4.4: Buffer Solutions
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Learning Objectives
- Recognize common ions from various salts, acids, and bases.
- Calculate concentrations involving common ions.
- Calculate ion concentrations involving chemical equilibrium.
The common-ion effect is used to describe the effect on an equilibrium involving a substance that adds an ion that is a part of the equilibrium.
Introduction
The solubility products K sp 's are equilibrium constants in heterogeneous equilibria (i.e., between two different phases). If several salts are present in a system, they all ionize in the solution. If the salts contain a common cation or anion, these salts contribute to the concentration of the common ion. Contributions from all salts must be included in the calculation of concentration of the common ion. For example, a solution containing sodium chloride and potassium chloride will have the following relationship:
\[\mathrm{[Na^+] + [K^+] = [Cl^-]} \label{1} \]
Consideration of charge balance or mass balance or both leads to the same conclusion.
Common Ions
When \(\ce{NaCl}\) and \(\ce{KCl}\) are dissolved in the same solution, the \(\mathrm{ {\color{Green} Cl^-}}\) ions are common to both salts. In a system containing \(\ce{NaCl}\) and \(\ce{KCl}\), the \(\mathrm{ {\color{Green} Cl^-}}\) ions are common ions.
\[\mathrm{NaCl \rightleftharpoons Na^+ + {\color{Green} Cl^-}} \nonumber \]
\[\mathrm{KCl \rightleftharpoons K^+ + {\color{Green} Cl^-}} \nonumber \]
\[\mathrm{CaCl_2 \rightleftharpoons Ca^{2+} + {\color{Green} 2 Cl^-}} \nonumber \]
\[\mathrm{AlCl_3 \rightleftharpoons Al^{3+} + {\color{Green} 3 Cl^-}} \nonumber \]
\[\mathrm{AgCl \rightleftharpoons Ag^+ + {\color{Green} Cl^-}} \nonumber \]
For example, when \(\ce{AgCl}\) is dissolved into a solution already containing \(\ce{NaCl}\) (actually \(\ce{Na+}\) and \(\ce{Cl-}\) ions), the \(\ce{Cl-}\) ions come from the ionization of both \(\ce{AgCl}\) and \(\ce{NaCl}\). Thus, \(\ce{[Cl- ]}\) differs from \(\ce{[Ag+]}\). The following examples show how the concentration of the common ion is calculated.
Example \(\PageIndex{1}\)
What are \(\ce{[Na+]}\), \(\ce{[Cl- ]}\), \(\ce{[Ca^2+]}\), and \(\ce{[H+]}\) in a solution containing 0.10 M each of \(\ce{NaCl}\), \(\ce{CaCl2}\), and \(\ce{HCl}\)?
Solution
Due to the conservation of ions, we have
\[\mathrm{[Na^+] = [Ca^{2+}] = [H^+] = 0.10\: \ce M} \nonumber \]
but
\(\begin{alignat}{3}
\nonumber &\ce{[Cl- ]} &&= && && \:\textrm{0.10 (due to NaCl)}\\
\nonumber & && && + &&\mathrm{\:0.20\: (due\: to\: CaCl_2)}\\
\nonumber & && && + &&\mathrm{\:0.10\: (due\: to\: HCl)}\\
\nonumber & &&= && &&\mathrm{\:0.40\: M}
\nonumber \end{alignat}\)
Exercise \(\PageIndex{1}\)
John poured 10.0 mL of 0.10 M \(\ce{NaCl}\), 10.0 mL of 0.10 M \(\ce{KOH}\), and 5.0 mL of 0.20 M \(\ce{HCl}\) solutions together and then he made the total volume to be 100.0 mL. What is \(\ce{[Cl- ]}\) in the final solution?
Solution
\[\mathrm{[Cl^-] = \dfrac{0.1\: M\times 10\: mL+0.2\: M\times 5.0\: mL}{100.0\: mL} = 0.020\: M} \nonumber \]
Le Ch â telier's Principle states that if an equilibrium becomes unbalanced, the reaction will shift to restore the balance. If a common ion is added to a weak acid or weak base equilibrium, then the equilibrium will shift towards the reactants, in this case the weak acid or base.
Example \(\PageIndex{2}\): Solubility of Lead Chloride
Consider the lead(II) ion concentration in this saturated solution of PbCl 2 . The balanced reaction is
\[ PbCl_{2 (s)} \rightleftharpoons Pb^{2+} _{(aq)} + 2Cl^-_{(aq)} \nonumber \]
Defining \(s\) as the concentration of dissolved lead(II) chloride, then:
\[[Pb^{2+}] = s \nonumber \]
\[[Cl^- ] = 2s \nonumber \]
These values can be substituted into the solubility product expression, which can be solved for \(s\):
\[\begin{eqnarray} K_{sp} &=& [Pb^{2+}] [Cl^-]^2 \\ &=& s \times (2s)^2 \\ 1.7 \times 10^{-5} &=& 4s^3 \\ s^3 &=& \frac{1.7 \times 10^{-5}}{4} \\ &=& 4.25 \times 10^{-6} \\ s &=& \sqrt[3]{4.25 \times 10^{-6}} \\ &=& 1.62 \times 10^{-2}\ mol\ dm^{-3} \end{eqnarray} \nonumber \]The concentration of lead(II) ions in the solution is 1.62 x 10 -2 M. Consider what happens if sodium chloride is added to this saturated solution. Sodium chloride shares an ion with lead(II) chloride. The chloride ion is common to both of them; this is the origin of the term "common ion effect".
Look at the original equilibrium expression again:
\[ PbCl_2 \; (s) \rightleftharpoons Pb^{2+} \; (aq) + 2Cl^- \; (aq) \nonumber \]
What happens to that equilibrium if extra chloride ions are added? According to Le Ch â telier , the position of equilibrium will shift to counter the change, in this case, by removing the chloride ions by making extra solid lead(II) chloride.
Of course, the concentration of lead(II) ions in the solution is so small that only a tiny proportion of the extra chloride ions can be converted into solid lead(II) chloride. The lead(II) chloride becomes even less soluble , and the concentration of lead(II) ions in the solution decreases . This type of response occurs with any sparingly soluble substance: it is less soluble in a solution which contains any ion which it has in common. This is the common ion effect.
Example \(\PageIndex{3}\)
If an attempt is made to dissolve some lead(II) chloride in some 0.100 M sodium chloride solution instead of in water, what is the equilibrium concentration of the lead(II) ions this time? As before, define s to be the concentration of the lead(II) ions.
\[[Pb^{2+}] = s \label{2} \]
The calculations are different from before. This time the concentration of the chloride ions is governed by the concentration of the sodium chloride solution. The number of ions coming from the lead(II) chloride is going to be tiny compared with the 0.100 M coming from the sodium chloride solution.
In calculations like this, it can be assumed that the concentration of the common ion is entirely due to the other solution. This simplifies the calculation.
So we assume:
\[[Cl^- ] = 0.100\; M \label{3} \]
The rest of the mathematics looks like this:
\begin{equation} \begin{split} K_{sp}& = [Pb^{2+}][Cl^-]^2 \\ & = s \times (0.100)^2 \\ 1.7 \times 10^{-5} & = s \times 0.00100 \end{split} \nonumber \end{equation}
therefore:
\begin{equation} \begin{split} s & = \dfrac{1.7 \times 10^{-5}}{0.0100} \\ & = 1.7 \times 10^{-3} \, \text{M} \end{split} \label{4} \end{equation}
Finally, compare that value with the simple saturated solution:
Original solution:
\[[Pb^{2+}] = 0.0162 \, M \label{5} \]
Solution in 0.100 M NaCl solution:
\[ [Pb^{2+}] = 0.0017 \, M \label{6} \]
The concentration of the lead(II) ions has decreased by a factor of about 10. If more concentrated solutions of sodium chloride are used, the solubility decreases further.
A Video Discussing Finding the Solubility of a Salt: Finding the Solubility of a Salt(opens in new window) [youtu.be]
Common Ion Effect with Weak Acids and Bases
Adding a common ion prevents the weak acid or weak base from ionizing as much as it would without the added common ion. The common ion effect suppresses the ionization of a weak acid by adding more of an ion that is a product of this equilibrium.
Adding a common ion to a system at equilibrium affects the equilibrium composition, but not the ionization constant.
The common ion effect of H 3 O + on the ionization of acetic acid
The common ion effect suppresses the ionization of a weak base by adding more of an ion that is a product of this equilibrium. Now consider the common ion effect of \(\ce{OH^{-}}\) on the ionization of ammonia
Adding the common ion of hydroxide shifts the reaction towards the left to decrease the stress (in accordance with Le Cha telier's Principle ), forming more reactants. This decreases the reaction quotient, because the reaction is being pushed towards the left to reach equilibrium. The equilibrium constant, \(K_b=1.8 \times 10^{-5}\), does not change. The reaction is put out of balance, or equilibrium.
\[Q_a = \frac{\ce{[NH_4^{+}][OH^{-}]}}{\ce{[NH3]}} \nonumber \]
At first, when more hydroxide is added, the quotient is greater than the equilibrium constant. The reaction then shifts right, causing the denominator to increase, decreasing the reaction quotient and pulling towards equilibrium and causing Q to decrease towards K.
Common Ion Effect on Solubility
When a slightly soluble ionic compound is added to water, some of it dissolves to form a solution, establishing an equilibrium between the pure solid and a solution of its ions. For the dissolution of calcium phosphate, one of the two main components of kidney stones, the equilibrium can be written as follows, with the solid salt on the left:
\[\ce{Ca3(PO4)2(s) <=> 3Ca^{2+}(aq) + 2PO^{3−}4(aq)} \label{17.4.1} \]
As you will discover in more advanced chemistry courses, basic anions, such as S 2− , PO 4 3 − , and CO 3 2 − , react with water to produce OH − and the corresponding protonated anion. Consequently, their calculated molarities, assuming no protonation in aqueous solution, are only approximate. The equilibrium constant for the dissolution of a sparingly soluble salt is the solubility product ( K sp ) of the salt. Because the concentration of a pure solid such as Ca 3 (PO 4 ) 2 is a constant, it does not appear explicitly in the equilibrium constant expression. The equilibrium constant expression for the dissolution of calcium phosphate is therefore
\[K=\dfrac{[\mathrm{Ca^{2+}}]^3[\mathrm{PO_4^{3-}}]^2}{[\mathrm{Ca_3(PO_4)_2}]} \label{17.4.2a} \]
\[[\mathrm{Ca_3(PO_4)_2}]K=K_{\textrm{sp}}=[\mathrm{Ca^{2+}}]^3[\mathrm{PO_4^{3-}}]^2 \label{17.4.2b} \]
At 25°C and pH 7.00, \(K_{sp}\) for calcium phosphate is 2.07 × 10 −33 , indicating that the concentrations of Ca 2 + and PO 4 3 − ions in solution that are in equilibrium with solid calcium phosphate are very low. The values of K sp for some common salts vary dramatically for different compounds (Table E3). Although \(K_{sp}\) is not a function of pH in Equation \(\ref{17.4.2a}\) , changes in pH can affect the solubility of a compound.
The solubility product expression tells us that the equilibrium concentrations of the cation and the anion are inversely related. That is, as the concentration of the anion increases, the maximum concentration of the cation needed for precipitation to occur decreases—and vice versa—so that \(K_{sp}\) is constant. Consequently, the solubility of an ionic compound depends on the concentrations of other salts that contain the same ions. This dependency is another example of the common ion effect where adding a common cation or anion shifts a solubility equilibrium in the direction predicted by Le Chatelier’s principle. As a result, the solubility of any sparingly soluble salt is almost always decreased by the presence of a soluble salt that contains a common ion.
Consider, for example, the effect of adding a soluble salt, such as CaCl 2 , to a saturated solution of calcium phosphate [Ca 3 (PO 4 ) 2 ]. We have seen that the solubility of Ca 3 (PO 4 ) 2 in water at 25°C is 1.14 × 10 −7 M ( K sp = 2.07 × 10 −33 ). Thus a saturated solution of Ca 3 (PO 4 ) 2 in water contains
- \(3 \times (1.14 \times 10^{−7}\; M) = 3.42 \times 10^{−7} M \; of \; Ca^{2+}\)
- \(2 \times (1.14 \times 10^{−7} M) = 2.28 \times 10^{−7} M \; of \; PO_4^{3−}\)
according to the stoichiometry shown in Equation \(\ref{17.4.2a}\) (neglecting hydrolysis to form HPO 4 2 − ). If CaCl 2 is added to a saturated solution of Ca 3 (PO 4 ) 2 , the Ca 2 + ion concentration will increase such that [Ca 2 + ] > 3.42 × 10 −7 M, making Q > K sp . The only way the system can return to equilibrium is for the reaction in Equation \(\ref{17.4.2a}\) to proceed to the left, resulting in precipitation of Ca 3 (PO 4 ) 2 . This will decrease the concentration of both Ca 2 + and PO 4 3 − until Q = K sp .
Note
Adding a common ion decreases solubility, as the reaction shifts toward the left to relieve the stress of the excess product. Adding a common ion to a dissociation reaction causes the equilibrium to shift left, toward the reactants, causing precipitation.
Example \(\PageIndex{5}\)
Consider the reaction:
\[ PbCl_2(s) \rightleftharpoons Pb^{2+}(aq) + 2Cl^-(aq) \nonumber \]
What happens to the solubility of PbCl 2 (s) when 0.1 M NaCl is added?
Solution
\[K_{sp}=1.7 \times 10^{-5} \nonumber \]
\[Q_{sp}= 1.8 \times 10^{-5}\nonumber \]
Identify the common ion: Cl -
Notice: Q sp > K sp The addition of NaCl has caused the reaction to shift out of equilibrium because there are more dissociated ions. Typically, solving for the molarities requires the assumption that the solubility of PbCl 2 is equivalent to the concentration of Pb 2 + produced because they are in a 1:1 ratio.
Because K sp for the reaction is 1.7×10 -5 , the overall reaction would be (s)(2s) 2 = 1.7×10 -5 . Solving the equation for s gives s= 1.62×10 -2 M. The coefficient on Cl - is 2, so it is assumed that twice as much Cl - is produced as Pb 2 + , hence the '2s.' The solubility equilibrium constant can be used to solve for the molarities of the ions at equilibrium.
The molarity of Cl - added would be 0.1 M because Na + and Cl - are in a 1:1 ration in the ionic salt, NaCl. Therefore, the overall molarity of Cl - would be 2s + 0.1, with 2s referring to the contribution of the chloride ion from the dissociation of lead chloride.
\[\begin{eqnarray} Q_{sp} &=& [Pb^{2+}][Cl^-]^2 \\ 1.8 \times 10^{-5} &=& (s)(2s + 0.1)^2 \\ s &=& [Pb^{2+}] \\ &=& 1.8 \times 10^{-3} M \\ 2s &=& [Cl^-] \\ &\approx & 0.1 M \end{eqnarray} \nonumber \]
Notice that the molarity of Pb 2 + is lower when NaCl is added. The equilibrium constant remains the same because of the increased concentration of the chloride ion. To simplify the reaction, it can be assumed that [Cl - ] is approximately 0.1M since the formation of the chloride ion from the dissociation of lead chloride is so small. The reaction quotient for PbCl 2 is greater than the equilibrium constant because of the added Cl - . This therefore shift the reaction left towards equilibrium, causing precipitation and lowering the current solubility of the reaction. Overall, the solubility of the reaction decreases with the added sodium chloride.
Exercise \(\PageIndex{5}\)
Calculate the solubility of silver carbonate in a 0.25 M solution of sodium carbonate. The solubility of silver carbonate in pure water is 8.45 × 10 −12 at 25°C.
Answer
2.9 × 10 −6 M (versus 1.3 × 10 −4 M in pure water)
A Video Discussing the Common Ion Effect in Solubility Products: The Common Ion Effect in Solubility Products (opens in new window) [youtu.be]
References
- Harwood, William S., F. G. Herring, Jeffry D. Madura, and Ralph H. Petrucci. General Chemistry Principles and Modern Applications. 9th ed. New Jersey: Prentice Hall, 2007.
Contributors and Attributions
- Emmellin Tung, Mahtab Danai (UCD)
- Jim Clark (ChemGuide)
-
Chung (Peter) Chieh (Professor Emeritus, Chemistry @ University of Waterloo)
Learning Objectives
- To understand how adding a common ion affects the position of an acid–base equilibrium.
- To know how to use the Henderson-Hasselbalch approximation to calculate the pH of a buffer.
Buffers are solutions that maintain a relatively constant pH when an acid or a base is added. They therefore protect, or “buffer,” other molecules in solution from the effects of the added acid or base. Buffers contain either a weak acid (\(HA\)) and its conjugate base \((A^−\)) or a weak base (\(B\)) and its conjugate acid (\(BH^+\)), and they are critically important for the proper functioning of biological systems. In fact, every biological fluid is buffered to maintain its physiological pH.
The Common Ion Effect: Weak Acids Combined with Conjugate Bases
To understand how buffers work, let’s look first at how the ionization equilibrium of a weak acid is affected by adding either the conjugate base of the acid or a strong acid (a source of \(\ce{H^{+}}\)). Le Chatelier’s principle can be used to predict the effect on the equilibrium position of the solution. A typical buffer used in biochemistry laboratories contains acetic acid and a salt such as sodium acetate. The dissociation reaction of acetic acid is as follows:
\[\ce{CH3COOH (aq) <=> CH3COO^{−} (aq) + H^{+} (aq)} \label{Eq1} \]
and the equilibrium constant expression is as follows:
\[K_a=\dfrac{[\ce{H^{+}}][\ce{CH3COO^{-}}]}{[\ce{CH3CO2H}]} \label{Eq2} \]
Sodium acetate (\(\ce{CH_3CO_2Na}\)) is a strong electrolyte that ionizes completely in aqueous solution to produce \(\ce{Na^{+}}\) and \(\ce{CH3CO2^{−}}\) ions. If sodium acetate is added to a solution of acetic acid, Le Chatelier’s principle predicts that the equilibrium in Equation \ref{Eq1} will shift to the left, consuming some of the added \(\ce{CH_3COO^{−}}\) and some of the \(\ce{H^{+}}\) ions originally present in solution.
Because \(\ce{Na^{+}}\) is a spectator ion , it has no effect on the position of the equilibrium and can be ignored. The addition of sodium acetate produces a new equilibrium composition, in which \([\ce{H^{+}}]\) is less than the initial value. Because \([\ce{H^{+}}]\) has decreased, the pH will be higher. Thus adding a salt of the conjugate base to a solution of a weak acid increases the pH. This makes sense because sodium acetate is a base, and adding any base to a solution of a weak acid should increase the pH.
If we instead add a strong acid such as \(\ce{HCl}\) to the system, \([\ce{H^{+}}]\) increases. Once again the equilibrium is temporarily disturbed, but the excess \(\ce{H^{+}}\) ions react with the conjugate base (\(\ce{CH_3CO_2^{−}}\)), whether from the parent acid or sodium acetate, to drive the equilibrium to the left. The net result is a new equilibrium composition that has a lower [\(\ce{CH_3CO_2^{−}}\)] than before. In both cases, only the equilibrium composition has changed; the ionization constant \(K_a\) for acetic acid remains the same. Adding a strong electrolyte that contains one ion in common with a reaction system that is at equilibrium, in this case \(\ce{CH3CO2^{−}}\), will therefore shift the equilibrium in the direction that reduces the concentration of the common ion. The shift in equilibrium is via the common ion effect.
Adding a common ion to a system at equilibrium affects the equilibrium composition, but not the ionization constant.
Example \(\PageIndex{1}\)
A 0.150 M solution of formic acid at 25°C (pKa = 3.75) has a pH of 2.28 and is 3.5% ionized.
- Is there a change to the pH of the solution if enough solid sodium formate is added to make the final formate concentration 0.100 M (assume that the formic acid concentration does not change)?
- What percentage of the formic acid is ionized if 0.200 M HCl is added to the system?
Given : solution concentration and pH, \(pK_a\), and percent ionization of acid; final concentration of conjugate base or strong acid added
Asked for : pH and percent ionization of formic acid
Strategy :
- Write a balanced equilibrium equation for the ionization equilibrium of formic acid. Tabulate the initial concentrations, the changes, and the final concentrations.
- Substitute the expressions for the final concentrations into the expression for Ka. Calculate \([\ce{H^{+}}]\) and the pH of the solution.
- Construct a table of concentrations for the dissociation of formic acid. To determine the percent ionization, determine the anion concentration, divide it by the initial concentration of formic acid, and multiply the result by 100.
Solution :
A Because sodium formate is a strong electrolyte, it ionizes completely in solution to give formate and sodium ions. The \(\ce{Na^{+}}\) ions are spectator ions, so they can be ignored in the equilibrium equation. Because water is both a much weaker acid than formic acid and a much weaker base than formate, the acid–base properties of the solution are determined solely by the formic acid ionization equilibrium:
\[\ce{HCO2H (aq) <=> HCO^{−}2 (aq) + H^{+} (aq)} \nonumber \]
The initial concentrations, the changes in concentration that occur as equilibrium is reached, and the final concentrations can be tabulated.
| ICE | \([HCO_2H (aq) ]\) | \([H^+ (aq) ]\) | \([HCO^−_2 (aq) ]\) |
|---|---|---|---|
| Initial | 0.150 | \(1.00 \times 10^{−7}\) | 0.100 |
| Change | −x | +x | +x |
| Equilibrium | (0.150 − x) | x | (0.100 + x) |
B We substitute the expressions for the final concentrations into the equilibrium constant expression and make our usual simplifying assumptions, so
\[\begin{align*} K_a=\dfrac{[H^+][HCO_2^−]}{[HCO_2H]} &=\dfrac{(x)(0.100+x)}{0.150−x} \\[4pt] &\approx \dfrac{x(0.100)}{0.150} \\[4pt] &\approx 10^{−3.75} \\[4pt] &\approx 1.8 \times 10^{−4} \end{align*} \nonumber \]
Rearranging and solving for \(x\),
\[\begin{align*} x &=(1.8 \times 10^{−4}) \times \dfrac{0.150 \;M}{ 0.100 \;M} \\[4pt] &=2.7 \times 10^{−4}\\[4pt] &=[H^+] \end{align*} \nonumber \]
The value of \(x\) is small compared with 0.150 or 0.100 M, so our assumption about the extent of ionization is justified. Moreover,
\[K_aC_{HA} = (1.8 \times 10^{−4})(0.150) = 2.7 \times 10^{−5} \nonumber \]
which is greater than \(1.0 \times 10^{−6}\), so again, our assumption is justified. The final pH is:
\[pH= −\log(2.7 \times 10^{−4}) = 3.57 \nonumber \]
compared with the initial value of 2.29. Thus adding a salt containing the conjugate base of the acid has increased the pH of the solution, as we expect based on Le Chatelier’s principle; the stress on the system has been relieved by the consumption of \(\ce{H^{+}}\) ions, driving the equilibrium to the left.
C Because \(HCl\) is a strong acid, it ionizes completely, and chloride is a spectator ion that can be neglected. Thus the only relevant acid–base equilibrium is again the dissociation of formic acid, and initially the concentration of formate is zero. We can construct a table of initial concentrations, changes in concentration, and final concentrations.
\[HCO_2H (aq) \leftrightharpoons H^+ (aq) +HCO^−_2 (aq) \nonumber \]
| \([HCO_2H (aq) ]\) | \([H^+ (aq) ]\) | \([HCO^−_2 (aq) ]\) | |
|---|---|---|---|
| initial | 0.150 | 0.200 | 0 |
| change | −x | +x | +x |
| final | (0.150 − x) | (0.200 + x) | x |
To calculate the percentage of formic acid that is ionized under these conditions, we have to determine the final \([\ce{HCO2^{-}}]\). We substitute final concentrations into the equilibrium constant expression and make the usual simplifying assumptions, so
\[K_a=\dfrac{[H^+][HCO_2^−]}{[HCO_2H]}=\dfrac{(0.200+x)(x)}{0.150−x} \approx \dfrac{x(0.200)}{0.150}=1.80 \times 10^{−4} \nonumber \]
Rearranging and solving for \(x\),
\[\begin{align*} x &=(1.80 \times 10^{−4}) \times \dfrac{ 0.150\; M}{ 0.200\; M} \\[4pt] &=1.35 \times 10^{−4}=[HCO_2^−] \end{align*} \nonumber \]
Once again, our simplifying assumptions are justified. The percent ionization of formic acid is as follows:
\[\text{percent ionization}=\dfrac{1.35 \times 10^{−4} \;M} {0.150\; M} \times 100\%=0.0900\% \nonumber \]
Adding the strong acid to the solution, as shown in the table, decreased the percent ionization of formic acid by a factor of approximately 38 (3.45%/0.0900%). Again, this is consistent with Le Chatelier’s principle: adding \(\ce{H^{+}}\) ions drives the dissociation equilibrium to the left.
Exercise \(\PageIndex{1}\)
A 0.225 M solution of ethylamine (\(\ce{CH3CH2NH2}\) with \(pK_b = 3.19\)) has a pH of 12.08 and a percent ionization of 5.4% at 20°C. Calculate the following:
- the pH of the solution if enough solid ethylamine hydrochloride (\(\ce{EtNH3Cl}\)) is added to make the solution 0.100 M in \(\ce{EtNH3^{+}}\)
- the percentage of ethylamine that is ionized if enough solid \(\ce{NaOH}\) is added to the original solution to give a final concentration of 0.050 M \(\ce{NaOH}\)
- Answer a
-
11.16
- Answer b
-
1.3%
A Video Discussing the Common Ion Effect: The Common Ion Effecr(opens in new window) [youtu.be]
The Common Ion Effect: Weak Bases Combined with Conjugate Acids
Now let’s suppose we have a buffer solution that contains equimolar concentrations of a weak base (\(B\)) and its conjugate acid (\(BH^+\)). The general equation for the ionization of a weak base is as follows:
\[B (aq) +H_2O (l) \leftrightharpoons BH^+ (aq) +OH^− (aq) \label{Eq3} \]
If the equilibrium constant for the reaction as written in Equation \(\ref{Eq3}\) is small, for example \(K_b = 10^{−5}\), then the equilibrium constant for the reverse reaction is very large: \(K = \dfrac{1}{K_b} = 10^5\). Adding a strong base such as \(OH^-\) to the solution therefore causes the equilibrium in Equation \(\ref{Eq3}\) to shift to the left, consuming the added \(OH^-\). As a result, the \(OH^-\) ion concentration in solution remains relatively constant, and the pH of the solution changes very little. Le Chatelier’s principle predicts the same outcome: when the system is stressed by an increase in the \(OH^-\) ion concentration, the reaction will proceed to the left to counteract the stress.
If the \(pK_b\) of the base is 5.0, the \(pK_a\) of its conjugate acid is
\[pK_a = pK_w − pK_b = 14.0 – 5.0 = 9.0. \nonumber \]
Thus the equilibrium constant for ionization of the conjugate acid is even smaller than that for ionization of the base. The ionization reaction for the conjugate acid of a weak base is written as follows:
\[BH^+ (aq) +H_2O (l) \leftrightharpoons B (aq) +H_3O^+ (aq) \label{Eq4} \]
Again, the equilibrium constant for the reverse of this reaction is very large: K = 1/K a = 10 9 . If a strong acid is added, it is neutralized by reaction with the base as the reaction in Equation \(\ref{Eq4}\) shifts to the left. As a result, the \(H^+\) ion concentration does not increase very much, and the pH changes only slightly. In effect, a buffer solution behaves somewhat like a sponge that can absorb \(H^+\) and \(OH^-\) ions, thereby preventing large changes in pH when appreciable amounts of strong acid or base are added to a solution.
Buffers are characterized by the pH range over which they can maintain a more or less constant pH and by their buffer capacity, the amount of strong acid or base that can be absorbed before the pH changes significantly. Although the useful pH range of a buffer depends strongly on the chemical properties of the weak acid and weak base used to prepare the buffer (i.e., on \(K\)), its buffer capacity depends solely on the concentrations of the species in the buffered solution. The more concentrated the buffer solution, the greater its buffer capacity. As illustrated in Figure \(\PageIndex{1}\), when \(NaOH\) is added to solutions that contain different concentrations of an acetic acid/sodium acetate buffer, the observed change in the pH of the buffer is inversely proportional to the concentration of the buffer. If the buffer capacity is 10 times larger, then the buffer solution can absorb 10 times more strong acid or base before undergoing a significant change in pH.
A buffer maintains a relatively constant pH when acid or base is added to a solution. The addition of even tiny volumes of 0.10 M \(NaOH\) to 100.0 mL of distilled water results in a very large change in pH. As the concentration of a 50:50 mixture of sodium acetate/acetic acid buffer in the solution is increased from 0.010 M to 1.00 M, the change in the pH produced by the addition of the same volume of \(NaOH\) solution decreases steadily. For buffer concentrations of at least 0.500 M, the addition of even 25 mL of the \(NaOH\) solution results in only a relatively small change in pH.
Calculating the pH of a Buffer
The pH of a buffer can be calculated from the concentrations of the weak acid and the weak base used to prepare it, the concentration of the conjugate base and conjugate acid, and the \(pK_a\) or \(pK_b\) of the weak acid or weak base. The procedure is analogous to that used in Example \(\PageIndex{1}\) to calculate the pH of a solution containing known concentrations of formic acid and formate.
An alternative method frequently used to calculate the pH of a buffer solution is based on a rearrangement of the equilibrium equation for the dissociation of a weak acid. The simplified ionization reaction is \(HA \leftrightharpoons H^+ + A^−\), for which the equilibrium constant expression is as follows:
\[K_a=\dfrac{[H^+][A^-]}{[HA]} \label{Eq5} \]
This equation can be rearranged as follows:
\[[H^+]=K_a\dfrac{[HA]}{[A^−]} \label{Eq6} \]
Taking the logarithm of both sides and multiplying both sides by −1,
\[ \begin{align} −\log[H^+] &=−\log K_a−\log\left(\dfrac{[HA]}{[A^−]}\right) \\[4pt] &=−\log{K_a}+\log\left(\dfrac{[A^−]}{[HA]}\right) \label{Eq7} \end{align} \]
Replacing the negative logarithms in Equation \(\ref{Eq7}\),
\[pH=pK_a+\log \left( \dfrac{[A^−]}{[HA]} \right) \label{Eq8} \]
or, more generally,
\[pH=pK_a+\log\left(\dfrac{[base]}{[acid]}\right) \label{Eq9} \]
Equation \(\ref{Eq8}\) and Equation \(\ref{Eq9}\) are both forms of the Henderson-Hasselbalch approximation, named after the two early 20th-century chemists who first noticed that this rearranged version of the equilibrium constant expression provides an easy way to calculate the pH of a buffer solution. In general, the validity of the Henderson-Hasselbalch approximation may be limited to solutions whose concentrations are at least 100 times greater than their \(K_a\) values.
There are three special cases where the Henderson-Hasselbalch approximation is easily interpreted without the need for calculations:
- \([base] = [acid]\): Under these conditions, \[\dfrac{[base]}{[acid]} = 1 \nonumber \] in Equation \ref{Eq9}. Because \(\log 1 = 0\), \[pH = pK_a \nonumber \] regardless of the actual concentrations of the acid and base. Recall that this corresponds to the midpoint in the titration of a weak acid or a weak base.
- \([base]/[acid] = 10\): In Equation \(\ref{Eq9}\), because \(\log 10 = 1\), \[pH = pK_a + 1. \nonumber \]
- \([base]/[acid] = 100\): In Equation \(\ref{Eq9}\), because \(\log 100 = 2\), \[pH = pK_a + 2. \nonumber \]
Each time we increase the [base]/[acid] ratio by 10, the pH of the solution increases by 1 pH unit. Conversely, if the [base]/[acid] ratio is 0.1, then pH = \(pK_a\) − 1. Each additional factor-of-10 decrease in the [base]/[acid] ratio causes the pH to decrease by 1 pH unit.
If [base] = [acid] for a buffer, then pH = \(pK_a\). Changing this ratio by a factor of 10 either way changes the pH by ±1 unit.
Example \(\PageIndex{2}\)
What is the pH of a solution that contains
- 0.135 M \(\ce{HCO2H}\) and 0.215 M \(\ce{HCO2Na}\)? (The \(pK_a\) of formic acid is 3.75.)
- 0.0135 M \(\ce{HCO2H}\) and 0.0215 M \(\ce{HCO2Na}\)?
- 0.119 M pyridine and 0.234 M pyridine hydrochloride? (The \(pK_b\) of pyridine is 8.77.)
Given : concentration of acid, conjugate base, and \(pK_a\); concentration of base, conjugate acid, and \(pK_b\)
Asked for : pH
Strategy :
Substitute values into either form of the Henderson-Hasselbalch approximation (Equations \ref{Eq8} or \ref{Eq9}) to calculate the pH.
Solution :
According to the Henderson-Hasselbalch approximation (Equation \ref{Eq8}), the pH of a solution that contains both a weak acid and its conjugate base is
\[pH = pK_a + \log([A−]/[HA]). \nonumber \]
A
Inserting the given values into the equation,
\[\begin{align*} pH &=3.75+\log\left(\dfrac{0.215}{0.135}\right) \\[4pt] &=3.75+\log 1.593 \\[4pt] &=3.95 \end{align*} \nonumber \]
This result makes sense because the \([A^−]/[HA]\) ratio is between 1 and 10, so the pH of the buffer must be between the \(pK_a\) (3.75) and \(pK_a + 1\), or 4.75.
B
This is identical to part (a), except for the concentrations of the acid and the conjugate base, which are 10 times lower. Inserting the concentrations into the Henderson-Hasselbalch approximation,
\[\begin{align*} pH &=3.75+\log\left(\dfrac{0.0215}{0.0135}\right) \\[4pt] &=3.75+\log 1.593 \\[4pt] &=3.95 \end{align*} \nonumber \]
This result is identical to the result in part (a), which emphasizes the point that the pH of a buffer depends only on the ratio of the concentrations of the conjugate base and the acid, not on the magnitude of the concentrations. Because the [A − ]/[HA] ratio is the same as in part (a), the pH of the buffer must also be the same (3.95).
C
In this case, we have a weak base, pyridine (Py), and its conjugate acid, the pyridinium ion (\(HPy^+\)). We will therefore use Equation \(\ref{Eq9}\), the more general form of the Henderson-Hasselbalch approximation, in which “base” and “acid” refer to the appropriate species of the conjugate acid–base pair. We are given [base] = [Py] = 0.119 M and \([acid] = [HPy^{+}] = 0.234\, M\). We also are given \(pK_b = 8.77\) for pyridine, but we need \(pK_a\) for the pyridinium ion. Recall from Equation 16.23 that the \(pK_b\) of a weak base and the \(pK_a\) of its conjugate acid are related:
\[pK_a + pK_b = pK_w. \nonumber \]
Thus \(pK_a\) for the pyridinium ion is \(pK_w − pK_b = 14.00 − 8.77 = 5.23\). Substituting this \(pK_a\) value into the Henderson-Hasselbalch approximation,
\[\begin{align*} pH=pK_a+\log \left(\dfrac{[base]}{[acid]}\right) \\[4pt] &=5.23+\log\left(\dfrac{0.119}{0.234}\right) \\[4pt] & =5.23 −0.294 \\[4pt] &=4.94 \end{align*} \nonumber \]
Once again, this result makes sense: the \([B]/[BH^+]\) ratio is about 1/2, which is between 1 and 0.1, so the final pH must be between the \(pK_a\) (5.23) and \(pK_a − 1\), or 4.23.
Exercise \(\PageIndex{2}\)
What is the pH of a solution that contains
- 0.333 M benzoic acid and 0.252 M sodium benzoate?
- 0.050 M trimethylamine and 0.066 M trimethylamine hydrochloride?
The \(pK_a\) of benzoic acid is 4.20, and the \(pK_b\) of trimethylamine is also 4.20.
- Answer a
-
4.08
- Answer b
-
9.68
A Video Discussing Using the Henderson Hasselbalch Equation: Using the Henderson Hasselbalch Equation(opens in new window) [youtu.be] (opens in new window)
The Henderson-Hasselbalch approximation ((Equation \(\ref{Eq8}\)) can also be used to calculate the pH of a buffer solution after adding a given amount of strong acid or strong base, as demonstrated in Example \(\PageIndex{3}\).
Example \(\PageIndex{3}\)
The buffer solution in Example \(\PageIndex{2}\) contained 0.135 M \(\ce{HCO2H}\) and 0.215 M \(\ce{HCO2Na}\) and had a pH of 3.95.
- What is the final pH if 5.00 mL of 1.00 M \(HCl\) are added to 100 mL of this solution?
- What is the final pH if 5.00 mL of 1.00 M \(NaOH\) are added?
Given : composition and pH of buffer; concentration and volume of added acid or base
Asked for : final pH
Strategy:
- Calculate the amounts of formic acid and formate present in the buffer solution using the procedure from Example \(\PageIndex{1}\). Then calculate the amount of acid or base added.
- Construct a table showing the amounts of all species after the neutralization reaction. Use the final volume of the solution to calculate the concentrations of all species. Finally, substitute the appropriate values into the Henderson-Hasselbalch approximation (Equation \ref{Eq9}) to obtain the pH.
Solution :
The added \(\ce{HCl}\) (a strong acid) or \(\ce{NaOH}\) (a strong base) will react completely with formate (a weak base) or formic acid (a weak acid), respectively, to give formic acid or formate and water. We must therefore calculate the amounts of formic acid and formate present after the neutralization reaction.
A We begin by calculating the millimoles of formic acid and formate present in 100 mL of the initial pH 3.95 buffer:
\[ 100 \, \cancel{mL} \left( \dfrac{0.135 \, mmol\; \ce{HCO2H}}{\cancel{mL}} \right) = 13.5\, mmol\, \ce{HCO2H} \nonumber \]
\[ 100\, \cancel{mL } \left( \dfrac{0.215 \, mmol\; \ce{HCO2^{-}}}{\cancel{mL}} \right) = 21.5\, mmol\, \ce{HCO2^{-}} \nonumber \]
The millimoles of \(\ce{H^{+}}\) in 5.00 mL of 1.00 M \(\ce{HCl}\) is as follows:
\[ 5.00 \, \cancel{mL } \left( \dfrac{1.00 \,mmol\; \ce{H^{+}}}{\cancel{mL}} \right) = 5\, mmol\, \ce{H^{+}} \nonumber \]
B Next, we construct a table of initial amounts, changes in amounts, and final amounts:
\[\ce{HCO^{2−}(aq) + H^{+} (aq) <=> HCO2H (aq)} \nonumber \]
| \(HCO^{2−} (aq) \) | \(H^+ (aq) \) | \(HCO_2H (aq) \) | |
|---|---|---|---|
| Initial | 21.5 mmol | 5.00 mmol | 13.5 mmol |
| Change | −5.00 mmol | −5.00 mmol | +5.00 mmol |
| Final | 16.5 mmol | ∼0 mmol | 18.5 mmol |
The final amount of \(H^+\) in solution is given as “∼0 mmol.” For the purposes of the stoichiometry calculation, this is essentially true, but remember that the point of the problem is to calculate the final \([H^+]\) and thus the pH. We now have all the information we need to calculate the pH. We can use either the lengthy procedure of Example \(\PageIndex{1}\) or the Henderson–Hasselbach approximation. Because we have performed many equilibrium calculations in this chapter, we’ll take the latter approach. The Henderson-Hasselbalch approximation requires the concentrations of \(HCO_2^−\) and \(HCO_2H\), which can be calculated using the number of millimoles (\(n\)) of each and the total volume (\(VT\)). Substituting these values into the Henderson-Hasselbalch approximation (Equation \(\ref{Eq9}\)):
\[\begin{align*} pH &=pK_a+\log \left( \dfrac{[HCO_2^−]}{[HCO_2H]} \right) \\[4pt] &=pK_a+\log\left(\dfrac{n_{HCO_2^−}/V_f}{n_{HCO_2H}/V_f}\right) \\[4pt] &=pK_a+\log \left(\dfrac{n_{HCO_2^−}}{n_{HCO_2H}}\right) \end{align*} \nonumber \]
Because the total volume appears in both the numerator and denominator, it cancels. We therefore need to use only the ratio of the number of millimoles of the conjugate base to the number of millimoles of the weak acid. So
\[\begin{align*} pH &=pK_a+\log\left(\dfrac{n_{HCO_2^−}}{n_{HCO_2H}}\right) \\[4pt] &=3.75+\log\left(\dfrac{16.5\; mmol}{18.5\; mmol}\right) \\[4pt] &=3.75 −0.050=3.70 \end{align*} \nonumber \]
Once again, this result makes sense on two levels. First, the addition of \(HCl \)has decreased the pH from 3.95, as expected. Second, the ratio of \(HCO_2^−\) to \(HCO_2H\) is slightly less than 1, so the pH should be between the \(pK_a\) and \(pK_a\) − 1.
A The procedure for solving this part of the problem is exactly the same as that used in part (a). We have already calculated the numbers of millimoles of formic acid and formate in 100 mL of the initial pH 3.95 buffer: 13.5 mmol of \(HCO_2H\) and 21.5 mmol of \(HCO_2^−\). The number of millimoles of \(OH^-\) in 5.00 mL of 1.00 M \(NaOH\) is as follows:
B With this information, we can construct a table of initial amounts, changes in amounts, and final amounts.
\[\ce{HCO2H (aq) + OH^{−} (aq) <=> HCO^{−}2 (aq) + H2O (l)} \nonumber \]
| \(HCO_2H (aq) \) | \(OH^−\) | \(HCO^−_2 (aq) \) | |
|---|---|---|---|
| Initial | 13.5 mmol | 5.00 mmol | 21.5 mmol |
| Change | −5.00 mmol | −5.00 mmol | +5.00 mmol |
| Final | 8.5 mmol | ∼0 mmol | 26.5 mmol |
The final amount of \(OH^-\) in solution is not actually zero; this is only approximately true based on the stoichiometric calculation. We can calculate the final pH by inserting the numbers of millimoles of both \(HCO_2^−\) and \(HCO_2H\) into the simplified Henderson-Hasselbalch expression used in part (a) because the volume cancels:
\[\begin{align*} pH &=pK_a+\log \left(\dfrac{n_{HCO_2^−}}{n_{HCO_2H}}\right) \\[4pt] &=3.75+\log \left(\dfrac{26.5\; mmol}{8.5\; mmol} \right) \\[4pt] &=3.75+0.494 =4.24 \end{align*} \nonumber \]
Once again, this result makes chemical sense: the pH has increased, as would be expected after adding a strong base, and the final pH is between the \(pK_a\) and \(pK_a\) + 1, as expected for a solution with a \(HCO_2^−/HCO_2H\) ratio between 1 and 10.
Exercise \(\PageIndex{3}\)
The buffer solution from Example \(\PageIndex{2}\) contained 0.119 M pyridine and 0.234 M pyridine hydrochloride and had a pH of 4.94.
- What is the final pH if 12.0 mL of 1.5 M \(\ce{NaOH}\) are added to 250 mL of this solution?
- What is the final pH if 12.0 mL of 1.5 M \(\ce{HCl}\) are added?
- Answer a
-
5.30
- Answer b
-
4.42
Only the amounts (in moles or millimoles) of the acidic and basic components of the buffer are needed to use the Henderson-Hasselbalch approximation, not their concentrations.
A Video Discussing the Change in pH with the Addition of a Strong Acid to a Buffer: The Change in pH with the Addition of a Strong Acid to a Buffer(opens in new window) [youtu.be]
The Change in pH with the Addition of a Strong Base to a Buffer:
The Change in pH with the Addition of a Strong Base to a Buffer (opens in new window) [youtu.be]
The results obtained in Example \(\PageIndex{3}\) and its corresponding exercise demonstrate how little the pH of a well-chosen buffer solution changes despite the addition of a significant quantity of strong acid or strong base. Suppose we had added the same amount of \(HCl\) or \(NaOH\) solution to 100 mL of an unbuffered solution at pH 3.95 (corresponding to \(1.1 \times 10^{−4}\) M HCl). In this case, adding 5.00 mL of 1.00 M \(HCl\) would lower the final pH to 1.32 instead of 3.70, whereas adding 5.00 mL of 1.00 M \(NaOH\) would raise the final pH to 12.68 rather than 4.24. (Try verifying these values by doing the calculations yourself.) Thus the presence of a buffer significantly increases the ability of a solution to maintain an almost constant pH.
The most effective buffers contain equal concentrations of an acid and its conjugate base.
A buffer that contains approximately equal amounts of a weak acid and its conjugate base in solution is equally effective at neutralizing either added base or added acid. This is shown in Figure \(\PageIndex{2}\) for an acetic acid/sodium acetate buffer. Adding a given amount of strong acid shifts the system along the horizontal axis to the left, whereas adding the same amount of strong base shifts the system the same distance to the right. In either case, the change in the ratio of \(CH_3CO_2^−\) to \(CH_3CO_2H\) from 1:1 reduces the buffer capacity of the solution.
Figure \(\PageIndex{2}\): Distribution Curve Showing the Fraction of Acetic Acid Molecules and Acetate Ions as a Function of pH in a Solution of Acetic Acid. The pH range over which the acetic acid/sodium acetate system is an effective buffer (the darker shaded region) corresponds to the region in which appreciable concentrations of both species are present (pH 3.76–5.76, corresponding to \(pH = pK_a \pm 1\)). (CC BY-SA-NC; Anonymous by request)
A Video Discussing The Buffer Region: The Buffer Region (opens in new window) [youtu.be]
The Relationship between Titrations and Buffers
There is a strong correlation between the effectiveness of a buffer solution and the titration curves discussed in Section 16.5. Consider the schematic titration curve of a weak acid with a strong base shown in Figure \(\PageIndex{3}\). As indicated by the labels, the region around \(pK_a\) corresponds to the midpoint of the titration, when approximately half the weak acid has been neutralized. This portion of the titration curve corresponds to a buffer: it exhibits the smallest change in pH per increment of added strong base, as shown by the nearly horizontal nature of the curve in this region. The nearly flat portion of the curve extends only from approximately a pH value of 1 unit less than the \(pK_a\) to approximately a pH value of 1 unit greater than the \(pK_a\), which is why buffer solutions usually have a pH that is within ±1 pH units of the \(pK_a\) of the acid component of the buffer.
This schematic plot of pH for the titration of a weak acid with a strong base shows the nearly flat region of the titration curve around the midpoint, which corresponds to the formation of a buffer. At the lower left, the pH of the solution is determined by the equilibrium for dissociation of the weak acid; at the upper right, the pH is determined by the equilibrium for reaction of the conjugate base with water.
In the region of the titration curve at the lower left, before the midpoint, the acid–base properties of the solution are dominated by the equilibrium for dissociation of the weak acid, corresponding to \(K_a\). In the region of the titration curve at the upper right, after the midpoint, the acid–base properties of the solution are dominated by the equilibrium for reaction of the conjugate base of the weak acid with water, corresponding to \(K_b\). However, we can calculate either \(K_a\) or \(K_b\) from the other because they are related by \(K_w\).
 This is not to say that the pH is uniform throughout all cells and tissues of a mammal. The internal pH of a red blood cell is about 7.2, but the pH of most other kinds of cells is lower, around 7.0. Even within a single cell, different compartments can have very different pH values. For example, one intracellular compartment in white blood cells has a pH of around 5.0.
Because no single buffer system can effectively maintain a constant pH value over the entire physiological range of approximately pH 5.0 to 7.4, biochemical systems use a set of buffers with overlapping ranges. The most important of these is the \(\ce{CO2}/\ce{HCO3^{−}}\) system, which dominates the buffering action of blood plasma.
The acid–base equilibrium in the \(\ce{CO2}/\ce{HCO3^{−}}\) buffer system is usually written as follows:
\[\ce{H2CO3 (aq) <=> H^{+} (aq) + HCO^{-}3 (aq)} \label{Eq10} \]
with \(K_a = 4.5 \times 10^{−7}\) and \(pK_a = 6.35\) at 25°C. In fact, Equation \(\ref{Eq10}\) is a grossly oversimplified version of the \(\ce{CO2}/\ce{HCO3^{-}}\) system because a solution of \(\ce{CO2}\) in water contains only rather small amounts of \(H_2CO_3\). Thus Equation \(\ref{Eq10}\) does not allow us to understand how blood is actually buffered, particularly at a physiological temperature of 37°C.
As shown in Equation \(\ref{Eq11}\), \(\ce{CO2}\) is in equilibrium with \(\ce{H2CO3}\), but the equilibrium lies far to the left, with an \(\ce{H2CO3}/\ce{CO2}\) ratio less than 0.01 under most conditions:
\[\ce{CO2 (aq) + H2O (l) <=> H2CO3 (aq)} \label{Eq11} \]
with \(K′ = 4.0 \times 10^{−3}\) at 37°C. The true \(pK_a\) of carbonic acid at 37°C is therefore 3.70, not 6.35, corresponding to a \(K_a\) of \(2.0 \times 10^{−4}\), which makes it a much stronger acid than Equation \ref{Eq10} suggests. Adding Equation \ref{Eq10} and Equation \ref{Eq11} and canceling \(\ce{H2CO3}\) from both sides give the following overall equation for the reaction of \(\ce{CO2}\) with water to give a proton and the bicarbonate ion:
\[\ce{CO2 (aq) + H2O (l) <=> H2CO3 (aq)} \label{16.65a} \]
with \(K'=4.0 \times 10^{−3} (37°C)\)
\[\ce{H2CO3 (aq) <=> H^{+} (aq) + HCO^{-}3 (aq)} \label{16.65b} \]
with \(K_a=2.0 \times 10^{−4} (37°C)\)
\[\ce{CO2 (aq) + H2O (l) <=> H^{+} (aq) + HCO^{-}3 (aq)} \label{16.65c} \]
with \(K=8.0 \times 10^{−7} (37°C)\)
The \(K\) value for the reaction in Equation \ref{16.65c} is the product of the true ionization constant for carbonic acid (\(K_a\)) and the equilibrium constant (K) for the reaction of \(\ce{CO2 (aq)} \) with water to give carbonic acid. The equilibrium equation for the reaction of \(\ce{CO2}\) with water to give bicarbonate and a proton is therefore
\[K=\dfrac{[\ce{H^{+}}][\ce{HCO3^{-}}]}{[\ce{CO2}]}=8.0 \times 10^{−7} \label{eq13} \]
The presence of a gas in the equilibrium constant expression for a buffer is unusual. According to Henry’s law ,
\[[\ce{CO2}]=k P_{\ce{CO2}} \nonumber \]
where \(k\) is the Henry’s law constant for \(\ce{CO2}\), which is \(3.0 \times 10^{−5} \;M/mmHg\) at 37°C. Substituting this expression for \([\ce{CO2}]\) in Equation \ref{eq13},
\[K=\dfrac{[\ce{H^{+}}][\ce{HCO3^{-}}]}{(3.0 \times 10^{−5}\; M/mmHg)(P_{\ce{CO2}})} \nonumber \]
where \(P_{\ce{CO2}}\) is in mmHg. Taking the negative logarithm of both sides and rearranging,
\[pH=6.10+\log \left( \dfrac{ [\ce{HCO3^{−}}]}{(3.0 \times 10^{−5} M/mm \;Hg)\; (P_{\ce{CO2}}) } \right) \label{Eq15} \]
Thus the pH of the solution depends on both the \(\ce{CO2}\) pressure over the solution and \([\ce{HCO3^{−}}]\). Figure \(\PageIndex{4}\) plots the relationship between pH and \([\ce{HCO3^{−}}]\) under physiological conditions for several different values of \(P_{\ce{CO2}}\), with normal pH and \([\ce{HCO3^{−}}]\) values indicated by the dashed lines.
According to Equation \ref{Eq15}, adding a strong acid to the \(\ce{CO2}/\ce{HCO3^{−}}\) system causes \([\ce{HCO3^{−}}]\) to decrease as \(\ce{HCO3^{−}}\) is converted to \(\ce{CO2}\). Excess \(\ce{CO2}\) is released in the lungs and exhaled into the atmosphere, however, so there is essentially no change in \(P_{\ce{CO2}}\). Because the change in \([\ce{HCO3^{−}}]/P_{CO_2}\) is small, Equation \ref{Eq15} predicts that the change in pH will also be rather small. Conversely, if a strong base is added, the \(\ce{OH^{-}}\) reacts with \(\ce{CO2}\) to form \(\ce{HCO3^{−}}\), but \(\ce{CO2}\) is replenished by the body, again limiting the change in both \([\ce{HCO3^{−}}]/P_{\ce{CO2}}\) and pH. The \(\ce{CO2}/\ce{HCO3^{−}}\) buffer system is an example of an open system, in which the total concentration of the components of the buffer change to keep the pH at a nearly constant value.
If a passenger steps out of an airplane in Denver, Colorado, for example, the lower \(P_{\ce{CO2}}\) at higher elevations (typically 31 mmHg at an elevation of 2000 m versus 40 mmHg at sea level) causes a shift to a new pH and \([\ce{HCO3^{-}}]\). The increase in pH and decrease in \([\ce{HCO3^{−}}]\) in response to the decrease in \(P_{\ce{CO2}}\) are responsible for the general malaise that many people experience at high altitudes. If their blood pH does not adjust rapidly, the condition can develop into the life-threatening phenomenon known as altitude sickness.
A Video Summary of the pH Curve for a Strong Acid/Strong Base Titration:
Summary of the pH Curve for a Strong Acid/Strong Base Titration(opens in new window) [youtu.be]
Summary
Buffers are solutions that resist a change in pH after adding an acid or a base. Buffers contain a weak acid (\(HA\)) and its conjugate weak base (\(A^−\)). Adding a strong electrolyte that contains one ion in common with a reaction system that is at equilibrium shifts the equilibrium in such a way as to reduce the concentration of the common ion. The shift in equilibrium is called the common ion effect. Buffers are characterized by their pH range and buffer capacity. The useful pH range of a buffer depends strongly on the chemical properties of the conjugate weak acid–base pair used to prepare the buffer (the \(K_a\) or \(K_b\)), whereas its buffer capacity depends solely on the concentrations of the species in the solution. The pH of a buffer can be calculated using the Henderson-Hasselbalch approximation, which is valid for solutions whose concentrations are at least 100 times greater than their \(K_a\) values. Because no single buffer system can effectively maintain a constant pH value over the physiological range of approximately 5 to 7.4, biochemical systems use a set of buffers with overlapping ranges. The most important of these is the \(CO_2/HCO_3^−\) system, which dominates the buffering action of blood plasma.


15.8: Introduction to Writing Ethically
What you’ll learn to do: evaluate concerns about plagiarism and how to demonstrate academic integrity
Parts of Melania Trump’s speech at the Republican National Convention in July 2016 were strikingly similar to Michelle Obama’s speech from eight years ago. How does this happen? Plagiarism at this level is typically the result of sloppiness, not outright theft, but it’s still just as egregious a result no matter the underlying cause. If you don’t want this to happen to you, then you need to be aware of plagiarism and how to avoid it.
In 2008, here’s what Michelle Obama said, in part (most of the highlights in these passages, which show the similar parts, are from a Wall Street Journal article about the plagiarism).
Like my family, they scrimped and saved so that he could have opportunities they never had themselves. And Barack and I were raised with so many of the same values: that you work hard for what you want in life ; that your word is your bond and you do what you say you’re going to do; that you treat people with dignity and respect , even if you don’t know them, and even if you don’t agree with them. And Barack and I set out to build lives guided by these values, and pass them on to the next generation. Because we want our children—and all children in this nation—to know that the only limit to the height of your achievements is the reach of your dreams and your willingness to work for them .
And here’s the similar passage from Melania Trump’s speech:
From a young age, my parents impressed on me the values that you work hard for what you want in life , that your word is your bond and you do what you say and keep your promise, that you treat people with respect . They taught and showed me values and morals in their daily lives. That is a lesson that I continue to pass along to our son. And we need to pass those lessons on to the many generations to follow. Because we want our children in this nation to know that the only limit to your achievements is the strength of your dreams and your willingness to work for them.
You can watch a video of the similar snippets from their speeches here .
How plagiarism really happens
According to author Josh Bernoff, Melania’s Trump plagiarism is probably not the obvious kind of cut-and-paste theft you might associate with cheating on a research paper, but it’s still clearly plagiarism. He imagines that Melania Trump and her speechwriters talked about what she wanted to communicate and what’s important to her. They probably discussed ideas, developed a theme, and researched it. They also researched past, successful speeches from prospective first ladies. The result of that research was a bunch of fragments from all over the place. One of those fragments was the piece of Michelle Obama’s speech. As the fragments coalesced into a speech, that one got included, because the writers lost track of its provenance, or Melania Trump did.
You may want to believe a more nefarious explanation, but carelessness is far more likely. For example, the famous primatologist Jane Goodall blamed “chaotic note taking” for plagiarized passages in her book, Seeds of Hope . Unfortunately, this stuff happens all the time.
Not much happened to Melania Trump after her “mistake.” The consequences for students who make these kinds of mistakes, however, can be severe.
How you can avoid this mistake
How do you gather your notes? Do you bookmark Web passages, use Evernote or Zotero, or create index cards or sticky notes? No matter your method, you should have a consistent and clear method to keep track of your sources. You wouldn’t walk around without clothes on; notes shouldn’t get around unless they’re clothed in source attributions. And you should make those attributions habitually and consistently, using the same format every time. Your mind can run free, your text can flow, but your attributions must be as fastidious as an accountant’s.
While this is a pain, it’s not nearly as painful as the repercussions of getting caught for plagiarizing. You will get caught. Do you really want to explain whether you were dishonest or just sloppy?
In this section, we’ll examine what exactly constitutes plagiarism, ways to avoid it, and best practices for demonstrating academic integrity.
Contributors and Attributions
- Outcome: Writing Ethically. Provided by : Lumen Learning. License : CC BY-NC-SA: Attribution-NonCommercial-ShareAlike
- Lessons from the plagiarism in Melania Trump's speech. Authored by : Josh Bernoff. Located at : http://withoutbullshit.com/blog/lessons-plagiarism-melania-trump-speech/ . License : CC BY-NC-SA: Attribution-NonCommercial-ShareAlike
- Melania Trump image. Located at : https://commons.wikimedia.org/wiki/File:Melania_Trump_2016.jpg . License : CC BY: Attribution


10.2: Huntington-Hill Method
The Huntington-Hill Method attempts to minimize the percent differences (unfairness) of how many people each representative will represent.
If states X and Y have already been allotted x and y representatives respectively, then state X should be given an additional representative over state Y if:
\[\begin{align}\frac{(\text{population of Y})^2}{y\times (y+1)}<\frac{(\text{population of X})^2}{x\times (x+1)}\end{align}\]
\[\begin{align}\frac{(\text{population of X})^2}{x\times (x+1)}\end{align}\] is called State X's Huntington-Hill Number .
- Each state gets one representative to begin with. Now calculate the Huntington-Hill number for each state with a Current Representation value of 1.
- Create a table of Huntington-Hill numbers for each state.
- Give the next representative to the state with the highest Huntington-Hill number.
- Recalculate the Huntington-Hill number for that state with the next Current Representation value.
- Repeat steps 3 and 4 until all the representatives have been given out.
- State the results in a complete sentence.
In a hypothetical world, State X has 750 people, State Y has 300 people, and State Z has 390 people. Use Huntington-Hill Method to apportion 9 representatives.
Solution
Step 1) State X, Y, and Z all get one representative. Note: The denominator will be 1(2).
State X has a population of 750 and 1 representative so the calculation for the Huntington-Hill number for State X is: \[\begin{align}\frac{750^2}{1\times 2}\end{align}\]
You can calculate the Huntington-Hill number for State Y and State Z by replacing the 750 with 300 or 390 respectively.
Step 2) Create a table where X, Y, and Z have one representative.
\(\begin{array}{lrrccc}
\text { Current Representation } & \text { State X } & \text{ State Y } & \text{ State Z } \\
\hline 1 & 281,250 & 45,000 & 76,050\\
\end{array}\)
Step 3) Give the 4th representative to X because it is the highest Huntington-Hill number. Recalculate State X's Huntington-Hill number for Current Representation of 2.
Step 4) Recalculate the Huntington-Hill number for X with 2 representatives. Note: The denominator will now be 2(3)
\(\begin{array}{lrrccc}
\text { Current Representation } & \text { State X } & \text{ State Y } & \text{ State Z } \\
\hline1 & 281,250 & 45,000 & 76,050\\2 & 93,750&&\\
\end{array}\)
Step 5) Repeat steps 3 and 4 until all 9 representatives have been assigned.
Give the 5th representative to X because it is the highest Huntington-Hill number. Recalculate State X's Huntington-Hill number for Current Representation of 3.
\(\begin{array}{lrrccc}
\text { Current Representation } & \text { State X } & \text{ State Y } & \text{ State Z } \\
\hline1 & 281,250 & 45,000 & 76,050\\2 & 93,750&&\\3&46,875&&\\
\end{array}\)
Give the 6th representative to Z and recalculate with Current Representation of 2.
\(\begin{array}{lrrccc}
\text { Current Representation } & \text { State X } & \text{ State Y } & \text{ State Z } \\
\hline1 & 281,250 & 45,000 & 76,050\\2 & 93,750&&25,350\\3&46,875&&\\
\end{array}\)
Give the 7th representative to X
\(\begin{array}{lrrccc}
\text { Current Representation } & \text { State X } & \text{ State Y } & \text{ State Z } \\
\hline1 & 281,250 & 45,000 & 76,050\\2 & 93,750&&25,350\\3&46,875&&\\4&28,125&&\\
\end{array}\)
Give the 8th representative to Y
\(\begin{array}{lrrccc}
\text { Current Representation } & \text { State X } & \text{ State Y } & \text{ State Z } \\
\hline1 & 281,250 & 45,000 & 76,050\\2 & 93,750&15,000&25,350\\3&46,875&&\\4&28,125&&\\
\end{array}\)
Give the 9th representative to X and there is no need to recalculate because that was the last representative.
Step 6) The representatives were assigned in order to State X, Y, Z, X, X, Z, X, Y, X. The final allocation is State X has 5 representatives, State Y has 2 representatives, and State Z has 2 representatives.
Three Universities will be sending representatives from the debate team to Washington D.C. for a special hearing. University P has 103 students on the debate team, University Q has 396 students, and University R has 247 students. Use Huntington-Hill Method to apportion 11 representatives.
Solution
The final table for Huntington-Hill numbers is below. Each University gets one representative to begin with. The number in the parentheses is the order to give out the representatives 4 through 11.
\(\begin{array}{lrrccc}
\text { Current Representation } & \text { University P } & \text{ University Q } & \text{ University R } \\
\hline1&5,304.5(10)&78,408.0(4)&30,504.5(5)\\2&1,768.2&26,136.0(6)&10,168.2(8)\\3&&13,068.0(7)&5,084.1\\4&&7,840.8(9)&\\5&&5,227.2(11)&\\
\end{array}\)
The delegates were assigned in order to University P, Q, R, Q, R, Q, Q, R, Q, P, Q. The final allocation is University P has 2 representatives, University Q has 6 representatives, and University R has 3 representatives.


8.3: Context Can Play a Role in Identifying Confirming and Disconfirming Responses
High- versus Low-Context Communication
Anthropologist Edward T. Hall applied the concept of context to different cultures and how people communicated. He introduced the idea of high-context cultures and low-context cultures (Hall, 1976).
So what is context and why is it important to interpersonal communication? The Department of State’s resource titled So You’re an American? A Guide to Answering Difficult Questions Abroad provides a good definition of context in communication between people from different cultures:
Context literally means “with text”—it’s all of the information surrounding what is being said, from the setting to the people involved and their standing within a given culture. The context of any interpersonal exchange can impact much of what is said and meant. Cultural norms regarding context vary greatly. In many ways, these cultural differences related to context influence how people communicate. (Foreign Service Institute, n.d.)
High-context communication typically occurs when, because of common history, a group of people have shared meanings, routines, and rituals. Because of this, communication tends to be indirect and implicit, and people rely on their intuition in interpreting what is being said and not said. Nonverbal behavior takes a central role in communication in high-context interactions.
Low-context communication occurs when messages must be direct and explicit because there is a lack of shared meaning. Ideas are expressed by individuals verbally with a good amount of explanation and reasoning. Things must be stated in an articulate, unambiguous manner without relying on intuition because there is little or no perceived common history.
In-Groups versus Out-Groups
This idea of high-context and low-context communication can be used to discuss the characteristics of groups. An in-group is a social group that people identify with and feel that they belong to. Members share the interests, narratives, and language (both verbal and nonverbal) of the in-group. The members of an in-group would typically communicate with each other toward the high context end of the high-context and low-context continuum.
An example of an in-group is Major League Baseball (MLB) fans. Members of the MLB fan in-group most likely know quite a bit about the history, statistics, language, and logos of the sport, especially for the team that they support and root for. One of our authors provides an example:
I remember an informative speech that a student gave on the history of the San Francisco Giants in my public speaking class. They delivered a well-organized speech in which they did a great job of outlining and explaining the history of the team. Since I am a big fan of major league baseball, I understood all of the lingo and examples that they used in their speech. I was reminded that I was a member of the MLB fan in-group when several of the students in the audience asked “What is an ‘RBI’?” and “Could you explain what a ‘tablesetter’ and a ‘southpaw’ are?” I had understood what the speaker was referring to, but the students in the class who did not identify as major league baseball fans had no clue about what was being talked about when these terms were used. Those students were part of the out-group and did not share the same knowledge that the speaker and I had about baseball. The speaker and I both understood the high-context aspects of the informative speech. The students who were not baseball fans needed a more low-context approach to understand the speaker’s message and feel confirmed as audience members interested in learning more about the San Francisco Giants.
The High Context–Low Context Communication Continuum
Context plays a major role in our interpretation of the communication climate of a situation and whether we feel confirmed or disconfirmed by our conversation partner. High-context communication and low-context communication can be depicted at the opposite ends of a continuum, as shown in Figure \(\PageIndex{1}\). An accessible text version of the infographic is linked in the figure caption.
How might a person from a high-context culture or in-group view the direct, explicit, detailed verbal communication style of someone using low-context communication? They might feel that the low-context person is talking down to them because they do not need all of that explanation or detailed instructions. They might think: “Why does that person think that I don’t get it? Do they think I can’t figure it out?  Do they not trust me?” These interpretations certainly lead to feelings of disconfirmation.
A person from a low-context culture could also experience feelings of disconfirmation when interacting with someone who is from a high-context culture or in-group. They might view the high context communicator as evasive, ambiguous, deceitful, and maybe even dishonest. They might think: “Why is this person so vague in their responses? Are they being lazy or sneaky? Maybe they just don’t get it. I just don’t know what they want or need. Why don’t they get to the point?”
High- and Low-Context Communication Interactions
As you can see, context can be a powerful factor affecting how people feel about an interpersonal interaction. In reality, most of us go back and forth between low-context and high-context communication all day long in our interpersonal interactions. Think about Figure \(\PageIndex{2}\), which shows a few examples of high-context and low-context culture (an accessible text version is linked in the figure caption). Then think about different examples of communication interactions that you have had today and consider where you might plot them along the high context–low context continuum.
Here is an example for you to think about as you reflect on your communication interactions. Let’s say your professor tells you to: “Write an essay about communication climate,” and that’s all they say. For many students, that may be considered very vague, ambiguous communication because no details or direct instructions were given for guidance. The directive could be interpreted as a disconfirming response by those students because they do not know what the expectations are nor how they will be graded. On the other hand, let’s say that some of the students in the class have taken multiple classes from that professor and know what the expectations are because of their previous experiences. They know the number of pages required, how to organize the five-paragraph essay, how much description and analysis to go into, how many specific sources to cite in APA 7 Style, etc. They already have a good idea of what the professor is looking for in a paper on communication climate without the professor having to go into details about their one-sentence direction: “Write an essay about communication climate.” For these students, they understand the implicit high-context directive. It would not be seen as a disconfirming response to them.
As stated earlier, high-context communication involves shared communication norms within an in-group because of familiarity and a common history. Think about some in-groups you belong to: your work group, different circles of friends, and your family. Because of shared experiences, you may even develop your own jargon or slang. This can certainly happen within a family. Families often have their own words and phrases for things that would confuse someone outside of that household. One of our professors provides this example:
Our minister’s wife came over to babysit my very young daughter one evening. When I returned home, the minister’s wife was very concerned about something that my daughter kept repeating over and over during the evening: “Fire. Fire. Fire.” I was confused that the minister’s wife did not understand what she was saying. Why? Because I knew that my daughter had not been saying "fire." She had been saying “fier” as in “paci fier ,” her word for her pacifier. My daughter’s communication was clear to me because we all understood what she meant within our family culture, but to someone outside of the family, her communication was unclear. Being a Communication professor, I reflected on this incident because it was a great example of how high-context communication within a family could lead to a misunderstanding by someone outside of the family. After that, I gave my babysitters a list of words and phrases that my daughter might say that were particular to our family communication patterns. I used a low-context way of informing our babysitters about our high-context family communication style. That low-context glossary of my family’s vocabulary seemed to do the trick!
Let’s explore high- and low-context communication within an in-group and between an in-group and an out-group.


4.2: Syngamy and Meiosis
Sexual Process and the Syngamy
A sexual process is important to the survival of a species. First, it makes the population more diverse, which allows more flexibility to adapt via natural selection. Natural selection means that all organisms are different, but if environmental conditions change, only most adapted will survive. If the population is uniform, it has less chance of survival. Second, it prevents lethal mutations from being transferred to the offspring, because those with the mutations will die instead of passing on these genes. The later happens when the mutated gene is either duplicated or alone in genotype.
( Genotype is a gene content of the organism. A gene is a piece of DNA, which is equal to one protein. A mutation is a “mistake” in DNA. A protein (many of them are enzymes) is made up of amino acids chained together. A population is a group of organisms which potentially may interbreed and have no isolation barriers.)
Figure \(\PageIndex{1}\) Cell cycle: I interphase, D cell division, G\(_1\) pre-synthetic stage, S synthetic stage, G\(_2\) post-synthetic stage, M mitosis (karyokinesis), C cyto
To make populations more diverse, organisms need to exchange DNA. One way that cells exchange genes is through syngamy. Syngamy (frequently labeled by “Y!”), is the fusion of two cells, resulting in a cell that has twice as many chromosomes. The two cells which are fused together are called gametes , and the resulting cell is a zygote . The goal of syngamy is the renovation of genetic material. The new cells have genotype different from the gametes. Continuous syngamy will increase the amount of DNA, so cells use meiosis (frequently labeled by “R!”) to counterbalance this side-effect of syngamy: \[\mbox{Y!} \rightarrow \mbox{R!}\]
Syngamy results in diploid cell:
\[X + X \rightarrow XX\]
In diploid organisms, chromosomes form pairs (these paired chromosomes are known as homologous ), whereas in halploid organisms they remain single.
There are three types of syngamy (Figure \(\PageIndex{2}\)): isogamy, heterogamy, and oogamy.
- Isogamy happens when the gametes that fuse together are similar. To avoid self-fertilization, they must have an advanced system of recognition. Different genotypes (mating types) recognize each other with the help of surface proteins, like cells of immune system.
- Heterogamy is when the gametes are of two different sizes. This difference makes rec848ognition easier, but even more important is division of labor : the female is larger because it has resources to care for the offspring, whereas the males are smaller and can increase in number to allow competition and make fertilization more likely.
- Oogamy is when the gametes also have different mobility. In oogamy, the non-motile female is known as the oocyte , and the flagellate male as the spermatozoon , which is only one mobile gamete here. In some organisms (red algae, sponges, crustaceans, most seed plants), spermatozoon become non-motile spermatium so it will need external agents to move it. Both spermatozoa and spermatia are called sperms .
Figure \(\PageIndex{2}\) Three types of syngamy.
Meiosis
Syngamy is the way for organisms to become more genetically diverse, but since it increases the amount of chromosomes, it needs to be balanced by meiosis. Meiosis reduces the number of chromosomes, recombines the chromosomes, and allows chromosomes to exchange their genetic material. Meiosis is a reductive form of cell division, where each new cell receives half of the original cell’s chromosomes. Unlike mitosis, meiosis does change the genotype of cells because whole chromosomes are recombined and also exchanged their genetic material. Another difference is that in mitosis, ploidy (“twoness” of chromosomes) stays constant, while in meiosis, ploidy halves.
There are two problems of meiosis: first, how to find out which chromosomes are homologous; and second, how to split chromosomes which were already duplicated in S-phase. First problem is solving with “gluing” homologous chromosomes together; this happens because similar chains of DNA can attach each other. Second problem is usually solving with the second stage of meiosis which is quite similar to ordinary mitosis.
There are two stages of meiosis: a reductive division (meiosis I, unique) and an equal division (meiosis II, similar to mitosis). Each of these stages are divided into prophase, metaphase, anaphase, and telophase.
In prophase I, chromosomes conjugate (form synapses ), and start to exchange DNA (crossing-over).
In anaphase I, chromosomes from each pair will go independently to different poles . Independence means that if we label “mother” and “father” chromosomes with, saying \(a\) and \(b\), then two variants are possible:
\[X{_a}X{_b} + Y{_a}Y{_b} \rightarrow (X{_a} + Y{_a}) + ({X_b} + Y{_b})\]
or
\[X{_a}X{_b} + Y{_a}Y{_b} \rightarrow (X{_a} + Y{_b}) + ({X_a} + Y{_b})\]
because chromosomes do not know which is “father’s” and which is “mother’s”.
Telophase I usually flows into prophase II. This second division of meiosis is very similar to mitosis without synthetic stage before it. Frequently, nuclei do not form until telophase II (Figure \(\PageIndex{3}\)).
In the first division, cell needs to split pairs of homologs to reduce ploidy. The second division of meiosis is necessary because DNA was already duplicated in the synthetic stage of the cell cycle. Consequently, every \(X\)-like chromosome needs to be split into two \(I\)-like chromosomes: \[XX \rightarrow X + X \rightarrow I + I + I + I\]
This is why there are two divisions and four cells in the end (sometimes, however, only one of these four survives). If DNA would not be duplicated before, it is also possible for meiosis to happen in one stage instead of two. This kind of meiosis is described in some protists. Inverted meiosis, when reductive division is the second and equal the first, is rare but also exists in nature (e.g., in some rushes, bugs and butterflies).
Figure \(\PageIndex{3}\) Principal scheme of meiosis. Only one of two telophase I variants is shown.
It is possible that meiosis won’t work properly, which results in a cell receiving a double set of chromosomes. If, in turn, that cell goes to syngamy, the resulting zygote will have 3 sets of chromosomes. Cells with more than two sets of chromosomes are called polyploids . Rarely, only some chromosome pairs do not want to split. In this case, after the syngamy, some chromosomes will be “triplicated” ( trisomy ). This is aneuploidy . One example of frequent (1/ 800 births) aneuploidy in humans is Down syndrome.


3.3: Military Service- District of Columbia v. Heller 554 U.S. 570 (2008) Annotate this Case
OCTOBER TERM, 2007
DISTRICT OF COLUMBIA V. HELLER
SUPREME COURT OF THE UNITED STATES
DISTRICT OF COLUMBIA et al. v . HELLER
certiorari to the united states court of appeals for the district of columbia circuit
No. 07–290. Argued March 18, 2008—Decided June 26, 2008
District of Columbia law bans handgun possession by making it a crime to carry an unregistered firearm and prohibiting the registration of handguns; provides separately that no person may carry an unlicensed handgun, but authorizes the police chief to issue 1-year licenses; and requires residents to keep lawfully owned firearms unloaded and dissembled or bound by a trigger lock or similar device. Respondent Heller, a D. C. special policeman, applied to register a handgun he wished to keep at home, but the District refused. He filed this suit seeking, on Second Amendment grounds, to enjoin the city from enforcing the bar on handgun registration, the licensing requirement insofar as it prohibits carrying an unlicensed firearm in the home, and the trigger-lock requirement insofar as it prohibits the use of functional firearms in the home. The District Court dismissed the suit, but the D. C. Circuit reversed, holding that the Second Amendment protects an individual’s right to possess firearms and that the city’s total ban on handguns, as well as its requirement that firearms in the home be kept nonfunctional even when necessary for self-defense, violated that right.
Held:
1. The Second Amendment protects an individual right to possess a firearm unconnected with service in a militia, and to use that arm for traditionally lawful purposes, such as self-defense within the home. Pp. 2–53.
(a) The Amendment’s prefatory clause announces a purpose, but does not limit or expand the scope of the second part, the operative clause. The operative clause’s text and history demonstrate that it connotes an individual right to keep and bear arms. Pp. 2–22.
(b) The prefatory clause comports with the Court’s interpretation of the operative clause. The “militia” comprised all males physically capable of acting in concert for the common defense. The Antifederalists feared that the Federal Government would disarm the people in order to disable this citizens’ militia, enabling a politicized standing army or a select militia to rule. The response was to deny Congress power to abridge the ancient right of individuals to keep and bear arms, so that the ideal of a citizens’ militia would be preserved. Pp. 22–28.
(c) The Court’s interpretation is confirmed by analogous arms-bearing rights in state constitutions that preceded and immediately followed the Second Amendment. Pp. 28–30.
(d) The Second Amendment’s drafting history, while of dubious interpretive worth, reveals three state Second Amendment proposals that unequivocally referred to an individual right to bear arms. Pp. 30–32.
(e) Interpretation of the Second Amendment by scholars, courts and legislators, from immediately after its ratification through the late 19th century also supports the Court’s conclusion. Pp. 32–47.
(f) None of the Court’s precedents forecloses the Court’s interpretation. Neither United States v. Cruikshank , 92 U. S. 542 , 553, nor Presser v. Illinois , 116 U. S. 252 , 264–265, refutes the individual-rights interpretation. United States v. Miller , 307 U. S. 174 , does not limit the right to keep and bear arms to militia purposes, but rather limits the type of weapon to which the right applies to those used by the militia, i.e. , those in common use for lawful purposes. Pp. 47–54.
2. Like most rights, the Second Amendment right is not unlimited. It is not a right to keep and carry any weapon whatsoever in any manner whatsoever and for whatever purpose: For example, concealed weapons prohibitions have been upheld under the Amendment or state analogues. The Court’s opinion should not be taken to cast doubt on longstanding prohibitions on the possession of firearms by felons and the mentally ill, or laws forbidding the carrying of firearms in sensitive places such as schools and government buildings, or laws imposing conditions and qualifications on the commercial sale of arms. Miller’ s holding that the sorts of weapons protected are those “in common use at the time” finds support in the historical tradition of prohibiting the carrying of dangerous and unusual weapons. Pp. 54–56.
3. The handgun ban and the trigger-lock requirement (as applied to self-defense) violate the Second Amendment. The District’s total ban on handgun possession in the home amounts to a prohibition on an entire class of “arms” that Americans overwhelmingly choose for the lawful purpose of self-defense. Under any of the standards of scrutiny the Court has applied to enumerated constitutional rights, this prohibition—in the place where the importance of the lawful defense of self, family, and property is most acute—would fail constitutional muster. Similarly, the requirement that any lawful firearm in the home be disassembled or bound by a trigger lock makes it impossible for citizens to use arms for the core lawful purpose of self-defense and is hence unconstitutional. Because Heller conceded at oral argument that the D. C. licensing law is permissible if it is not enforced arbitrarily and capriciously, the Court assumes that a license will satisfy his prayer for relief and does not address the licensing requirement. Assuming he is not disqualified from exercising Second Amendment rights, the District must permit Heller to register his handgun and must issue him a license to carry it in the home. Pp. 56–64.
478 F. 3d 370, affirmed.
Scalia, J., delivered the opinion of the Court, in which Roberts, C. J., and Kennedy, Thomas, and Alito, JJ., joined. Stevens, J., filed a dissenting opinion, in which Souter, Ginsburg, and Breyer, JJ., joined. Breyer, J., filed a dissenting opinion, in which Stevens, Souter, and Ginsburg, JJ., joined.
Contributors and Attributions
- Authored by : US Government . Located at : https://supreme.justia.com/cases/federal/us/554/570/ . License : Public Domain: No Known Copyright


2.3: Reading- The Building Blocks of Matter
Why study the chemistry of minerals? Minerals are made of atoms , which have an impact on the behavior and characteristics of the mineral. Thus, to understand, explain, and predict the behavior of minerals, and rocks—which are made of minerals—we must understand some basic facts about atoms and how they behave. This requires a basic understanding of some chemistry. We will begin by constructing atoms in our thinking in terms of the three sub-atomic particles of which atoms are made.
Atoms
Atoms consist of protons, neutrons, and electrons. Protons have a positive (+) electrical charge. Electrons have a negative (−) charge that is exactly equal and opposite to the electrical charge of a proton. Neutrons are electrically neutral.
Most of the mass of an atom is packed into its tiny nucleus. An atomic nucleus is made of protons and neutrons, which have approximately the same mass (about 1.67 × 10 −24 grams). Electrons, on the other hand, are arranged in specific orbitals around the nucleus of an atom; they are also much smaller in mass than protons and neutrons, weighing only 9.11 × 10 −28 grams, or about 1/1800 the weight of protons and neutrons.
Even though the mass of an electron is a tiny mass compared to the mass of a proton or a neutron, the electrons occupy most of the volume of an atom (see Figure 1).
Ions
A neutral atom has the same number of electrons as it does protons. An atom that has lost or gained any electrons is no longer an electrically neutral atom. That type of atom, which is not electrically neutral and has an electrical charge associated with it, is called an ion . Atoms that have gained electrons are negatively (−) charged ions, or anions . Atoms that have lost electrons are positively (+) charged ions, or cations .
It is also possible to have ions that are actually small groups of atoms bonded together. These are known as polyatomic ions. One example of a polyatomic ion is the carbonate ion, (CO 3 ) 2− , which has two extra electrons, giving it the net electrical charge of 2−.
We’ve just seen that a carbonate ion can also be called (CO 3 ) 2− . But what exactly does this mean?
First let’s look at the letters: CO. Atoms have chemical symbols; each element has been assigned one or two letters to represent it. Thus, C stands in for carbon, and O stands in for oxygen (all of these chemical symbols can be seen in the periodic table in figure 2 below).
As we read above, the 2− means that the ion has two extra electrons. But what about the 3 ? This means there are three oxygen atoms in the ion. The number of atoms in a particular formula is always notated in subscript. Charge is always written in superscript at the end of the formula (a superscript at the beginning of the formula means something else—we’ll get to this when we discuss isotopes below). The parentheses around CO 3 indicate that the charge belongs to the whole polyatomic unit rather than just the O 3 .
Thus the carbonate ion is one carbon atom (C), three oxygen atoms (O 3 ), and two extra electrons ( 2− ), which charge the whole polyatomic ion.
The Periodic Table
Naturally occurring atoms found in the earth range from hydrogen, with just one proton in its nucleus, to uranium, with 92 protons in its nucleus. These are the naturally occurring chemical elements, which includes such commonly known elements as carbon, oxygen, iron, and so on. The periodic table lists all the chemical elements in a way that tells us how many protons each of them has, how its electrons are arranged, and what the general chemical behavior of each element is.
As shown in Figure 2, the periodic table consists of eighteen groups and seven periods. Two additional rows of elements, known as the lanthanides and actinides, are placed beneath the main table. These elements are placed separately to make the table more compact. All the elements in a group have a similar chemical behavior. This is because all the elements in a group have a similar arrangement of electrons in their atoms, and it is the electron arrangement that determines the chemical behavior of an element.
For each element, the name, atomic symbol, atomic number, and atomic mass are provided. The atomic number is a whole number that represents the number of protons: each chemical element is distinguished by the number of protons in its nucleus. For example, every atom of the element oxygen has eight protons in its nucleus. That is why the atomic number of oxygen is 8. If an atom has greater or fewer than eight protons in its nucleus, it is not oxygen, it is some other chemical element. In the periodic table, the atomic number of each element is listed above the chemical symbol of the element.
The atomic mass, which is the average mass of different isotopes, is estimated to two decimal places. For example, hydrogen has the atomic symbol H, the atomic number 1, and an atomic mass of 1.01. The atomic mass is always larger that the atomic number. For most small elements, the atomic mass is approximately double the atomic number, as the number of protons and neutrons is about equal.
The elements are divided into three categories: metals, nonmetals and metalloids. These form a diagonal line from period two, group thirteen to period seven, group sixteen. All elements to the left of the metalloids are metals, and all elements to the right are nonmetals.
The periodic table was created to help chemists better understand elements and how they function. It is a map to elemental behavior.
Isotopes
Every atom of a specific element must have the same number of protons in its nucleus. This number is its atomic number. However, there is a range of possible numbers of neutrons it can have in its nucleus. The fact that atoms of a chemical element may have different numbers of neutrons results in each chemical element having several isotopes. Isotopes are atoms of a given chemical element that have different numbers of neutrons in their nuclei.
For example, while all atoms of the element oxygen have eight protons in their nuclei, those oxygen atoms may have eight, nine, or ten neutrons. The different numbers of neutrons in the nucleus distinguishes the three isotopes of oxygen. Oxygen-16 is the isotope of oxygen with 8 neutrons in its nucleus. The number 16 is called the atomic mass number. The atomic mass number is the total number of protons and neutrons in the nucleus of an isotope. From this definition, and knowing that all oxygen atoms have 8 protons in the nucleus, you can deduce that oxygen-17 is the oxygen isotope with 9 neutrons and oxygen-18 is the oxygen isotope with 10 neutrons. Abbreviated into symbols, the three isotopes of oxygen are written as 16 O, 17 O and 18 O.
Isotopes are not very important for understanding minerals, but are important in understanding how to apply chemistry and nuclear physics to geology, such as how to use measurements of radioactive isotopes to measure the ages of rocks and minerals and how to use oxygen isotopes from layers of glacial ice to determine what the temperature of the earth was during an ice age.
Contributors and Attributions
- Revision and Adaptation. Provided by : Lumen Learning. License : CC BY: Attribution
- Basics - Minerals. Authored by : Ralph L. Dawes and Cheryl D. Dawes. Provided by : Wenatchee Valley College. Located at : http://commons.wvc.edu/rdawes/G101OCL/Basics/minerals.html . Project : Geology 101 - Introduction to Physical Geology. License : CC BY: Attribution
- Periodic Table of Elements. Provided by : OpenStax. Located at : http://cnx.org/contents/185cbf87-c72e-48f5-b51e-f14f21b5eabd@9.74:6/Biology . License : CC BY: Attribution
- The Periodic Table: Crash Course Chemistry #4. Authored by : CrashCourse. Located at : https://youtu.be/0RRVV4Diomg . License : All Rights Reserved . License Terms : Standard YouTube License


<IP_ADDRESS>: Functions of Sugar
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As mentioned, sugar is the third most used ingredient in the bakeshop. Sugar has several functions in baking. The most recognized purpose is, of course, to sweeten food, but there are many other reasons sugar is used in cooking and baking:
- Maillard reaction: It can be used for browning effect on everything from breads to cookies to cakes. Browning gives a pleasant color and flavor to the finished product. Caramelization results from the action of heat on sugars. At high temperatures, the chemical changes associated with melting sugars result in a deep brown Color and new flavors. The Maillard reaction results from chemical interactions between sugars and proteins at high heat. An amino group from a protein combines with a reducing sugar to produce a brown colors in a variety of foods (e.g., brewed coffee, fried foods, and breads).
- Tenderizing: It acts as the most important tenderizing agent in all baked goods, and one of the factors responsible for the spread in cookies. It helps delay the formation of gluten, which is essential for maintaining a soft or tender product.
- Texture: It makes an important contribution to the way we perceive the texture of food. For example, adding sugar to ice cream provides body and texture, which is perceived as smoothness. This addition helps prevent lactose crystallization and thus reduces sugar crystal formation that otherwise causes a grainy texture sometimes associated with frozen dairy products.
- Preservative: It preserves food when used in sufficient quantity. Use the link bellow to learn more.
- Leavening: In baking, it increases the effectiveness of yeast by providing an immediate and more usable source of nourishment for the yeast’s growth. This hastens the leavening process by producing more carbon dioxide, which allows the dough to rise at a quicker and more consistent rate. Use the link bellow to learn more.
- crystallization: Use the link bellow to learn about crystalization:
- Solubility: Use the link bellow to learn about solubility:
- Hygroscopicity: Use the link bellow to learn about hygroscopicity:
- Fermentation: Use the link bellow to learn about fermentation:
Just as there are many functions of sugar in the bakeshop, there are different uses for the various types of sugar as well:
- Fine granulated sugar is most used by bakers. It generally dissolves easily in mixes and is pure enough for sugar crafters to boil for “pulled” sugar decorations.
- Coarse granulated sugar may be used for a topping on sugar cookies, puff pastry, and Danish pastries as it doesn’t liquefy or caramelize so readily. In some European countries, an extra coarse sugar (called hail — a literal translation) is used for this purpose.
- Icing or powdered sugar is used in icings and fillings and in sifted form as a top decoration on many baked goods.
- Brown or yellow sugars are used where their unique flavor is important, or in bakeries where an old-fashioned or rustic image is projected. Brown sugar can usually be substituted for white sugar without technical problems in sugar/batter mixes such as cakes and muffins, and in bread dough.
The link bellow provides a full overview of all functions of sugar:
***Calories: There is 4 kcal/gram of sugar/carbohydrate.