentary particles and use this knowledge to make predictions about how these properties will influence fluid motion and sediment transport.
- Describe, identify, and explain the differences between the three types of fluid flow transport
- Describe, identify, and explain the differences between the four types of sediment gravity flows
-
- 2.3: Fluid-Flow Transport
- There are three different ways that moving fluid can carry sediment. Dissolved load is the movement of dissolved ions, suspended load is when particles are held aloft by turbulence, and bedload transport happens when the fluid causes particles to move through rolling, tumbling, saltation, etc.
Chapter thumbnail shows a boulder split by frost wedging ( Dominicus Johannes Bergsma via Wikimedia Commons ; CC BY-SA 4.0 ).


29.1: Quantization of Energy
Learning Objectives
By the end of this section, you will be able to:
- Explain Max Planck’s contribution to the development of quantum mechanics.
- Explain why atomic spectra indicate quantization.
Planck’s Contribution
Energy is quantized in some systems, meaning that the system can have only certain energies and not a continuum of energies, unlike the classical case. This would be like having only certain speeds at which a car can travel because its kinetic energy can have only certain values. We also find that some forms of energy transfer take place with discrete lumps of energy. While most of us are familiar with the quantization of matter into lumps called atoms, molecules, and the like, we are less aware that energy, too, can be quantized. Some of the earliest clues about the necessity of quantum mechanics over classical physics came from the quantization of energy.
Where is the quantization of energy observed? Let us begin by considering the emission and absorption of electromagnetic (EM) radiation. The EM spectrum radiated by a hot solid is linked directly to the solid’s temperature (Figure \(\PageIndex{1}\)) An ideal radiator is one that has an emissivity of 1 at all wavelengths and, thus, is jet black. Ideal radiators are therefore called blackbodies , and their EM radiation is called blackbody radiation . It was discussed that the total intensity of the radiation varies as \(T^4\), the fourth power of the absolute temperature of the body, and that the peak of the spectrum shifts to shorter wavelengths at higher temperatures. All of this seems quite continuous, but it was the curve of the spectrum of intensity versus wavelength that gave a clue that the energies of the atoms in the solid are quantized. In fact, providing a theoretical explanation for the experimentally measured shape of the spectrum was a mystery at the turn of the century. When this “ultraviolet catastrophe” was eventually solved, the answers led to new technologies such as computers and the sophisticated imaging techniques described in earlier chapters. Once again, physics as an enabling science changed the way we live.
The German physicist Max Planck (1858–1947) used the idea that atoms and molecules in a body act like oscillators to absorb and emit radiation. The energies of the oscillating atoms and molecules had to be quantized to correctly describe the shape of the blackbody spectrum. Planck deduced that the energy of an oscillator having a frequency \(f\) is given by
\[E = \left(n + \dfrac{1}{2} \right) hf.\]
Here \(n\) is any nonnegative integer (0, 1, 2, 3, …). The symbol \(h\) stands for Planck’s constant , given by
\[h = 6.626 \times 10^{-34} \, J \cdot s.\] The equation \(E = (n + \frac{1}{2}) hf\) means that an oscillator having a frequency \(f\) (emitting and absorbing EM radiation of frequency \(f\)) can have its energy increase or decrease only in discrete steps of size
\[\Delta E = hf.\]
It might be helpful to mention some macroscopic analogies of this quantization of energy phenomena. This is like a pendulum that has a characteristic oscillation frequency but can swing with only certain amplitudes. Quantization of energy also resembles a standing wave on a string that allows only particular harmonics described by integers. It is also similar to going up and down a hill using discrete stair steps rather than being able to move up and down a continuous slope. Your potential energy takes on discrete values as you move from step to step.
Using the quantization of oscillators, Planck was able to correctly describe the experimentally known shape of the blackbody spectrum. This was the first indication that energy is sometimes quantized on a small scale and earned him the Nobel Prize in Physics in 1918. Although Planck’s theory comes from observations of a macroscopic object, its analysis is based on atoms and molecules. It was such a revolutionary departure from classical physics that Planck himself was reluctant to accept his own idea that energy states are not continuous. The general acceptance of Planck’s energy quantization was greatly enhanced by Einstein’s explanation of the photoelectric effect (discussed in the next section), which took energy quantization a step further. Planck was fully involved in the development of both early quantum mechanics and relativity. He quickly embraced Einstein’s special relativity, published in 1905, and in 1906 Planck was the first to suggest the correct formula for relativistic momentum, \(p = \gamma mu\).
Note that Planck’s constant \(h\) is a very small number. So for an infrared frequency of \(10^{14}\) being emitted by a blackbody, for example, the difference between energy levels is only \(\Delta E = hf = (6.63 \times 10^{-34} \, J \cdot s)(10^{14} \, Hz) = 6.63 \times 10^{-20} \, J\), or about 0.4 eV. This 0.4 eV of energy is significant compared with typical atomic energies, which are on the order of an electron volt, or thermal energies, which are typically fractions of an electron volt. But on a macroscopic or classical scale, energies are typically on the order of joules. Even if macroscopic energies are quantized, the quantum steps are too small to be noticed. This is an example of the correspondence principle. For a large object, quantum mechanics produces results indistinguishable from those of classical physics.
Atomic Spectra
Now let us turn our attention to the emission and absorption of EM radiation by gases . The Sun is the most common example of a body containing gases emitting an EM spectrum that includes visible light. We also see examples in neon signs and candle flames. Studies of emissions of hot gases began more than two centuries ago, and it was soon recognized that these emission spectra contained huge amounts of information. The type of gas and its temperature, for example, could be determined. We now know that these EM emissions come from electrons transitioning between energy levels in individual atoms and molecules; thus, they are called atomic spectra . Atomic spectra remain an important analytical tool today. Figure \(\PageIndex{3}\) shows an example of an emission spectrum obtained by passing an electric discharge through a material. One of the most important characteristics of these spectra is that they are discrete. By this we mean that only certain wavelengths, and hence frequencies, are emitted. This is called a line spectrum. If frequency and energy are associated as \(\Delta E = hf\), the energies of the electrons in the emitting atoms and molecules are quantized. This is discussed in more detail later in this chapter.
It was a major puzzle that atomic spectra are quantized. Some of the best minds of 19th-century science failed to explain why this might be. Not until the second decade of the 20th century did an answer based on quantum mechanics begin to emerge. Again a macroscopic or classical body of gas was involved in the studies, but the effect, as we shall see, is due to individual atoms and molecules.
PHET EXPLORATIONS: MODELS OF THE HYDROGEN ATOM
How did scientists figure out the structure of atoms without looking at them? Try out different models by shooting light at the atom. Check how the prediction of the model matches the experimental results.
Summary
- The first indication that energy is sometimes quantized came from blackbody radiation, which is the emission of EM radiation by an object with an emissivity of 1.
- Planck recognized that the energy levels of the emitting atoms and molecules were quantized, with only the allowed values of \(E = (n + \frac{1}{2})hf\), where \(n\) is any non-negative integer (0, 1, 2, 3, …).
- \(h\) is Planck’s constant, whose value is \(h = 6.626 \times 10^{-34} \, J \cdot s\).
- Thus, the oscillatory absorption and emission energies of atoms and molecules in a blackbody could increase or decrease only in steps of size \(\Delta E = hf\) where \(f\) is the frequency of the oscillatory nature of the absorption and emission of EM radiation.
- Another indication of energy levels being quantized in atoms and molecules comes from the lines in atomic spectra, which are the EM emissions of individual atoms and molecules.
Glossary
- blackbody
- an ideal radiator, which can radiate equally well at all wavelengths
- blackbody radiation
- the electromagnetic radiation from a blackbody
- Planck’s constant
- \(6.626 \times 10^{-34} \, J \cdot s\)
- atomic spectra
- the electromagnetic emission from atoms and molecules


3.5E: Measures of the Location of the Data (Exercises)
Listed are 29 ages for Academy Award winning best actors in order from smallest to largest.
18; 21; 22; 25; 26; 27; 29; 30; 31; 33; 36; 37; 41; 42; 47; 52; 55; 57; 58; 62; 64; 67; 69; 71; 72; 73; 74; 76; 77
- Find the 40 th percentile.
- Find the 78 th percentile.
Answer
- The 40 th percentile is 37 years.
- The 78 th percentile is 70 years.
Listed are 32 ages for Academy Award winning best actors in order from smallest to largest.
18; 18; 21; 22; 25; 26; 27; 29; 30; 31; 31; 33; 36; 37; 37; 41; 42; 47; 52; 55; 57; 58; 62; 64; 67; 69; 71; 72; 73; 74; 76; 77
- Find the percentile of 37.
- Find the percentile of 72.
Jesse was ranked 37 th in his graduating class of 180 students. At what percentile is Jesse’s ranking?
Answer
Jesse graduated 37 th out of a class of 180 students. There are 180 – 37 = 143 students ranked below Jesse. There is one rank of 37.
\(x = 143\) and \(y = 1\). \(\dfrac{x + 0.5y}{n}(100) = \dfrac{143 + 0.5(1)}{180}(100) = 79.72\). Jesse’s rank of 37 puts him at the 80 th percentile.
- For runners in a race, a low time means a faster run. The winners in a race have the shortest running times. Is it more desirable to have a finish time with a high or a low percentile when running a race?
- The 20 th percentile of run times in a particular race is 5.2 minutes. Write a sentence interpreting the 20 th percentile in the context of the situation.
- A bicyclist in the 90 th percentile of a bicycle race completed the race in 1 hour and 12 minutes. Is he among the fastest or slowest cyclists in the race? Write a sentence interpreting the 90 th percentile in the context of the situation.
- For runners in a race, a higher speed means a faster run. Is it more desirable to have a speed with a high or a low percentile when running a race?
- The 40 th percentile of speeds in a particular race is 7.5 miles per hour. Write a sentence interpreting the 40 th percentile in the context of the situation.
Answer
- For runners in a race it is more desirable to have a high percentile for speed. A high percentile means a higher speed which is faster.
- 40% of runners ran at speeds of 7.5 miles per hour or less (slower). 60% of runners ran at speeds of 7.5 miles per hour or more (faster).
On an exam, would it be more desirable to earn a grade with a high or low percentile? Explain.
Mina is waiting in line at the Department of Motor Vehicles (DMV). Her wait time of 32 minutes is the 85 th percentile of wait times. Is that good or bad? Write a sentence interpreting the 85 th percentile in the context of this situation.
Answer
When waiting in line at the DMV, the 85 th percentile would be a long wait time compared to the other people waiting. 85% of people had shorter wait times than Mina. In this context, Mina would prefer a wait time corresponding to a lower percentile. 85% of people at the DMV waited 32 minutes or less. 15% of people at the DMV waited 32 minutes or longer.
In a survey collecting data about the salaries earned by recent college graduates, Li found that her salary was in the 78 th percentile. Should Li be pleased or upset by this result? Explain.
In a study collecting data about the repair costs of damage to automobiles in a certain type of crash tests, a certain model of car had $1,700 in damage and was in the 90 th percentile. Should the manufacturer and the consumer be pleased or upset by this result? Explain and write a sentence that interprets the 90 th percentile in the context of this problem.
Answer
The manufacturer and the consumer would be upset. This is a large repair cost for the damages, compared to the other cars in the sample. INTERPRETATION: 90% of the crash tested cars had damage repair costs of $1700 or less; only 10% had damage repair costs of $1700 or more.
The University of California has two criteria used to set admission standards for freshman to be admitted to a college in the UC system:
- Students' GPAs and scores on standardized tests (SATs and ACTs) are entered into a formula that calculates an "admissions index" score. The admissions index score is used to set eligibility standards intended to meet the goal of admitting the top 12% of high school students in the state. In this context, what percentile does the top 12% represent?
- Students whose GPAs are at or above the 96 th percentile of all students at their high school are eligible (called eligible in the local context), even if they are not in the top 12% of all students in the state. What percentage of students from each high school are "eligible in the local context"?
Suppose that you are buying a house. You and your realtor have determined that the most expensive house you can afford is the 34 th percentile. The 34 th percentile of housing prices is $240,000 in the town you want to move to. In this town, can you afford 34% of the houses or 66% of the houses?
Answer
You can afford 34% of houses. 66% of the houses are too expensive for your budget. INTERPRETATION: 34% of houses cost $240,000 or less. 66% of houses cost $240,000 or more.
Use Exercise to calculate the following values:
First quartile = _______
Second quartile = median = 50 th percentile = _______
Answer
4
Third quartile = _______
Interquartile range ( IQR ) = _____ – _____ = _____
Answer
\(6 - 4 = 2\)
10 th percentile = _______
70 th percentile = _______
Answer
6


4.8: "There" Construction
“There” Construction Video
Instructor Video: There Construction
English is not a pro-drop language, in other words, you cannot have an English sentence in which subject is not available. This’s why, we need dummy subjects like ‘there’ and ‘it’ in English.
For examples:
- There is a car.
- There are ten students.
Unlike English though, Hindi is a pro-drop language hence subject can be dropped, or it is not needed in ‘there’ or ‘it’ sentences. In these sentences, the verb agrees with the object.
For Examples:
| 1. | English | In my town, there is a museum. |
| Hindi Word order | My town in + (there) + a museum + is | |
| Hindi | मेरे शहर में (___) एक संग्रहालय है | | |
| 2. | English | In my room, there are ten books. |
| Hindi Word order | My room in + (there) + ten books + are | |
| Hindi | मेरे कमरे में (___) दस किताबें हैं | | |
| 3. | English | There is a computer on my desk. |
| Hindi Word order | My desk on + (there) + a computer + is | |
| Hindi | मेरी मेज़ पर (___) एक कंप्यूटर है | |
Activities
With “there construction,” please form five sentences for describing places given below.
| Place | Sentences | |
|
मेरी रसोई में,
In my kitchen, |
1 | Example: मेरी रसोई में एक चूल्हा है “There is a stove in my kitchen.” |
| 2 | ||
| 3 | ||
| 4 | ||
| 5 | ||
|
मेरे बैठक में,
In my living room, |
1 | |
| 2 | ||
| 3 | ||
| 4 | ||
| 5 | ||
|
मेरे थैले में,
In my bag, |
1 | |
| 2 | ||
| 3 | ||
| 4 | ||
| 5 | ||
|
मेरे स्कूल में,
In my school, |
1 | |
| 2 | ||
| 3 | ||
| 4 | ||
| 5 |
धन्यवाद!


15.9: Excel Chart in PowerPoint
Learning Objectives
- Create PowerPoint slide with integrated Excel chart.
Option 1 for Chart Integration
The next slide in the presentation is the budget distribution chart. Cameron will create a chart from scratch in a new slide. He opens the New Slide options and selects Content with Caption, then types in the title in the box. He moves the cursor over the large box and clicks on the Insert Chart icon and decides on the type of chart.
For this slide, he selects a 3-D Pie chart. A small Excel spreadsheet pops up above the pie chart for Cameron to enter the data for the chart.
Cameron begins to type in the data by double-clicking on a cell and replacing the text or numbers in it. He changes the numbers cells to percentages. There is an extra row he doesn’t need, so he selects it, right-clicks, and selects the Delete Row from the menu. Cameron now clicks the X in the small window and closes it.
Cameron decides to show the budget present age on the chart and needs to modify a few other aspects about the chart to make it easier to see.
He takes the following steps:
- Chart Elements Button . This + button allows Cameron to change aspects of the chart.
- Elements List. Cameron clicks the Data Labels box to change the labels.
- Arrow Button. Cameron clicks the small right arrow, and a new menu appears.
- Data Label List. Cameron moves the chart numbers to the center of their respective areas.
Next, Cameron changes the area numbers to percentages and increases the font size of the chart text and the Store label sizes.
The chart is now too close to the store labels. Cameron clicks on the chart and hovers over the border until the four-arrow cursor appears. He drags the chart up, closer to the title. This centers the chart and makes it easier to read the labels.
Option 2 for Chart Integration
What if Cameron had an existing sales budget chart? He could copy and paste the chart into his presentation. First he opens the Excel file with the chart in it. Since he has learned to copy and paste from the previous table, he puts the PowerPoint and Excel windows side by side then takes the following steps:
- Select Excel . Cameron makes sure Excel is the current window and opens the chart file.
- Select Chart . Cameron selects the chart in the Excel file by right-clicking on the chart.
- Select Copy. Cameron sees a menu appear and selects the Copy choice. He can also use Ctrl+C to copy the chart.
Practice Question
- PowerPoint Presentation Paste Chart. With the desired chart copied, Cameron moves over to his PowerPoint presentation and places the curser in the correct spot and right-clicks again to bring up the menu. In the menu he selects the Paste option and the chart is placed in the presentation. He can also use the Ctrl+V keys to paste the chart.
- Resize Chart. Cameron now grabs one of the chart border area spots and moves it to the desired size in the presentation.
- Change Chart. If Cameron needs to change something about the chart, such as the type of chart, the data in the chart, the labels, and so on, he can right-click on the chart itself. A menu will appear that allows him to change aspects of that chart. Any changes made will not affect the original Excel file chart.
Practice Question
Contributors and Attributions
- Excel Chart in PowerPoint. Authored by : Sherri Pendleton. Provided by : Lumen Learning. License : CC BY: Attribution


1.1: Chapter 1 Flat Pan
We will start out with a flat pan template build. This will be one the simplest template we will construct, but there are key components of this simple template that will follow through to more complex templates.
Here we have both the perspective view and an orthographic drawing. Using this format it is easy to see what the finish product will look like. What we have to do here is reverse the process to see what shape we have to start with to build the pan. The one thing we need now are dimensions.
Now that we have the parts overall size we can determine how to go about laying this part flat and work with the dimensions given so we can create the pan.
By inspecting the drawing we can see that the over flat portion of the pan or the inside dimension is 8” x 12”. Each side has a total inside height of 1” each. By adding these together we can tell that what we will need to material to build the pan. This can commonly be called the STRETCHOUT. 8” + 1” + 1” + 12” + 1” +1” gives us an overall sheet size needed of 10” X 14”.

A simple rectangle that measure 8” x12”.
Now we need to add the 1” to all four corners that will become the edges of the pan. First you need to extend each line by 1” at each corner.
Now we simply connect a line on each side for each 1” extended line.
Now you can see in each corner there is a square notch. These notches allow 1” portion of the sheet to be bent up 90° to form the pan.
After sides are bent at 90°, the finish product should looks like this.
Notes about flat sheet layout and dimensions.
You need to pay attention to how the parts are dimensioned, inside or outside. If parts are dimensioned to the outside you will need to account for material thickness for each formed edge to properly determine Stretch-out . The Stretch-out will be based off of the inside surface. For each 90° bend .12” will be gained to overall length.
For example:
The outside dimension are 6.24” x 8.24” x 1” tall. To figure out what inside dimension is for bottom of pan for layout we need to take 2 x the sheet thickness minus the given outside dimension. The inside dimension of pan is 6” x 8” with an inside height of .88”.
To figure the stretch-out , we will need to add 6” + .88” + .88” = 7.76” and 8” + .88” +.88”=9.76”
The flat Stretch-out will look like this:


1: What is Human Sexuality Last updated Save as PDF Page ID 167162 Susan Rahman with Nathan Bowman, Dahmitra Jackson, Lushtak, Remi Newman, & Prateek Sunder California Consortium for Equitable Change in Hispanic Serving Institutions Open Educational Resources (CC ECHO) 1.1: Learning Objectives 1.2: What is Human Sexuality? 1.3: A Picture of Sara Baartman 1.4: Sexual Attitudes and Practices 1.5: Why is it important to Study Sexuality? 1.6: Sexuality Around the World 1.7: Sexuality in the United States 1.8: Sex Education 1.9: Sociological Perspectives on Sex and Sexuality 1.10: Conclusion 1.11: Discussion Questions 1.12: For Further Exploration 1.13: Glossary 1.14: References Thumbnail: Finger heart illustration ( Freepick Premium license ; Freepick via freepick )


7.6: Rights of the Arrestee
As with other criminal justice processes, the arrestee also has specific rights upon arrest. The rights of a person being arrested fall under the Fifth, Sixth and Eighth Amendment.
- "No person…shall be compelled in any criminal case to be a witness against himself, nor be deprived of life, liberty, or property, without due process of law...." (Fifth Amendment).
- "In all criminal prosecutions, the accused shall enjoy the right to a speedy and public trial, by an impartial jury…and to be informed of the nature and cause of the accusation; to be confronted with the witnesses against him; to have compulsory process for obtaining witnesses in his favor, and to have the assistance of counsel for his defense." (Sixth Amendment).
- "Excessive bail shall not be required, nor excessive fines imposed, nor shall cruel and unusual punishments be inflicted." (Eighth Amendment).
Once a person is arrested, the officer is required to provide the person taken into custody their rights. Commonly known as the Miranda Rights, the following admonishment must be provided.
- You have the right to remain silent
- Anything you say can and will be held against you in a court of law.
- You have the right to an attorney, if you cannot afford an attorney, one will be appointed to you free of charge.
- Do you understand these rights? (The arrestee must verbally provide they understand the rights as explained)
If these rights are not provided to the arrestee, any statements made could be inadmissible in a criminal trial. Officers must also respect the rights. Officers cannot force or coerce statements from the arrestee. If they decline to talk, and invoke their right to remain silent, the officer must cease questioning. An arrested person has the right to be informed about the grounds for arrest and about the factual circumstances and legal classification of the crime he or she is suspected of committing.
After being taken into custody, the arrestee also has the right to make a phone call to an attorney and a family member within a reasonable amount of time. If custody status changes or the arrestee is moved to a new facility, they must be allowed to make a phone call to notify their attorney and family member.
The arrestee must be arraigned within 48 hours from arrest, excluding weekends. If a person is arrested on Monday, they must be arraigned by Wednesday. However, if the arrest occurs on Thursday, they will be arraigned on Monday.


19.4: Oxidation of Aldehydes and Ketones
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Objectives
After completing this section, you should be able to
-
write an equation for the oxidation of an aldehyde using
- CrO 3 /sulphuric acid.
- Tollens reagent.
- explain the difference in structure which makes aldehydes susceptible to oxidation and ketones difficult to oxidize.
- identify the carboxylic acid produced when a given aldehyde is oxidized.
- identify the aldehyde, the oxidizing agent, or both, needed to prepare a given carboxylic acid.
Key Terms
Make certain that you can define, and use in context, the key term below.
- Tollens reagent
Study Notes
An important difference between aldehydes and ketones is the ease with which the latter can be oxidized. Tollen’s reagent is a classical organic laboratory technique to test for the presence of an aldehyde. The reagent consists of silver(I) ions dissolved in dilute ammonia. When the aldehyde is oxidized, the silver(I) ions are reduced to silver metal. When the reaction is carried out in a test‑tube, the metallic silver is deposited on the walls of the tube, giving it a mirrorlike appearance. This characteristic accounts for the term “silver mirror test” which is applied when this reaction is used to distinguish between aldehydes and ketones—the latter, of course, do not react.
Why do aldehydes and ketones behave differently?
Aldehydes have a proton attached to the carbonyl carbon which can be abstracted, allowing them to be easily oxidized to form carboxylic acids. The lack of this hydrogen, makes ketones generally inert to these oxidation conditions. Nevertheless, ketones can be oxidized but only under extreme conditions.
A significant distinction between aldehydes and ketones lies in their susceptibility to oxidation, with aldehydes being more easily oxidized than ketones. One of the classic methods for identifying aldehydes in the laboratory is through the use of Tollen’s reagent which consists of silver(I) ions dissolved in dilute ammonia. When an aldehyde is present, it undergoes oxidation, causing the silver(I) ions to be reduced to metallic silver. This reduction results in the deposition of silver on the surface of the test tube, forming a reflective, mirror-like coating. This reaction is commonly referred to as the “silver mirror test,” and it serves as a reliable way to distinguish aldehydes from ketones. Ketones, in contrast, do not react with Tollen's reagent, making this test specific for aldehydes.
Oxidation of hexanal to form hexanoic acid using Tollens Reagent
Oxidation of Aldehydes
There are a wide variety of reagents which can cause the oxidation of aldehydes to carboxylic acids. The most common reagent for this conversi on is \(\ce{CrO3}\) in aq ueous acid also called Jones Reagent . This reaction generally gives good yields at room temperature.
Example of the oxidation of hexanal to form hexanoic acid using Jones Reagent
Mechanism
The oxidation of aldehydes occur through the reversible nucleophilic addition of water to the carbonyl to form a gem-diol functional group. This addition reaction is discussed in greater detail in Section 19.5 . One of the OH groups of the gem -diol is oxidized to create a carbonyl (C=O) thereby forming a carboxylic acid.
Oxidation of Ketones and Baeyer-Villiger Oxidation
Because ketones do not have hydrogen atom attached to their carbonyl, they are resistant to oxidation. Only very strong oxidizing agents such as potassium manganate(VII) (potassium permanganate) solution oxidize ketones. However, this type of powerful oxidation occurs with cleavage, breaking carbon-carbon bonds and forming two carboxylic acids. Because of this destructive nature this reaction is rarely used.
Oxidation of cyclopentanone to form pentanedioic acid
Peroxycarboxylic acids , such as meta-chloroperoxybenzoic acid (mCPBA) , are capable of oxidizing ketones to esters in a reaction known as the Baeyer-Villiger oxidation . Baeyer-Villiger oxidation has considerable synthetic utility because ketones normally are difficult to oxidize without degrading the structure to smaller fragments.
Baeyer-Villiger oxidations can be used with both straight chain ketones and cyclic ketones as shown in the following examples.
Baeyer-Villiger Oxidation of 2-octanone to form hexyl ethanoate
Baeyer-Villiger Oxidation of cyclohexanone to 6-hexanolactone
Mechanism
The mechanism of the Baeyer-Villiger oxidation has been studied extensively and is of interest because it involves a rearrangement step in which a substituent group (R) moves from a carbon to an oxygen. In the first step, one oxygen from the peroxy carboxylic acid adds to the carbonyl group of the ketone. The adduct has multiple oxygen atoms on which protons can reside. An intramolecular proton transfer followed by protonation allows for generation of the Criegee intermediate. Migration of an alkyl group and elimination of a carboxylic acid, R 1 CO 2 H, then occur in the fourth step. This generates a protonated form of the ester product, which is deprotonated in the final step of the mechanism.
Step 1: Nucleophilic attack on the carbonyl
Step 2 : Intramolecular Proton Transfer
Step 3 : Protonation of the alkoxide
Step 4: Migration of an Alkyl Group
Step 5: Deprotonation