4.1: Plate Tectonics and Volcanism
The relationships between plate tectonics and volcanism are shown on Figure \(\PageIndex{1}\). As summarized in Chapter 3, magma is formed at three main plate-tectonic settings: divergent boundaries (decompression melting), convergent boundaries (flux melting), and mantle plumes (decompression melting).
The mantle and crustal processes that take place in areas of volcanism are illustrated in Figure \(\PageIndex{2}\). At a spreading ridge, hot mantle rock moves slowly upward by convection (centimetre/year), and within about 60 kilometers (km) of the surface, partial melting starts because of decompression. Over the triangular area shown in Figure \(\PageIndex{2}\)a, about 10% of the ultramafic mantle rock melts, producing mafic magma that moves upward toward the axis of spreading (where the two plates are moving away from each other). The magma fills vertical fractures produced by the spreading and spills out onto the sea floor to form basaltic pillows (more on that later) and lava flows. There is spreading-ridge volcanism taking place about 200 km offshore from the west coast of Vancouver Island.
Figure \(\PageIndex{2}\)a shows a triangular zone about 60 km thick; within this zone, approximately 10% of the mantle rock melts to form oceanic crust. Based on this information, approximately how thick do you think the resulting oceanic crust should be?
See Appendix 3 for Exercise 4.1 answers .
 At an ocean-ocean convergent boundary, oceanic crust is being subducted beneath another oceanic-crust plate.[/footnote] (Figure \(\PageIndex{2}\)b). In both situations the oceanic crust is heated up, and while there isn’t enough heat to melt the subducting crust, there is enough heat to force the water out of some of its minerals. This released water rises into the overlying mantle where it contributes to flux melting of the mantle rock. The mafic magma produced rises through the mantle to the base of the crust. There it contributes to partial melting of crustal rock, and thus it assimilates much more felsic material. That magma, now likely intermediate in composition, continues to rise and assimilate crustal material. In the upper part of the crust, it accumulates into plutons. From time to time, the magma from the plutons rises toward surface, leading to volcanic eruptions. Mount Garibaldi (Figures 4.0.1 and 4.0.2) is an example of subduction-related volcanism.
A mantle plume is an ascending column of hot rock (not magma) that originates deep in the mantle, possibly just above the core-mantle boundary. Mantle plumes are thought to rise approximately 10 times faster than the rate of mantle convection. The ascending column may be on the order of kilometers to tens of kilometers across, but near the surface it spreads out to create a mushroom-style head that is several tens to over 100 km across. Near the base of the lithosphere (the rigid part of the mantle), the mantle plume (and possibly some of the surrounding mantle material) partially melts to form mafic magma that rises to feed volcanoes. Since most mantle plumes are beneath the oceans, the early stages of volcanism typically take place on the sea floor. Over time, islands may form like those in Hawaii.
Volcanism in northwestern B.C. (Figures 4.1.3 and 4.1.4) is related to continental rifting. This area is not at a divergent or convergent boundary, and there is no evidence of an underlying mantle plume. A likely explanation is that the crust of northwestern B.C. is being stressed by the northward movement of the Pacific Plate against the North America Plate, and the resulting crustal fracturing provides a conduit for the flow of magma from the mantle. This may, or may not, be an early stage of continental rifting, such as that found in eastern Africa.
Media Attributions
- Figure \(\PageIndex{1}\): Understanding Plate Motions by USGS. Public domain. Modified by Steven Earle.
- Figure \(\PageIndex{2}\): By USGS. Public domain. Modified by Steven Earle.
- Figure \(\PageIndex{3}\): “ South-West Canada ” by USGS. Public domain. Modified by Steven Earle. Volcanic locations from Edwards, B. & Russell, J. (2000). Distribution, nature, and origin of Neogene-Quaternary magmatism in the northern Cordilleran volcanic province, Canada. Geological Society of America Bulletin. pp. 1280-1293 [Steven Earle] Cordillera Volcanic Province, B.C.
- Figure \(\PageIndex{4}\): © Steven Earle. CC BY.


8.1.3: What are Probabilities?
Lesson
Let's find out what's possible.
Exercise \(\PageIndex{1}\): Which Game Would You CHoose?
Which game would you choose to play? Explain your reasoning.
Game 1: You flip a coin and win if it lands showing heads.
Game 2: You roll a standard number cube and win if it lands showing a number that is divisible by 3.
Exercise \(\PageIndex{2}\): What's Possible?
-
For each situation, list the
sample space
and tell how many outcomes there are.
- Han rolls a standard number cube once.
- Clare spins this spinner once.
- Kiran selects a letter at random from the word “MATH.”
- Mai selects a letter at random from the alphabet.
- Noah picks a card at random from a stack that has cards numbered 5 through 20.
-
Next, compare the likelihood of these outcomes. Be prepared to explain your reasoning.
- Is Clare more likely to have the spinner stop on the red or blue section?
- Is Kiran or Mai more likely to get the letter T?
- Is Han or Noah more likely to get a number that is greater than 5?
- Suppose you have a spinner that is evenly divided showing all the days of the week. You also have a bag of papers that list the months of the year. Are you more likely to spin the current day of the week or pull out the paper with the current month?
Are you ready for more?
Are there any outcomes for two people in this activity that have the same likelihood? Explain or show your reasoning.
Exercise \(\PageIndex{3}\): What's in the Bag?
Your teacher will give your group a bag of paper slips with something printed on them. Repeat these steps until everyone in your group has had a turn.
- As a group, guess what is printed on the papers in the bag and record your guess in the table.
- Without looking in the bag, one person takes out one of the papers and shows it to the group.
- Everyone in the group records what is printed on the paper.
- The person who took out the paper puts it back into the bag, shakes the bag to mix up the papers, and passes the bag to the next person in the group.
| Guess the sample space. | What is printed on the paper? | |
|---|---|---|
| person 1 | ||
| person 2 | ||
| person 3 | ||
| person 4 |
- How was guessing the sample space the fourth time different from the first?
- What could you do to get a better guess of the sample space?
- Look at all the papers in the bag. Were any of your guesses correct?
- Are all of the possible outcomes equally likely? Explain.
- Use the sample space to determine the probability that a fifth person would get the same outcome as person 1.
Summary
The probability of an event is a measure of the likelihood that the event will occur. Probabilities are expressed using numbers from 0 to 1.
- If the probability is 0, that means the event is impossible. For example, when you flip a coin, the probability that it will turn into a bottle of ketchup is 0. The closer the probability of some event is to 0, the less likely it is.
- If the probability is 1, that means the event is certain. For example, when you flip a coin, the probability that it will land somewhere is 1. The closer the probability of some event is to 1, the more likely it is.
If we list all of the possible outcomes for a chance experiment, we get the sample space for that experiment. For example, the sample space for rolling a standard number cube includes six outcomes: 1, 2, 3, 4, 5, and 6. The probability that the number cube will land showing the number 4 is \(\frac{1}{6}\). In general, if all outcomes in an experiment are equally likely and there are \(n\) possible outcomes, then the probability of a single outcome is \(\frac{1}{n}\).
Sometimes we have a set of possible outcomes and we want one of them to be selected at random . That means that we want to select an outcome in a way that each of the outcomes is equally likely . For example, if two people both want to read the same book, we could flip a coin to see who gets to read the book first.
Glossary Entries
Definition: Chance Experiment
A chance experiment is something you can do over and over again, and you don’t know what will happen each time.
For example, each time you spin the spinner, it could land on red, yellow, blue, or green.
Definition: Event
An event is a set of one or more outcomes in a chance experiment. For example, if we roll a number cube, there are six possible outcomes.
Examples of events are “rolling a number less than 3,” “rolling an even number,” or “rolling a 5.”
Definition: Outcome
An outcome of a chance experiment is one of the things that can happen when you do the experiment. For example, the possible outcomes of tossing a coin are heads and tails.
Definition: Probability
The probability of an event is a number that tells how likely it is to happen. A probability of 1 means the event will always happen. A probability of 0 means the event will never happen.
For example, the probability of selecting a moon block at random from this bag is \(\frac{4}{5}\).
Definition: Random
Outcomes of a chance experiment are random if they are all equally likely to happen.
Definition: Sample Space
The sample space is the list of every possible outcome for a chance experiment.
For example, the sample space for tossing two coins is:
| heads-heads | tails-heads |
| heads-tails | tails-tails |
Practice
Exercise \(\PageIndex{4}\)
List the sample space for each chance experiment.
- Flipping a coin
- Selecting a random season of the year
- Selecting a random day of the week
Exercise \(\PageIndex{5}\)
A computer randomly selects a letter from the alphabet.
- How many different outcomes are in the sample space?
- What is the probability the computer produces the first letter of your first name?
Exercise \(\PageIndex{6}\)
What is the probability of selecting a random month of the year and getting a month that starts with the letter “J?” If you get stuck, consider listing the sample space.
Exercise \(\PageIndex{7}\)
\(E\) represents an object’s weight on Earth and \(M\) represents that same object’s weight on the Moon. The equation \(M=\frac{1}{6}E\) represents the relationship between these quantities.
- What does the \(\frac{1}{6}\) represent in this situation?
- Give an example of what a person might weigh on Earth and on the Moon.
(From Unit 2.2.1)
Exercise \(\PageIndex{8}\)
Here is a diagram of the base of a bird feeder which is in the shape of a pentagonal prism. Each small square on the grid is 1 square inch.
The distance between the two bases is 8 inches. What will be the volume of the completed bird feeder?
(From Unit 7.3.3)
Exercise \(\PageIndex{9}\)
Find the surface area of the triangular prism.
(From Unit 7.3.4)


5.1: Module Introduction
He burned his house down for the fire insurance
And spent the proceeds on a telescope
To satisfy a lifelong curiosity
About our place among the infinities
The strongest thing that’s given us to see with’s
A telescope. Someone in every town
Seems to me owes it to the town to keep one.
In Littleton it may as well be me.
Robert Frost
The Star-Splitter
This module introduces you to the primary instrument used by astronomers: the telescope, as well as how the telescope works. From its historical first use to today’s modern instruments, the telescope changed forever how we look at our Universe.
Objectives
Upon completion of this module, the student will be able to:
- Identify the types of telescopes and telescope mounts
- Describe the differences between the refracting telescope and reflecting telescope, including both positive and negative features
- Describe the types of instruments astronomers use with the telescope and the function of each type
- Explain why space-based telescopes are of such importance to today’s astronomers and the collection of information


3.1: Putting It Together- Planning and Mission
Let’s return our attention to Sun City Boards and its owner, Tom Wilson. Dani travelled to Tom’s operations on the West Coast and confirmed her suspicions that the problems the organization is facing are due to a lack of planning. Dani is meeting with Tom to report on her findings and to submit her recommendations.
“OK Tom, I’ve spent the last several days going through your operations and records, and I’m confident that there are actions you can take that will help turn around the business.”
Tom lets out a long sigh and says, “Dani, you have no idea how glad I am to hear that. So at a high level, what’s the problem?”
“In a nutshell, the problem with Sun City Boards is that you have lost your focus and there is no clear plan for moving forward.” Tom looks confused, so Dani explains. “When you started the business, what was your vision?”
Tom answers, “To set people free from the earth … free to ride the waves.”
“Perfect!” Dani responds. “And how were you going to do that?”
Tom thinks for a moment and then emphatically declares, “We will strive to provide the absolute best in high-end surfboard design and production!”
Dani tells Tom that he has just created the vision and mission statements for Sun City Boards. She further explains, “You have to start creating a focused plan that will help the organization deliver on its vision and mission. The reason your company is struggling is that there is no clear roadmap to success.”
Tom asks, “What do you mean?”
Dani continues, “When you started the company you were very successful because you focused on one key thing—making the best surfboards anyone had seen. As a result, everything your business did revolved around that core idea.”
Tom interrupts, “OK Dani, I think I see where you’re headed. When we started making money in the early days, I began to start expanding into other areas without a lot of thought. I listened to salespeople who came in and told me that I should branch out into low-end surf gear, then swimwear, then camping gear. Customers don’t know what we’re all about any more. And neither do my managers and employees.”
Dani can see Tom is getting his arms around the concept of vision and mission, so she moves on. Next, she presents her recommendation that Tom and his key leaders should come together regularly to create, review and update Sun City’s business plan. This is new to the team, so she gives them a quick example:
- Define Objectives : Sun City Boards should look to reduce noncore product lines and associated inventory. Minimum margin target is 35 percent across all remaining product lines.
- Develop Premises : Monitor competitor’s plans to implement a new polymer board and determine potential impact to Sun City’s sales.
- Evaluate Alternatives : Evaluate opportunity of opening new retail outlets along the West Coast. Also, explore international sales options for high-end surf boards.
- Identify Resources : Analyze the level of capital necessary to achieve expansion goals and incorporate plan to close financial gaps in the annual budget.
- Plan and Implement Tasks : Create a Gantt chart or other project planning tool that outlines the actions necessary to reach the goals outlined earlier.
- Determine Tracking and Evaluation Methods : Create SMART goals for both managers and employees that align with direction the overall business plan provides.
Finally, Dani reminds Tom that the planning process is only a tool and that the plan should be regularly reviewed and updated. Dani warns him that planning is no substitute for taking action, but it is a guide for him and his team as to what actions should be taken.
This consulting project has reminded Dani how important it is for an organization to have a clear vision and mission to know where it’s headed. Likewise, the planning process is critical to know how it’s going to get there. The alternative is like trying to take a cross-country trip without a map—you don’t know where you might end up!
Contributors and Attributions
- Putting It Together: Planning and Mission. Authored by : Jeff Heflin and Lumen Learning. License : CC BY: Attribution
- moldova-aerial-view-sunrise-morning. Authored by : tpsdave. Located at : https://pixabay.com/en/moldova-aerial-view-sunrise-morning-2553433/ . License : CC0: No Rights Reserved


5.4: Exercises
EXERCISE 5.1. Prove that the definitions of limit on pages 129 and 130 are the same.
EXERCISE 5.2. Prove Lemma 5.14, and the related assertion that \(|c|-|d| \leq|c+d|\) .
EXERCISE 5.3. For \(n \in \mathbb{N}^{+}, a_{i} \in \mathbb{R}(1 \leq i \leq n)\) , prove that \[\left|\sum_{i=1}^{n} a_{i}\right| \leq \sum_{i=1}^{n}\left|a_{i}\right| .\] ExERCISE 5.4. Prove Lemma \(5.15\) .
EXERCISE 5.5. Prove part (v) of Theorem 5.9.
EXERCISE 5.6. Give an example of two functions \(f\) and \(g\) that don’t have limits at a point \(a\) but such that \(f+g\) does. For the same pair of functions, can \(f-g\) also have a limit at \(a\) ? EXERCISE 5.7. Assume that \(f\) is a real function and \(\lim _{x \rightarrow a} f(x)=\) \(L\) . Prove that if \(X \subseteq \operatorname{Dom}(f)\) , then \[\lim _{X \ni x \rightarrow a} f(x)=L .\] EXERCISE 5.8. Use Archimedes’s method (the method of Riemann sums) to prove that \[\int_{0}^{1} x^{2} d x=\frac{1}{3}\] (You will need to know a formula for \(\sum_{k=0}^{n} k^{2}\) - see Proposition 4.6).
EXERCISE 5.9. Use Archimedes’s method to prove that \[\int_{0}^{1} x^{3} d x=\frac{1}{4} .\] (See Exercise 4.16).
EXERCISE 5.10. Prove that the Heaviside function has both left and right-hand limits at \(0 .\)
EXERCISE 5.11. Prove that a function has a limit at a point if and only if it has both left and right limits at that point and their values coincide.
EXERCISE 5.12. Prove that Theorem \(5.9\) applies to restricted limits.
EXERCISE 5.13. The point \(a\) is a limit point of the set \(X\) if, for every \(\delta>0\) , there exists a point \(x\) in \(X \backslash\{a\}\) with \(|x-a|<\delta\) . Let \(f\) be a real-valued function on \(X \subseteq \mathbb{R}\) . Prove that if \(a\) is a limit point of \(X\) , then if \(f\) has a restricted limit at \(a\) it is unique. Prove that if \(a\) is not a limit point of \(X\) , then every real number is a restricted limit of \(f\) at \(a\) .
EXERCISE 5.14. Prove that \(\lim _{x \rightarrow 0} \sin (x) / x=1\) .
EXERCISE 5.15. Prove Proposition 5.22.
EXERCISE 5.16. Prove that the function \(f(x)=x\) is continuous everywhere on \(\mathbb{R}\) . ExERCISE 5.17. A formula for the Fibonacci numbers is given in Exercise 4.12. Evaluate \(\lim _{n \rightarrow \infty} F_{n+1} / F_{n}\) .
EXERCISE 5.18. How large must \(n\) be to ensure that \(F_{n+1} / F_{n}\) is within \(10^{-1}\) of the limit in Exercise 5.17? Within \(10^{-2}\) ? Within \(10^{-k}\) ?
EXERCISE 5.19. Define the function \(\psi: \mathbb{R} \rightarrow \mathbb{R}\) by \[\psi(x):= \begin{cases}0 & x \notin \mathbb{Q} \\ 1 & x \in \mathbb{Q} .\end{cases}\] Prove that \(\psi\) is discontinuous everywhere.
EXERCISE \(5.20\) . Define the function \(\phi: \mathbb{R} \rightarrow \mathbb{R}\) by \[\phi(x):= \begin{cases}0 & x \notin \mathbb{Q} \\ \frac{1}{n} & x \in \mathbb{Q} \backslash\{0\}, x=\frac{m}{n}, \operatorname{gcd}(m, n)=1, n>0 \\ 1 & x=0 .\end{cases}\] Prove that \(\phi\) is continuous at every irrational number and discontinuous at every rational number.
EXERCISE 5.21. Prove that a real-valued function \(f\) on an open interval \(I\) is continuous at any point where its derivative exists, i.e. where \[\lim _{x \rightarrow a} \frac{f(x)-f(a)}{x-a}\] exists. What is the converse of this assertion? Prove that the converse is not true.
EXERCISE 5.22. Prove that if the function \(f\) has the limit \(L\) from the right at \(a\) , then the sequence \(f\left(a+\frac{1}{n}\right)\) has limit \(L\) as \(n \rightarrow \infty\) . Show that the converse is false in general.
EXERCISE 5.23. Let \(f\) and \(g\) be real functions. Let \(a \in \mathbb{R}\) and suppose that \[\lim _{x \rightarrow a} g(x)=L_{1}\] and \[\lim _{x \rightarrow L_{1}} f(x)=L_{2} .\] Prove that \[\lim _{x \rightarrow a} f \circ g=L_{2} .\] If \(g\) is continuous at \(a\) and \(f\) is continuous at \(g(a)\) , is \(f \circ g\) continuous at \(a\) ?
EXERCISE 5.24. Let \(f\) be a real function, \(a \in \mathbb{R}\) and \(\lim _{x \rightarrow a} f(x)=\) \(L\) . If \(\left\langle a_{n}\right\rangle\) converges to \(a\) , prove that \(\left\langle f\left(a_{n}\right)\right\rangle\) converges to \(L\) .
EXERCISE 5.25. Complete Example 5.26. That is, prove that \[\lim _{n \rightarrow \infty}\left(\frac{1}{n^{3}}\right) \frac{(n)(n+1)(2 n+1)}{6}=1 / 3 .\] EXERCISE 5.26. Evaluate \[\lim _{n \rightarrow \infty} \sum_{k=0}^{n}\left(\frac{k}{n}\right) \frac{1}{n} .\]
EXERCISE 5.27. Use induction to prove that every polynomial is continuous at every real number.
EXERCISE 5.28. Let \(-1<x<1\) . Prove that the geometric series with ratio \(x, \sum_{k=0}^{\infty} x^{k}\) , converges to \(\frac{1}{1-x}\) .
EXERCISE 5.29. Let the Fibonacci numbers \(F_{n}\) be defined as in Exercise 4.12. Consider the power series \(F(x)=\sum_{n=1}^{\infty} F_{n} x^{n}\) . Prove that the power series satisfies \[F(x)=x^{2} F(x)+x F(x)+x .\] Solve (5.32) for \(F(x)\) , decompose it by partial fractions, and use Exercise \(5.28\) to derive Formula \(4.20\) . This technique to find a formula for \(F_{n}\) by studying the function \(F\) is often fruitful. The function \(F\) is called the generating function for the sequence.
EXERCISE 5.30. Suppose one defines a sequence with the same recurrence relation as the Fibonacci numbers, \(F_{n+2}=F_{n+1}+F_{n}\) , but with different starting values for \(F_{1}\) and \(F_{2}\) . Find the generating function for the new sequence, and hence calculate a formula for the general term. Is \(\lim _{n \rightarrow \infty} F_{n+1} / F_{n}\) always the same?
EXERCISE 5.31. Prove that sine and cosine are continuous functions on all of \(\mathbb{R}\) .


21.8: Transplantation and Cancer Immunology
By the end of this section, you will be able to:
- Explain why blood typing is important and what happens when mismatched blood is used in a transfusion
- Describe how tissue typing is done during organ transplantation and the role of transplant anti-rejection drugs
- Show how the immune response is able to control some cancers and how this immune response might be enhanced by cancer vaccines
The immune responses to transplanted organs and to cancer cells are both important medical issues. With the use of tissue typing and anti-rejection drugs, transplantation of organs and the control of the anti-transplant immune response have made huge strides in the past 50 years. Today, these procedures are commonplace. Tissue typing is the determination of MHC molecules in the tissue to be transplanted to better match the donor to the recipient. The immune response to cancer, on the other hand, has been more difficult to understand and control. Although it is clear that the immune system can recognize some cancers and control them, others seem to be resistant to immune mechanisms.
The Rh Factor
Red blood cells can be typed based on their surface antigens. ABO blood type, in which individuals are type A, B, AB, or O according to their genetics, is one example. A separate antigen system seen on red blood cells is the Rh antigen. When someone is “A positive” for example, the positive refers to the presence of the Rh antigen, whereas someone who is “A negative” would lack this molecule.
An interesting consequence of Rh factor expression is seen in erythroblastosis fetalis , a hemolytic disease of the newborn (Figure \(\PageIndex{1}\)). This disease occurs when people negative for Rh antigen have multiple Rh-positive children. During the birth of a first Rh-positive child, the birth parent makes a primary anti-Rh antibody response to the fetal blood cells that enter the pregnant person's bloodstream. If the same parent has a second Rh-positive child, IgG antibodies against Rh-positive blood mounted during this secondary response cross the placenta and attack the fetal blood, causing anemia. This is a consequence of the fact that the fetus is not genetically identical to the birth parent, and thus the parent is capable of mounting an immune response against it. This disease is treated with antibodies specific for Rh factor. These are given to the pregnant person during the first and subsequent births, destroying any fetal blood that might enter their system and preventing the immune response.
Tissue Transplantation
Tissue transplantation is more complicated than blood transfusions because of two characteristics of MHC molecules. These molecules are the major cause of transplant rejection (hence the name “histocompatibility”). MHC polygeny refers to the multiple MHC proteins on cells, and MHC polymorphism refers to the multiple alleles for each individual MHC locus. Thus, there are many alleles in the human population that can be expressed (Table \(\PageIndex{1}\) and Table \(\PageIndex{2}\)). When a donor organ expresses MHC molecules that are different from the recipient, the latter will often mount a cytotoxic T cell response to the organ and reject it. Histologically, if a biopsy of a transplanted organ exhibits massive infiltration of T lymphocytes within the first weeks after transplant, it is a sign that the transplant is likely to fail. The response is a classical, and very specific, primary T cell immune response. As far as medicine is concerned, the immune response in this scenario does the patient no good at all and causes significant harm.
| Gene | # of alleles | # of possible MHC I protein components |
|---|---|---|
| A | 2132 | 1527 |
| B | 2798 | 2110 |
| C | 1672 | 1200 |
| E | 11 | 3 |
| F | 22 | 4 |
| G | 50 | 16 |
| Gene | # of alleles | # of possible MHC II protein components |
|---|---|---|
| DRA | 7 | 2 |
| DRB | 1297 | 958 |
| DQA1 | 49 | 31 |
| DQB1 | 179 | 128 |
| DPA1 | 36 | 18 |
| DPB1 | 158 | 136 |
| DMA | 7 | 4 |
| DMB | 13 | 7 |
| DOA | 12 | 3 |
| DOB | 13 | 5 |
Immunosuppressive drugs such as cyclosporine A have made transplants more successful, but matching the MHC molecules is still key. In humans, there are six MHC molecules that show the most polymorphisms, three class I molecules (A, B, and C) and three class II molecules called DP, DQ, and DR. A successful transplant usually requires a match between at least 3–4 of these molecules, with more matches associated with greater success. Family members, since they share a similar genetic background, are much more likely to share MHC molecules than unrelated individuals do. In fact, due to the extensive polymorphisms in these MHC molecules, unrelated donors are found only through a worldwide database. The system is not foolproof however, as there are not enough individuals in the system to provide the organs necessary to treat all patients needing them.
One disease of transplantation occurs with bone marrow transplants, which are used to treat various diseases, including SCID and leukemia. Because the bone marrow cells being transplanted contain lymphocytes capable of mounting an immune response, and because the recipient’s immune response has been destroyed before receiving the transplant, the donor cells may attack the recipient tissues, causing graft-versus-host disease . Symptoms of this disease, which usually include a rash and damage to the liver and mucosa, are variable, and attempts have been made to moderate the disease by first removing mature T cells from the donor bone marrow before transplanting it.
Immune Responses Against Cancer
It is clear that with some cancers, for example Kaposi’s sarcoma, a healthy immune system does a good job at controlling them (Figure \(\PageIndex{2}\)). This disease, which is caused by the human herpesvirus, is almost never observed in individuals with strong immune systems, such as the young and immunocompetent. Other examples of cancers caused by viruses include liver cancer caused by the hepatitis B virus and cervical cancer caused by the human papilloma virus. As these last two viruses have vaccines available for them, getting vaccinated can help prevent these two types of cancer by stimulating the immune response.
On the other hand, as cancer cells are often able to divide and mutate rapidly, they may escape the immune response, just as certain pathogens such as HIV do. There are three stages in the immune response to many cancers: elimination, equilibrium, and escape. Elimination occurs when the immune response first develops toward tumor-specific antigens specific to the cancer and actively kills most cancer cells, followed by a period of controlled equilibrium during which the remaining cancer cells are held in check. Unfortunately, many cancers mutate, so they no longer express any specific antigens for the immune system to respond to, and a subpopulation of cancer cells escapes the immune response, continuing the disease process.
This fact has led to extensive research in trying to develop ways to enhance the early immune response to completely eliminate the early cancer and thus prevent a later escape. One method that has shown some success is the use of cancer vaccines, which differ from viral and bacterial vaccines in that they are directed against the cells of one’s own body. Treated cancer cells are injected into cancer patients to enhance their anti-cancer immune response and thereby prolong survival. The immune system has the capability to detect these cancer cells and proliferate faster than the cancer cells do, overwhelming the cancer in a similar way as they do for viruses. Cancer vaccines have been developed for malignant melanoma, a highly fatal skin cancer, and renal (kidney) cell carcinoma. These vaccines are still in the development stages, but some positive and encouraging results have been obtained clinically.
It is tempting to focus on the complexity of the immune system and the problems it causes as a negative. The upside to immunity, however, is so much greater: The benefit of staying alive far outweighs the negatives caused when the system does sometimes go awry. Working on “autopilot,” the immune system helps to maintain your health and kill pathogens. The only time you really miss the immune response is when it is not being effective and illness results, or, as in the extreme case of HIV disease, the immune system is gone completely.
Everyday Connection
How Stress Affects the Immune Response: The Connections between the Immune, Nervous, and Endocrine Systems of the Body
The immune system cannot exist in isolation. After all, it has to protect the entire body from infection. Therefore, the immune system is required to interact with other organ systems, sometimes in complex ways. Thirty years of research focusing on the connections between the immune system, the central nervous system, and the endocrine system have led to a new science with the unwieldy name of called psychoneuroimmunology . The physical connections between these systems have been known for centuries: All primary and secondary organs are connected to sympathetic nerves. What is more complex, though, is the interaction of neurotransmitters, hormones, cytokines, and other soluble signaling molecules, and the mechanism of “crosstalk” between the systems. For example, white blood cells, including lymphocytes and phagocytes, have receptors for various neurotransmitters released by associated neurons. Additionally, hormones such as cortisol (naturally produced by the adrenal cortex) and prednisone (synthetic) are well known for their abilities to suppress T cell immune mechanisms, hence, their prominent use in medicine as long-term, anti-inflammatory drugs.
One well-established interaction of the immune, nervous, and endocrine systems is the effect of stress on immune health. In the human vertebrate evolutionary past, stress was associated with the fight-or-flight response, largely mediated by the central nervous system and the adrenal medulla. This stress was necessary for survival. The physical action of fighting or running, whichever the animal decides, usually resolves the problem in one way or another. On the other hand, there are no physical actions to resolve most modern day stresses, including short-term stressors like taking examinations and long-term stressors such as being unemployed or losing a spouse. The effect of stress can be felt by nearly every organ system, and the immune system is no exception (Table \(\PageIndex{3}\)).
| System | Stress-related illness |
|---|---|
| Integumentary system | Acne, rashes, irritation |
| Nervous system | Headaches, depression, anxiety, irritability, loss of appetite, lack of motivation, reduced mental performance |
| Muscular and skeletal systems | Muscle and joint pain, neck and shoulder pain |
| Circulatory system | Increased heart rate, hypertension, increased probability of heart attacks |
| Digestive system | Indigestion, heartburn, stomach pain, nausea, diarrhea, constipation, weight gain or loss |
| Immune system | Depressed ability to fight infections |
| Male reproductive system | Lowered sperm production, impotence, reduced sexual desire |
| Female reproductive system | Irregular menstrual cycle, reduced sexual desire |
At one time, it was assumed that all types of stress reduced all aspects of the immune response, but the last few decades of research have painted a different picture. First, most short-term stress does not impair the immune system in healthy individuals enough to lead to a greater incidence of diseases. However, older individuals and those with suppressed immune responses due to disease or immunosuppressive drugs may respond even to short-term stressors by getting sicker more often. It has been found that short-term stress diverts the body’s resources towards enhancing innate immune responses, which have the ability to act fast and would seem to help the body prepare better for possible infections associated with the trauma that may result from a fight-or-flight exchange. The diverting of resources away from the adaptive immune response, however, causes its own share of problems in fighting disease.
Chronic stress, unlike short-term stress, may inhibit immune responses even in otherwise healthy adults. The suppression of both innate and adaptive immune responses is clearly associated with increases in some diseases, as seen when individuals lose a spouse or have other long-term stresses, such as taking care of a spouse with a fatal disease or dementia. The new science of psychoneuroimmunology, while still in its relative infancy, has great potential to make exciting advances in our understanding of how the nervous, endocrine, and immune systems have evolved together and communicate with each other.


16.7: Suggested Reading
Allegaert, K., & Iacobelli, S. (2022). Editorial: The developing kidney: Perinatal aspects and relevance throughout life. Frontiers in Pediatrics , 10 , 990854. https://doi.org/10.3389/fped.2022.990854
Kovesdy, C. (2022). Epidemiology of chronic kidney disease: An update 2022. Kidney International Supplements , 12 (1), 7–11. https://doi.org/10.1016/j.kisu.2021.11.003
National Kidney Foundation. (n.d.). Nutrition . https://www.kidney.org/nutrition
UNICEF. (2022, July 6). The state of food security and nutrition in the world 2022 . Food and Agriculture Organization of the United Nations, International Fund for Agricultural Development, United Nations Children’s Fund, United Nations World Food Programme, World Health Organization. https://data.unicef.org/resources/sofi-2022/
UPMC. (2022). What to Eat on a Renal Diet . https://share.upmc.com/2022/05/renal-diet/
Verbitsky, M., Krishnamurthy, S., Krithivasan, P., Hughes, D., Khan, A., Marasa, M., Vena, N., Kholasa, P., Zhang, J., Lim, T., Glessner, J., Weng, C., Shang, N., Shen, Y., Hripcsak, G., Hakonarson, H., Ionita-Laza, I., Levy, B., Kenny, E., … Gharavi, A. (2023). Genomic disorders in CKD across the lifespan. Journal of the American Society of Nephrology , 34 (4), 607–618. https://doi.org/10.1681/ASN.2022060725


8.5: Inelastic Collisions in One Dimension
Learning Objectives
By the end of this section, you will be able to:
- Define inelastic collision.
- Explain perfectly inelastic collision.
- Apply an understanding of collisions to sports.
- Determine recoil velocity and loss in kinetic energy given mass and initial velocity.
We have seen that in an elastic collision, internal kinetic energy is conserved. An inelastic collision is one in which the internal kinetic energy changes (it is not conserved). This lack of conservation means that the forces between colliding objects may remove or add internal kinetic energy. Work done by internal forces may change the forms of energy within a system. For inelastic collisions, such as when colliding objects stick together, this internal work may transform some internal kinetic energy into heat transfer. Or it may convert stored energy into internal kinetic energy, such as when exploding bolts separate a satellite from its launch vehicle.
Definition: Inelastic Collisions
An inelastic collision is one in which the internal kinetic energy changes (it is not conserved).
Figure \(\PageIndex{1}\) shows an example of an inelastic collision. Two objects that have equal masses head toward one another at equal speeds and then stick together. Their total internal kinetic energy is initially
\[\dfrac{1}{2}mv^2 + \dfrac{1}{2}mv^2 = mv^2.\]
The two objects come to rest after sticking together, conserving momentum. But the internal kinetic energy is zero after the collision. A collision in which the objects stick together is sometimes called a perfectly inelastic collision because it reduces internal kinetic energy more than does any other type of inelastic collision. In fact, such a collision reduces internal kinetic energy to the minimum it can have while still conserving momentum.
Definition: Perfectly Inelastic Collisions
A collision in which the objects stick together is sometimes called “perfectly inelastic.”
Example \(\PageIndex{1}\): Calculating Velocity and Change in Kinetic Energy - Inelastic Collision of a Puck and a Goalie
- Find the recoil velocity of a 70.0-kg ice hockey goalie, originally at rest, who catches a 0.150-kg hockey puck slapped at him at a velocity of 35.0 m/s.
- How much kinetic energy is lost during the collision? Assume friction between the ice and the puck-goalie system is negligible (Figure \(\PageIndex{2}\))
Strategy
Momentum is conserved because the net external force on the puck-goalie system is zero. We can thus use conservation of momentum to find the final velocity of the puck and goalie system. Note that the initial velocity of the goalie is zero and that the final velocity of the puck and goalie are the same. Once the final velocity is found, the kinetic energies can be calculated before and after the collision and compared as requested.
Solution for (a)
Momentum is conserved because the net external force on the puck-goalie system is zero.
Conservation of momentum is
\[p_1 + p_2 = p'_1 + p'_2 \nonumber\]
or
\[m_1v_1 + m_2v_2 = m_1v'_1 + m_2v'_2. \nonumber\]
Because the goalie is initially at rest, we know \(v_2 = 0.\) Because the goalie catches the puck, the final velocities are equal, or \(v'_1 = v'_2 = v'.\) Thus, the conservation of momentum equation simplifies to
\[m_1v_1 = (m_1 + m_2)v'. \nonumber\]
Solving for \(v'\) yields
\[v' = \dfrac{m_1}{m_1 + m_2}v_1. \nonumber\]
Entering known values in this equation, we get
\[\begin{align*} v' &= \left( \dfrac{0.150 \, kg}{70.0 \, kg + 0.150 \, kg} \right)(35.0 \, m/s) \\[5pt] &= 7.48 \times 10^{-2} m/s)^2 \\[5pt] &= 0.196 \, J. \end{align*}\]
The change in internal kinetic energy is thus
\[\begin{align*} KE'_{int} - KE_{int} &= 0.196 \, J - 91.9 \, J \\[5pt] &= -91.7 \, J \end{align*}\]
where the minus sign indicates that the energy was lost.
Discussion for (b)
Nearly all of the initial internal kinetic energy is lost in this perfectly inelastic collision. \(KE_{int} \) is mostly converted to thermal energy and sound.
During some collisions, the objects do not stick together and less of the internal kinetic energy is removed—such as happens in most automobile accidents. Alternatively, stored energy may be converted into internal kinetic energy during a collision. Figure \(\PageIndex{3}\) shows a one-dimensional example in which two carts on an air track collide, releasing potential energy from a compressed spring. Example \(\PageIndex{2}\) deals with data from such a collision.
Collisions are particularly important in sports and the sporting and leisure industry utilizes elastic and inelastic collisions. Let us look briefly at tennis. Recall that in a collision, it is momentum and not force that is important. So, a heavier tennis racquet will have the advantage over a lighter one. This conclusion also holds true for other sports—a lightweight bat (such as a softball bat) cannot hit a hardball very far.
The location of the impact of the tennis ball on the racquet is also important, as is the part of the stroke during which the impact occurs. A smooth motion results in the maximizing of the velocity of the ball after impact and reduces sports injuries such as tennis elbow. A tennis player tries to hit the ball on the “sweet spot” on the racquet, where the vibration and impact are minimized and the ball is able to be given more velocity. Sports science and technologies also use physics concepts such as momentum and rotational motion and vibrations.
Take-Home Experiment—Bouncing of Tennis Ball
- Find a racquet (a tennis, badminton, or other racquet will do). Place the racquet on the floor and stand on the handle. Drop a tennis ball on the strings from a measured height. Measure how high the ball bounces. Now ask a friend to hold the racquet firmly by the handle and drop a tennis ball from the same measured height above the racquet. Measure how high the ball bounces and observe what happens to your friend’s hand during the collision. Explain your observations and measurements.
- The coefficient of restitution \((c)\) is a measure of the elasticity of a collision between a ball and an object, and is defined as the ratio of the speeds after and before the collision. A perfectly elastic collision has a \(c\) of 1. For a ball bouncing off the floor (or a racquet on the floor), \(c\) can be shown to be \(c = (h/H)^{1/2} \) where \(h\) is the height to which the ball bounces and \(H\) is the height from which the ball is dropped. Determine \(c\) for the cases in Part 1 and for the case of a tennis ball bouncing off a concrete or wooden floor (\(c = 0.85 \) for new tennis balls used on a tennis court).
Example \(\PageIndex{2}\): Calculating Final Velocity and Energy Release - Two Carts Collide
In the collision pictured in Figure \(\PageIndex{3}\), two carts collide inelastically. Cart 1 (denoted \(m_1\) carries a spring which is initially compressed. During the collision, the spring releases its potential energy and converts it to internal kinetic energy. The mass of cart 1 and the spring is 0.350 kg, and the cart and the spring together have an initial velocity of \(-0.500 \, m/s\). After the collision, cart 1 is observed to recoil with a velocity of \(-4.00 \, m/s\).
- What is the final velocity of cart 2?
- How much energy was released by the spring (assuming all of it was converted into internal kinetic energy)?
Strategy
We can use conservation of momentum to find the final velocity of cart 2, because \(F_{net} = 0\) (the track is frictionless and the force of the spring is internal). Once this velocity is determined, we can compare the internal kinetic energy before and after the collision to see how much energy was released by the spring.
Solution for (a)
As before, the equation for conservation of momentum in a two-object system is
\[m_1v_1 + m_2v_2 = m_1v'_1 + m_2v'_2. \nonumber\]
The only unknown in this equation is \(v'_2.\) Solving for \(v'_2\) and substituting known values into the previous equation fields
\[\begin{align*} v'_2 &= \dfrac{m_1v_1 + m_2v_2 - m_1v'_1}{m_2} \\[5pt] &= \dfrac{0.350 \, kg)(2.00 \, m/s) + (0.500 \, kg)(-0.500 \, m/s)}{0.500 \, kg} - \dfrac{(0.350 \, kg)(-4.00 \, m/s)}{0.500 \, kg} \\[5pt] &= 3.70 \, m/s.\end{align*}\]
Solution for (b)
The internal kinetic energy before the collision is
\[\begin{align*} KE_{int} &= \dfrac{1}{2}m_1v_1^2 + \dfrac{1}{2} m_2v_2^2 \\[5pt] &= \dfrac{1}{2}(0.350 \, kg)(2.00 \, m/s)^2 + \dfrac{1}{2}(0.500 \, kg)(-0.500 \, m/s)^2 \\[5pt] &= 0.763 \, J. \end{align*}\]
After the collision, the internal kinetic energy is
\[\begin{align*} KE'_{int} &= \dfrac{1}{2}m_1v_1^{'2} + \dfrac{1}{2} m_2v_2^{'2} \\[5pt] &= \dfrac{1}{2}(0.350 \, kg)(-4.00 \, m/s)^2 + \dfrac{1}{2}(0.500 \, kg)(0.370 \, m/s)^2 \\[5pt] &= 6.22 \, J. \end{align*}\]
The change in internal kinetic energy is thus
\[\begin{align*} KE' - KE &= 6.22 \, J - 0.763 \, J \\[5pt] &= 5.46 \, J. \end{align*}\]
Discussion
The final velocity of cart 2 is large and positive, meaning that it is moving to the right after the collision. The internal kinetic energy in this collision increases by 5.46 J. That energy was released by the spring.
Summary
- An inelastic collision is one in which the internal kinetic energy changes (it is not conserved).
- A collision in which the objects stick together is sometimes called perfectly inelastic because it reduces internal kinetic energy more than does any other type of inelastic collision.
- Sports science and technologies also use physics concepts such as momentum and rotational motion and vibrations.
Glossary
- inelastic collision
- a collision in which internal kinetic energy is not conserved
- perfectly inelastic collision
- a collision in which the colliding objects stick together


13.1: Introduction to Acid/Base Equilibria
Make sure you thoroughly understand the following essential concepts:
- Write a chemical equation for the autoprotolysis of a species such as HCO 3 – ; write an equation for the equilibrium constant for this reaction.
- Write equations that express the chemical reactions and equilibrium constant expressions that define the strength of an acid or base in terms of proton transfer processes (Brønsted-Lowry concept).
- Write expressions defining the equilibrium constants K a and K b .
- Given the value of any two of K a , K b , and K w , evaluate the third one. (This requires that you can show that K a × K b = K w .)
- Explain the significance of the value of pK a for a given monoprotic acid-base system — that is, what it tells us about that particular acid and its conjugate base.
- Sketch out a simple proton free energy (PFE) diagrams illustrating the distinction between monoprotic strong and weak acids .
- State the meaning of the leveling effect , and construct a simple PFE diagram that illustrates its action.
- Describe, in your own words, the meaning of PFE , how it relates to pH, and in its significance in terms of the relative concentrations of a conjugate acid-base pair in a solution.
Acid-base reactions, in which protons are exchanged between donor molecules (acids) and acceptors (bases), form the basis of the most common kinds of equilibrium problems which you will encounter in almost any application of chemistry. In order to thoroughly understand the material in this unit, you are expected to be familiar with the following topics which were covered in the separate unit Introduction to Acid-Base Chemistry:
- The Arrhenius concept of acids and bases
- The Brønsted-Lowry concept, conjugate acids and bases
- Definition of pH and the pH scale
- Strong vs. weak acids and bases
You should also have some familiarity with the principles of chemical equilibrium, and know how to write and use equilibrium constants.
Proton transfer reactions in acid-base equilibria
As you should recall from your earlier introduction to acids and bases, the +1 electric charge of the tiny proton (a bare hydrogen nucleus) is contained in such a miniscule volume of space that the resulting charge density is far too large to enable its independent existence in solution; it will always attach to, and essentially bury itself in, the non-bonding orbitals of a solvent. Thus in aqueous solution, what we commonly represent as the "hydrogen ion" H + is more accurately described as the hydronium ion H 3 O + .
Acid-base chemistry is a transactional process in which protons are exchanged between two chemical species. A molecule or ion that loses or "donates" a proton is acting as an acid ; a species that receives or "accepts" that proton plays the role of a base .
Defining an acid as a substance that simply "dissociates" into its component parts
\[HA \rightarrow H^+ + AB^–\]
fails to capture the concept outlined in the above box. It is therefore important that you become comfortable with the Brønsted-Lowry model of acid-base behavior that represents the generalized acid-base reaction as a proton transfer process :
\[\underbrace{HA}_{\text{acid}} + \underbrace{B}_{base} \rightarrow \underbrace{A^-}_{\text{conjugate base}} + \underbrace{HB^+}_{\text{conjugate acid}} \]
The actual electric charges of the reaction products will of course depend on the particular nature of the species \(A\), but the base will always have one more negative charge than the acid HA. Although anyone familiar with chemicals tends to regard certain substances as "acids" and others as "bases", it is important to bear in mind that these labels have meaning only in terms of the particular process being considered. For example, in the absence of water, ammonia, which we usually think of as a base, can lose a proton (and thus act as an acid) to a "stronger" base \(B\) to form the amide ion:
\[NH_3 + B → NH_2^– + HB^+\]
Similarly, anhydrous sulfuric acid can accept a proton from a stronger acid AH:
\[H_2SO_4 + AH \rightarrow H_3SO_4^+ + A^–\]
However, this is rather exotic stuff that is beyond the scope of this introductory lesson. The acid-base processes we describe in this set of lessons take place in aqueous solution, so we will always assume that \(H_2O\), which can act either as a proton donor or proton acceptor, plays an active role:
- A substance that acts as an acid in aqueous solution donates a proton to \(H_2O\), yielding the hydronium ion \(H_3O^+\).
\[HA + H_2O → H_3O^+ + A^–\]
- Similarly, a base will remove a proton from \(H_2O\), yielding the hydroxide ion \(OH^–\).
\[B + H_2O → BH^+ + OH^–\]
Autoprotolysis
A small number of acids are themselves amphiprotic, meaning that one molecule can transfer a proton to another molecule of the same kind. The best-known example of this process, known as autoprotolysis , is of course that of water:
\[2 H_2O_{(l)} \rightleftharpoons OH^– + H_3O^+ \label{1-1}\]
Pure liquid ammonia and sulfuric acid are other well-known examples:
\[2 NH_{3(l)} \rightleftharpoons NH_2^– + NH_4^+\]
\[2 H_2SO_{4(l)} \rightleftharpoons HSO_4^– + H_3SO_4^+\]
Autoprotolysis reactions such as these typically proceed to only a very small extent. The equilibrium constant for Equation \(\ref{1-1}\) is commonly written as
\[K_w = [H^+][OH^–] \label{1-2}\]
and is usually known as the dissociation constant or ion product of water. For most purposes in elementary courses, we can use the value 1.0 × 10 –14 for K w . But you should understand that this value is correct only for pure water at 25° C and 1 atm pressure. K w varies considerably with temperature, pressure, and the ionic content of water:
It should be obvious that anyone working in the fields of physiological or oceanographic chemistry cannot routinely use \(1.0 \times 10^{–14}\) for K w !
The equilibrium constants K a and K b
Acids and bases vary greatly in their "strength" — that is, their tendencies to donate or accept protons. As with any chemical reaction, we can define an equilibrium constant \(K_c\) whose numerical value reflects the extent of the reaction — that is, the relative concentrations of the products and reactants when the reaction reaches equilibrium. The strength of an acid \(HA\) in water can be defined by the equilibrium
\[HA + H_2O \rightleftharpoons H_3O^+ + A^–\]
\[K_a =\dfrac{[H_3O^+][A^-]}{[HA]} \label{1-3a}\]
Similarly, the strength of the base \(A^−\) in water is defined by
\[A^− + H_2O \rightleftharpoons HA + OH^–\]
\[K_b = \dfrac{'[HA][OH^-]}{[A^-]} \label{1-3b}\]
Note carefully that reaction \(\ref{1-3a}\) is not the reverse of \(\ref{1-3b}\).
K a and K b are inversely proportional
How are K a and K b related? The answer can found by multiplying the above two expressions for K a and K b :
(1-3)
Alternatively, we can simply add the two corresponding reaction equations:
and — as you should recall from your earlier study of equilibrium, the equilibrium constant of the sum of two reactions is the product of the two K's. So either way you figure it, the \(K_a\) for an acid and \(K_b\) for its conjugate base are related in a very simple way:
\[K_a K_b = K_w \label{1-4}\]
Clearly, as the strength of a series of acids increases, the strengths of their conjugate bases will decrease, this inverse relation between K a and K b is implicit in Equation \(\ref{1-4}\). You will recall that the pH scale serves as a convenient means of compressing a wide range of [H + ]-values into a small range of numbers. Just as we defined the pH as the negative logarithm of the hydrogen ion concentration, we can define
\[pK = −\log K \label{2-5}\]
for any equilibrium constant. Acid and base strengths are very frequently expressed in terms of \(pK_a\) and \(pK_b\). From Equation \(\ref{1-4}\), it should be apparent that
\[pK_a + pK_b = pK_w \;(14.0\; at\; 25^o C) \label{1-5}\]
pK values of some common acids and bases
In order to convert pK a into K a , simply take the negative antilog of pK a . Thus, for acetic acid with
\[pK_a = 4.7\]
\[K_a = 10^{–4.7} = 2.0 \times 10^{–5} \]
The following table gives the p Ka values for a number of commonly-encountered acid-base systems which are listed in order of decreasing acid strength. It's worth taking some time to study this table in some detail.
lease take special note of the following points:
- Acids (1-4) are the major strong acids. All dissociate in water to produce the hydronium ion (5) , the strongest acid that can exist in water.
- (6) H 2 SO 3 has never been detected; "sulfurous acid" and its formula refer to a solution of SO 2 in water.
- The names bisulfate (3, 7) , bisulfite (6, 16) , and bicarbonate (13) are still commonly used by chemists, but the preferred names are now hydrogen sulfate , hydrogen sulfite , and hydrogen carbonate , respectively. Similarly, bisulfide (14, 26) is also known as hydrosulfide .
- The ion (9) as it exists in aqueous solution is often represented as Fe 3+ and is known more commonly as the ferrous or iron(III) ion. The aqueous aluminum ion Al 3+ (12) as well as many transition metal ions have similarly complex structures and give acidic solutions in water because the high positive charge increases the acidity of the attached H 2 O units. For more on these aquo complexes, see this brief discussion, or this much more detailed one by Jim Cook of the UK.
- The molecule H 2 CO 3 (13) exists as a minority species in an aqueous solution of CO 2 and has never been isolated. In most contexts, the formula H 2 CO 3 , the name carbonic acid and the pK a 6.3 refer to "total dissolved CO 2 " in the water. "True" H 2 CO 3 is about a thousand times stronger than the pK a of its equilibrium mixture with CO 2 (aq) .
- The sulfide, amide and oxide ions (26-28) , being stronger bases than OH – , cannot exist in water.
| ref | acid | pK a | conjugate base | pK b | |
|---|---|---|---|---|---|
| 1 | HClO 4 | perchloric acid | ~ –7 | ClO 4 – chlorate | ~ 21 |
| 2 | HCl | hydrochloric acid | ~ –3 | Cl – chloride | ~ 17 |
| 3 | H 2 SO 4 | sulfuric acid | ~ –3 | HSO 4 – bisulfate | ~ 17 |
| 4 | HNO 3 | nitric acid | ~ –1 | NO 3 – nitrate | 15 |
| 5 | H 3 O + | hydronium ion | 0 | H 2 O water | 14.00 |
| 6 | H 2 SO 3 | sulfurous acid | 1.8 | HSO 3 – bisulfite | 12.2 |
| 7 | HSO 4 – | bisulfate ion | 1.9 | SO 4 2– sulfate | 12.1 |
| 8 | H 3 PO 4 | phosphoric acid | 2.12 | H 2 PO 4 – dihydrogen phosphate | 11.88 |
| 9 | [Fe(H 2 O) 6 ] 3+ | hexaaquoiron(III) | 2.10 | [Fe(H 2 O) 5 OH] 2+ | 11.90 |
| 10 | HF | hydrofluoric acid | 3.2 | F – fluoride | 10.8 |
| 11 | CH 3 COOH | acetic acid | 4.76 | CH 3 COO – acetate | 9.3 |
| 12 | [Al(H 2 O) 6 ] 3+ | hexaaquoaluminum(III) | 4.9 | [Al(H 2 O) 5 OH] 2+ | 9.1 |
| 13 | H 2 CO 3 | carbonic acid | 6.3 | HCO 3 – bicarbonate | 7.7 |
| 14 | H 2 S | hydrogen sulfide | 7.04 | HS – bisulfide | 6.96 |
| 15 | H 2 PO 4 – | dihydrogen phosphate | 7.2 | HPO 4 – monohydrogen phosphate | 6.8 |
| 16 | HSO 3 – | bisulfite ion | 7.21 | SO 3 2– sulfite | 6.79 |
| 17 | HOCl | hypochlorous acid | 8.0 | OCl – hypochlorite | 6.0 |
| 18 | HCN | hydrocyanic acid | 9.2 | CN – cyanide | 4.8 |
| 19 | H 3 BO 4 | boric acid | 9.30 | B(OH) 4 – borate | 4.79 |
| 20 | NH 4 + | ammonium ion | 9.25 | NH 3 ammonia | 4.75 |
| 21 | Si(OH) 4 | ortho -silicic acid | 9.50 | SiO(OH) 3 – silicate | 4.50 |
| 22 | HCO 3 – | bicarbonate ion | 10.33 | CO 3 2– carbonate | 3.67 |
| 23 | HPO 4 2– | monohydrogen phosphate ion | 12.32 | PO 4 3– phosphate | 1.67 |
| 24 | SiO(OH) 3 – | silicate ion | 12.6 | SiO 2 (OH) 2 2– | 1.4 |
| 25 | H 2 O | water | 14.00 | OH – hydroxide | 0 |
| 26 | HS – | bisulfide ion | ~ 19 | S 2– sulfide | ~ –5 |
| 27 | NH 3 | ammonia | ~ 23 | NH 2 – amide | ~ -9 |
| 28 | OH – | hydroxide ion | ~ 24 | O 2– oxide | ~ –10 |
The Fall of the proton
The concept of the "potential energy of the proton" (more properly, proton free energy ), is not nearly as intimidating as it might sound; it is easy to grasp, requires no mathematics, and will enable you to achieve a far better understanding of acid-base chemistry.
Most courses (and their textbooks) do a good job of explaining the concept of proton transfer, but few of them provide you with a clear basis for understanding and making useful predictions about the direction and extent this transfer will take. For example, when the gas HCl is added to water, you probably know that the reaction proceeds strongly to the right:
\[HCl + H_2O \rightleftharpoons H_3O^+ + Cl^– \label{2-1}\]
...but what principle determines this, and what is to prevent the hydronium ion (also an acid) from pushing one of its own protons back onto a chloride ion, and thus reversing the reaction?
The concept of the proton free energy level was introduced in R.W. Gurney's classic Ionic Processes in Solution (1953). The best detailed exposition can be found in the first (1970) edition of Aquatic Chemistry by W. Stumm and J.L. Morgan.
The answer is that a proton is never really "pushed"; it will spontaneously seek out the lowest potential energy state (that is, the strongest base) it can find. Think how a macroscopic object such as a book will tend to fall to a location where its [gravitational] potential energy is as low as possible. Of course, the potential energy of a proton has nothing to do with gravity, but rather with how tightly it can bind to a base. This kind of potential energy is known as proton free energy (PFE) — but in the context of first-year chemistry, we can also just call it "proton energy".
You are not sure what free energy is? Don’t worry about it for the time being; just think of it as you would any other form of potential energy: something that falls when chemical reactions take place. See here for the lesson on free energy.
For the present, all you need to know is that a proton will tend to "fall" to the lowest potential energy state it can find. In keeping with this idea, you can think of an acid as a proton source , and a base as a proton sink . Using this terminology, we can depict the generalized process HA + B → AB – + HB + as in this plot, which depicts the proton on HA "falling" to the lower-PFE "sink" provided by the base B, leaving AB – and HB + as products.
We will be making a lot of use of schematics like this farther on, so take a moment to familiarize yourself with information it depicts.
This "source/sink" nomenclature recalls the tendency of water to flow down from a high elevation to a lower one; this tendency (which is related to the amount of energy that can be extracted in the form of electrical work if the water flows through a power station at the bottom of the dam) will be directly proportional to the difference in elevation (difference in potential energy ) between the source (top of the dam) and the sink (bottom of the dam).
Thus you can regard proton free energy as entirely analogous to the gravitational potential energy of the water contained in a dam; in this case, we are dealing with chemical potential energy, more commonly known as free energy . Take particular note of the following:
- The ordinate represents the relative free energy of a mole of protons when attached to a given base; we label it simply "proton energy" for brevity.
- The gray vertical line at the center of the diagram separates the acidic form of each conjugate pair from its base form; the terms "proton sources" and "proton sinks" are there to remind us of the role these forms play in acid-base chemistry.
- The red diagonal arrow represents the actual process in which a mole of protons "falls" from the acid HA to the base B. The fact that the arrow slants downward indicates that this process is spontaneous, meaning that the reaction proceeds to completion. In other words, the equilibrium constant for this reaction is very large.
- The green horizontal arrows depict the changes (that is, the formation of the two new conjugate species) that result from the proton transfer as the protons "fall" (in free energy) from the acid HA to the base B.
Strong acids in water
When you were first introduced to the Brønsted-Lowry model, you learned that the proper way to represent the "dissociation" of a strong acid like hydrochloric is
\[HCl + H_2O \rightarrow H_3O^+ + Cl^- \label{2-1a}\]
Thus water is revealed as an active participant in the behavior of an acid, rather than as an inert spectator or solvent. Although analogous acid-base reactions can take place in other solvents such as liquid ammonia, virtually all of the acid-base chemistry we encounter in daily life is based on water. Since water is acting as the proton acceptor here, it must be fulfilling the role of the base B in the schematic given above. We can therefore construct a similar plot specifically for HCl (or, in fact, for any strong acid).
Recall that, for strong acids, this reaction is essentially complete. This means that a "solution of hydrochloric acid" is in reality a solution of hydronium and chloride ions; except in solutions much more concentrated than 1 M, the species HCl is only present in trace amounts.
Weak acids in water
When acetic acid CH 3 COOH is dissolved in water, the resulting solution displays only very mild acidic properties. For simplicity we will represent acetic acid and the acetate ion CH 3 COO – by HAc and Ac – , respectively:
\[HAc + H_2O \rightleftharpoons H_3O^+ + Ac^– \label{2-2}\]
But although this equation has the same form as (2-1) above, it hides an important qualitative difference: HAc is a weak acid, meaning that the proton transfer takes place to only a tiny extent; the equilibrium strongly favors the left side:
HAc + H 2 O H 3 O + + Ac –
The corresponding proton free energy diagram now shows the H 3 O + -H 2 O system at the top, with the proton-transfer arrow pointing upward, indicating that the protons get kicked up to a higher free-energy level in this process.
Where does this additional energy come from? Simply from the random thermal energy in the solution. As you might expect, this is not a thermodynamically favorable process, so it only happens to a very small extent. In contrast to the HCl example we described previously in which 99.99+ percent of the HCl molecules end up as hydronium ions, almost all of the acetic acid molecules remain unchanged, and only tiny amounts of H 3 O + and Ac – are produced.
Strong and weak acids both donate protons to water, yielding hydrogen ions and thus rendering the solution acidic.
What's different is that virtually all of the protons of the strong acid "fall" to the H 2 O level, whereas only a tiny fraction of the weak acid molecules acquire enough thermal energy to promote their protons up to the H 2 O level.
The fundamental difference between strong acids and weak acids is clearly evident when we combine the preceding two proton-free energy diagrams into one. Strong acids (strong "conjugate-pairs") are always above the PFE of the H 3 O + -H 2 O system, while the weak conjugate pairs are below it.
The other big difference with acetic acid is that water is acting as a base here, rather than as an acid as it did in the HCl example. So it's not surprising that water, the basis of almost all acid-base chemistry, should appear in the middle of these diagrams.
Water can swing both ways
Substances that can both donate and accept protons are said to be amphiprotic . Many oxides and hydroxides behave in this way, and since H 2 O falls into both of these classes, it is not surprising that water is amphiprotic:
- Water ("the-acid") : \[H_2O + B \rightleftharpoons BH^+ + OH^–\]
- Water ("the base") : \[H_2O + AH \rightleftharpoons H_3O^+ + A^–\]
("AH" above represents any acid; B stands for any base)
These two opposing tendencies can be represented by
In the case of water, the contest is a draw; not only does neither side win, but the fraction of
H
2
O molecules that break up in this way is miniscule: In pure water, only about one H
2
O molecule out of 10
9
is “dissociated” at any instant:
H 2 O H + + OH – (2-4)
The actual reaction, of course, is the proton transfer
(2-5)
for which the equilibrium constant
K w = [H 3 O + ][OH – ] (2-6)
is known as the ion product of water . The value of K w at room temperature is 1.008 x 10 –14 . We can also express this in logarithmic terms: pK w = 14.0.
As with any equilibrium constant, the value of K w is affected by the
- temperature : K w undergoes a 10-fold increase between 0°C and 60°C
- pressure : K w is about doubled at 1000 atm
- presence of ionic species in the solution .
Because most practical calculations involving K w refer to ionic solutions rather than to pure water, the common practice of using 10 –14 as if it were a universal constant is unwise; under the conditions commonly encountered in the laboratory, pK w can vary from about 11 to almost 15. In seawater, K w is 6.3 × 10 −12 . (See Stephen J. Hawkes: “ pK w is almost never 14.0 ”, J. Chem. Education 1995: 72(9) 799-802)
Notice that under conditions when K w differs significantly from 1.0 × 10 –14 , the pH of a neutral solution will not be 7.0.
For example, at a pressure of 93 kbar and 527°C, K w = 10 −3.05 , the pH of pure water would be 1.5. Similarly unusual conditions apply to deposits of water in geological formations and in undersea vents.
At 60° C, the ion product of water is \(9.6 \times 10^{-14}\). What is the pH of a neutral solution at this temperature?
Solution:
Under these conditions, [H + ][OH – ] = 9.6E–14. If the solution is neutral, [H + ] = [OH – ] = (9.6E–14) ½ = 3.1E–7, corresponding to a pH of log 3.1E-7 = 6.5 .
The leveling effect
The strengths of strong acids and bases are "leveled" in water. This means that all strong acids and strong bases are equally strong in water . To see this more clearly, notice that the free energies of protons in all of these acids (even nitric) are sufficiently above the H 3 O + /H 2 O level that their "dissociation" is essentially complete. Thus
The strongest acid that can exist in water is H
3
O
+
;
The strongest base that can exist in water is OH
–
Similarly, the hydroxide ion is the strongest base that can exist in water.
So what happens if you add a soluble oxide such as Na
2
O to water? Since O
2
–
is a proton sink to H
2
O, it will react with the solvent, leaving OH
–
as the strongest base present: Na
2
O + H
2
O → 2 OH
–
+ Na
+
.
Thus all bases stronger than OH – appear equally strong in water — simply because they are all converted to OH – .
PFE's are the key to understanding acid-base reactions
In the preceding section, we have employed simple PFE (proton free energy) diagrams to help you better understand the concepts of strong and weak acids, water as an acid or a base, and the leveling effect. But we can gain a more generalized and comprehensive view of acid-base chemistry with the aid of a PFE diagram that shows a whole series of acid-base systems, arranged in order of decreasing acid strengths.
How acid strengths relate to proton free energies
As we explained near the beginning of this lesson, we ordinarily express the strength of an acid by quoting its "dissociation constant" K a , which is the equilibrium constant for transfer of a proton to the base H 2 O.
HA + H 2 O → H 3 O + + A –
We also introduced the concept of PFE as an alternative way of expressing the strength of an acid. If both PFE and K a express the same thing, you may wonder how these two quantities are related.
If you have had an introduction to thermodynamics and specifically to free energy G , you might wish to look at the Appendix section of this lesson to review the relationship between the fall in free energy and the value of pK a .
A thorough answer to this question requires a bit of thermodynamics, which you may not have yet studied, but you don't really need any of the math here. It should suffice to show the results in the form of a plot, which yields a simple straight-line relationship when plotted on a semi-logarithmic scale.
Note that
- The extent of any chemical reaction is proportional to the fall in free energy when it takes place. So negative Δ G values correspond to larger equilibrium constants.
- For acid-base reactions specifically, K a is inversely proportional to Δ G , and thus to the fall in PFE.
- This means that the proton on a strong acid ( K a > 1) has a large negative PFE; it has a "long way to fall" to the H 2 O level.
- Conversely, a proton on a weak acid ( K a <1) can only move up to the H 2 O level when supplied with thermal energy from the surroundings.
- A K value of 10 0 = 1 corresponds to a zero-free energy change, meaning that the reaction is at equilibrum when [HA] = [A – ].
- Proton-free energy changes exceeding –20 kJ mol –1 result in K a values so large that the proton transfer can be considered "complete" for most practical purposes.
Understanding the PFE chart
On the PFE diagram shown below, each acid-base system is represented by a horizontal line connecting an acidic species with its conjugate base. Each line is placed at a height on the diagram that is proportional to its pK a , as measured on the logarithmic scale (labeled "pH") at the right. What we have, then, is a list of acid-base systems arranged in order of decreasing acid strength.
Think of this chart as an alternative view of the table of acid strengths displayed near the top of this page, in which some of the information that is "hidden" in the table is now revealed in an easily-grasped way.
Take particular note of the following points:
- The Δ G scale on the left measures the fall in potential energy when one mole of protons is transferred from a given acid to H 2 O. Negative values correspond to stronger acids.
-
Any acid shown on the left side of the vertical line running down the center of the diagram can donate protons to any base (on the right side of the line) that appears
below
it.
The value of the equilibrium constant for such a reaction will exceed unity; for Δ G separations of more than about 10 kJ/mol, the reaction can be considered to be complete. - The zero free energy assigned to the hydronium ion corresponds to the standard free energy of formation Δ G f ° of this ion.
- Notice how the two conjugate pairs H 3 O + / H 2 O and H 2 O/OH – bracket the 80-kJ/mol region that comprises the weak acids. The weakness of these acids reflects the energetic cost of moving a proton "up-hill" to the H 2 O level.
- The strong acids at the top of the table can all be regarded as totally "dissociated" in water. This means that these acids cannot exist in water, as they are totally transformed into H 3 O + . Of course, they will also react with any bases below the H 2 O level that may be available. (See "leveling effect" below.)
- Similarly, the two strong bases NH 2 – and O 2 – shown at the bottom of the table cannot exist in water. Thus if solid sodium oxide Na 2 O is added to water, the oxide ions are such a low-energy sink that will totally fill up with protons from the H 2 O acid level which defines the top of the strong-base range.
- Notice the pH scale along the right side, and note the range of pH values possible in water. We will say more about the significance of pH in these diagrams farther on.
The PFE chart helps you visualize (and understand!) acid-base chemistry
The virtue of a PFE chart is that you can see, at a glance, the relations between conjugate acid-base pairs that enables you to make quick, qualitative determinations of what is going on even in complex mixtures of acids and bases. The speed and convenience this affords enables you to build a much deeper conceptual understanding of acid-base chemistry.
In order to do the same thing by consulting a simple list of acid K a 's or pK a 's, you also need to be familiar with the principles of chemical equilibrium and thermodynamics — and then take the time to work through the arithmetic.
In the preceding sections we have shown how the PFE-chart approach can clarify the distinction between strong and weak acids. We will now go on to describe a variety of other principles that are easily grasped by consulting a PFE chart.
Significance of the pH scale
You will have noticed the scale labeled "pH" at the left side of the PFE plot.
First, please understand what this scale does NOT do; it does not tell you what the pH of the solution will be when you add an acid to pure water. This pH will depend on both the strength ( K a or pK a ) of the acid, and on its concentration in the resulting solution; you will learn how to carry out pH calculations in later lessons.
All but the very weakest acids can drive the pH down to near the bottom of the scale if their solutions are sufficiently concentrated. The only thing you can be sure of is that the pH of a solution of a solution of an acid in pure water will never be greater than 7.
pH sets the average proton free energy
When pH was first introduced in 1909, it was defined in terms of the "hydrogen ion concentration" [H + ]. This has since been amended to the hydrogen ion activity {H + }. The latter term refers to the "effective" concentration of these ions — that is to the "availability" of protons (regardless of whether they physically exist as H 3 O + units or in other forms) to react with bases. But from the standpoint of thermodynamics, "availability" is just a loose term for "free energy". So we can also say that
pH is a measure of the average PFE in an aqueous solution
The more negative the pH, the higher the proton free energy in the solution. (The inverse relation is a consequence of the negative-logarithm definition of pH.)
The pH has the dual advantages of being both dimensionless and readily observable. More importantly, we can easily alter the pH of a solution by adding some strong acid or base. And when we do this, we are in effect using the pH as a tool for controlling the average PFE in the solution.
More on this in a moment; first, let us introduce an important relation that you may have seen in an earlier chemistry course, and which we will discuss more in a later lesson in this group.
pH can control the relative abundance of conjugate acid-base pairs
Recalling the equilibrium expression for a weak acid
\[K_a = \dfrac{[H^+][A^-]}{[HA]} \label{3-1}\]
We can solve this for \([H^+]\):
\[H^+] = K_a \dfrac{[HA]}{[A^-]} \label{3-2}\]
Re-writing this in terms of negative logarithms, this becomes
(3-3)
or, since pK a = – log K a , we invert the ratio to preserve the positive sign:
We will make extensive use of this equation in a later lesson on acid-base buffering and titration.
(3-4)
This extremely important relation tells us that the pH of a solution containing a weak acid-base system controls the relative concentrations of the acid and base forms of that system. Of special interest is the case in which the pH of a solution of an acid-base system is set to the value of its
pK
a
. According to the above equation, when pH =
pK
a
, the log term becomes zero, so that the ratio
[AB
–
] / [HA] = 10
0
= 1, meaning that [HA] = [AB
–
]. In other words,
This pH adjustment can be easily made by adding appropriate amounts of a strong acid or strong base to the solution.
When the pH of a solution is set to the value of the pK a of an acid-base pair, the concentrations of the acid- and base forms will be identical. This condition can be represented schematically:
This is the basis for the arrangement of the various acid-base systems depicted in the PFE diagram; each conjugate pair is placed at a location on the vertical pH scale that corresponds to the acid's pK a . It is instructive to consider how sensitively the concentration ratio [A – ] / [HA] depends on the pH.
Hypochlorous acid, HOCl, has a pK a of 8.0.
- Estimate the ratio R = [OCl – ] / [HOCl] in a solution of sodium hypochlorite whose pH has been adjusted to 8.2.
- If the initial concentration of the hypochlorite solution was 0.10 M , what will be the concentrations of the acid and base forms at pH 8.2?
Solution
a)
Substituting in Eq 3-4, the log term becomes (8.2 – 8.0) = 0.2,
so
R
= 10
0.2
= 1.6.
Note that it does not matter whether the initial solution consists of HOCl or NaOCl or some mixture thereof; adjusting the pH forces the system composition to the ratio calculated above.)
b) In the resulting solution, the base form OCl – is the predominant species, the mole ratio R being 1.6/1. The mole fraction of OCl – will therefore be 1.6/2.6 = 0.62, making that of HOCl (1.0 – .62) = 0.38. The actual concentrations will be 0.1 times these values, reflecting the 0.10 M concentration of the solution.
To what pH must we adjust a solution of acetic acid ( pK a 4.7) in order to convert 20 percent of the acid into its base form?
Solution:
In the final solution, the mole fractions will be [AB – ] = .20, [HA] = 0.80; the ratio [AB – ]/[HA] = 0.20/0.80 = 0.25. Substituting into Eq 2-9, we have
\[pH = 4.7 + \log\' 0.25 = 4.7 – 0.6 = 4.1\]
We can explore the relation between the pH and the distribution of conjugate species between the acid and base forms by carrying out calculations similar to those in Problem Example 2 for an acid-base system at different pH values.
The orange bars and blue numbers show the mole fractions of the acid and base forms at pH values of the pK a and the pK a ± 1. The green numbers give the percent dissociation of the acid at each pH.
The apparent symmetry between the two extreme pH values suggests a simple relation between the log of the ratio
[AB
–
] / [HA] and the pH. This is seen very clearly when this ratio is plotted as a function of pH.
Solutions containing many different weak acid-base systems are very commonly encountered in nature, especially in biological fluids or natural waters, including the ocean. When the pH of such solutions is altered by the addition of strong acid or base (as can occur, for example, when wastes from industrial processes or mine drainage, or acidic rain, enter a lake or stream); the distribution of all conjugate acid-base systems will change.
It is important to note that the [AB – ] / [HA] ratios (calculated here from Eq 3-4) yield only a qualitative picture and do not take into account the concentrations of the different acid-base systems or proton transfers between them. The latter processes would cause the lowest proton-vacant levels (that is, the strongest bases) to be completely filled from the bottom up, depending on the number of protons available. An exact calculation would require solving a set of simultaneous equations and is beyond the scope of this lesson.
A . What do we mean by the "strength" of an acid ?
The Brønsted-Lowry model defines the strength of an acid HA as its tendency to donate protons to a base B. But bases vary in their abilities to accept protons, so the tendency for the complete transaction
HA + B → BH + + A – (A-1)
to occur depends on the two processes
HA → H + + A – K 1 (A-2)
B + H + → BH + K 2 (A-3)
K a 's are always expressed on a scale relative to the K b of the solvent
Because it is impossible to study either of these steps independently, what we might call the "absolute" strength of an acid (Eq A-2) cannot be measured. For this reason, we define a "standard" base, usually the solvent, and most commonly, water. Thus we replace Eq A-3 with
H 2 O + H + → H 3 O + K 3 (A-4)
so that Eq A – 1 now expresses the strength of HA relative to the base H 2 O :
HA + H 2 O → H 3 O + + A – (A-5)
which, for convenience, we write in its abbreviated form
HA → H + + A – (A-5)
whose equilibrium constant is
(A-6)
This value of K a is, of course, numerically identical to that for Eq A-5.
B . pH and proton free energy
This section may be of interest to readers who have studied elementary thermodynamics, and have some familiarity with Gibbs free energy . it is not required for understanding and using the concept of PFE.
From elementary thermodynamics, the driving force of a chemical reaction is given by the standard free energy change:
Δ G ° = –RT ln K (B-1)
Solving this equation for K yields
(B-2)
For a weak acid we know that
(B-3)
Substituting Eq B-2 into this expression, we get
(B-4)
Next, we take the negative log of each term, recalling that ln x = 2.3 log x :
(B-5)
or
(B-6)
For the special case in which the concentrations of the conjugate pairs [H + ] and [HA] are equal, pH = pK a , and the rightmost term above disappears, leaving
(B-7)
Thus the pH is a direct measure of the average proton free energy in a solution.
Construct a PFE diagram showing a) H 3 O + , b) acetic acid HAc ( pK a = 4.76), and c) H 2 O.
Solution
Solving Eq A – 13 for ΔG° yields ΔG° = (2.3RT) × pK a , or, at 25°C, ΔG° = (2.3 × 8.314 J K –1 mol –1 ) × pK a = 5698 J K –1 mol –1 × pK a
a) The proton in H 3 O + is already at the level of that in the standard base H 2 O (Eq A – 5), so for this transfer is [by definition] ΔG° = 5698 J K –1 mol –1 × 0 = 0
b) The increase in the free energy required to transfer a proton from its level in HAc to that in H 2 O is ΔG° = 5698 J K –1 mol –1 × 4.76 = 27.1 kJ mol –1
c) To transfer a proton from H 2 O to another H 2 O (autoprotolysis), its free energy must increase by 5698 J K –1 mol –1 × 14.0 = 79.8 kJ mol –1


6.9: Force expressions and projections
Now we express the forces drawn in the FBD in the form of equations. For the forces in Figure 6.4, assuming a kinetic friction coefficient \(\mu_{k}\) during the sliding motion we get:
\[\begin{align} \overrightarrow{\boldsymbol{F}}_{A, g} & =-m_{A} g \hat{\boldsymbol{\jmath}} \tag{6.11} \label{6.11}\\[4pt] \overrightarrow{\boldsymbol{F}}_{A, N} & =F_{N} \hat{\boldsymbol{n}} \tag{6.12} \label{6.12}\\[4pt] \overrightarrow{\boldsymbol{F}}_{A, f} & =-\mu_{k} F_{N} \hat{\boldsymbol{t}} \tag{6.13} \label{6.13}\end{align}\]
Projecting the force vectors
After having drawn the object and force vectors in the FBD, we need their scalar components to evaluate Newton’s second law along each axis. We thus project the forces on the coordinate system (CS) that was selected in the previous section and was drawn in the sketch and/or FBD. This is done using the method described in Sec. 3.2.5, and shown in Figure 6.5. For example we get the \(x\) component of the vector \(\overrightarrow{\boldsymbol{F}}_{A, f}\) using \(F_{A, f, x}=\overrightarrow{\boldsymbol{F}}_{A, f} \cdot \hat{\boldsymbol{\imath}}\).
As projected components of \(\overrightarrow{\boldsymbol{F}}_{g}\) we obtain:
\[\begin{align} F_{A, g, t} & =m_{A} g \sin \alpha \tag{6.14} \label{6.14}\\[4pt] F_{A, g, n} & =-m_{A} g \cos \alpha \tag{6.15} \label{6.15}\end{align}\]


7.4: Conservative Forces and Potential Energy
Learning Objectives
By the end of this section, you will be able to:
- Define conservative force, potential energy, and mechanical energy.
- Explain the potential energy of a spring in terms of its compression when Hooke’s law applies.
- Use the work-energy theorem to show how having only conservative forces implies conservation of mechanical energy.
Work is done by a force, and some forces, such as weight, have special characteristics. A conservative force is one, like the gravitational force, for which work done by or against it depends only on the starting and ending points of a motion and not on the path taken. We can define a potential energy (PE) for any conservative force, just as we did for the gravitational force. For example, when you wind up a toy, an egg timer, or an old-fashioned watch, you do work against its spring and store energy in it. (We treat these springs as ideal, in that we assume there is no friction and no production of thermal energy.) This stored energy is recoverable as work, and it is useful to think of it as potential energy contained in the spring. Indeed, the reason that the spring has this characteristic is that its force is conservative . That is, a conservative force results in stored or potential energy. Gravitational potential energy is one example, as is the energy stored in a spring. We will also see how conservative forces are related to the conservation of energy.
Potential Energy and Conservative Forces
Potential energy is the energy a system has due to position, shape, or configuration. It is stored energy that is completely recoverable.
A conservative force is one for which work done by or against it depends only on the starting and ending points of a motion and not on the path taken.
We can define a potential energy (PE) for any conservative force. The work done against a conservative force to reach a final configuration depends on the configuration, not the path followed, and is the potential energy added.
Potential Energy of a Spring
First, let us obtain an expression for the potential energy stored in a spring \((PE_s\). We calculate the work done to stretch or compress a spring that obeys Hooke’s law. (Hooke’s law was examined in Elasticity: Stress and Strain , and states that the magnitude of force \(F\) on the spring and the resulting deformation \(\Delta L\) are proportional, \(F = k\Delta L\)). (See Figure.) For our spring, we will replace \(\Delta L\) (the amount of deformation produced by a force \(F\)) by the distance \(x\) that the spring is stretched or compressed along its length. So the force needed to stretch the spring has magnitude \(F = kx\), where \(k\) is the spring’s force constant. The force increases linearly from 0 at the start to \(kx\) in the fully stretched position. The average force is \(kx/2\). Thus the work done in stretching or compressing the spring is
\[W_s = Fd = (\frac{kx}{2})x = \frac{1}{2}kx^2. \]
Alternatively, we noted in Kinetic Energy and the Work-Energy Theorem that the area under a graph of \(F\) vs. \(x\) is the work done by the force. In Figure (c) we see that this area is also \(\frac{1}{2}kx^2\). We therefore define the potential energy of a spring , \(PE_s|) to be
\[PE_s = \dfrac{1}{2}kx^2,\]
where \(k\) is the spring’s force constant and \(x\) is the displacement from its undeformed position. The potential energy represents the work done on the spring and the energy stored in it as a result of stretching or compressing it a distance \(x\). The potential energy of the spring \(PE_s\) does not depend on the path taken; it depends only on the stretch or squeeze \(x\) in the final configuration.
The equation \(PE_s = \frac{1}{2}kx^2\) has general validity beyond the special case for which it was derived. Potential energy can be stored in any elastic medium by deforming it. Indeed, the general definition of potential energy is energy due to position, shape, or configuration. For shape or position deformations, stored energy is \(PE_s = \frac{1}{2}kx^2\), where \(k\) is the force constant of the particular system and \(x\) is its deformation. Another example is seen in Figure for a guitar string.
Conservation of Mechanical Energy
Let us now consider what form the work-energy theorem takes when only conservative forces are involved. This will lead us to the conservation of energy principle. The work-energy theorem states that the net work done by all forces acting on a system equals its change in kinetic energy. In equation form, this is
\[W_{net} = \dfrac{1}{2}mv^2 - \dfrac{1}{2}mv_0^2 = \Delta KE.\]
If only conservative forces act, then
\[W_{net} = W_c\]
where \(W_c\) is the total work done by all conservative forces. Thus,
\[W_c = \Delta KE.\]
Now, if the conservative force, such as the gravitational force or a spring force, does work, the system loses potential energy. That is,
\[W_c = -\Delta PE\]
Therefore, \[-\Delta PE = KE\] or
\[\Delta PE + \Delta KE = 0. \]
This equation means that the total kinetic and potential energy is constant for any process involving only conservative forces. That is,
\[KE + PE = {constant}\]
or
\[KE_i + PE_i = KE_f + PE_f \]
\[(conservative \, forces \, only),\]
where i and f denote initial and final values. This equation is a form of the work-energy theorem for conservative forces; it is known as the conservation of mechanical energy principle . Remember that this applies to the extent that all the forces are conservative, so that friction is negligible. The total kinetic plus potential energy of a system is defined to be its mechanical energy , \((KE + PE)\). In a system that experiences only conservative forces, there is a potential energy associated with each force, and the energy only changes form between \(KE\) and the various types of \(PE\), with the total energy remaining constant.
Example \(\PageIndex{1}\): Using Conservation of Mechanical Energy to Calculate the Speed of a Toy Car
A 0.100-kg toy car is propelled by a compressed spring, as shown in Figure. The car follows a track that rises 0.180 m above the starting point. The spring is compressed 4.00 cm and has a force constant of 250.0 N/m. Assuming work done by friction to be negligible, find (a) how fast the car is going before it starts up the slope and (b) how fast it is going at the top of the slope.
Strategy
The spring force and the gravitational force are conservative forces, so conservation of mechanical energy can be used. Thus,
\[KE_i + PE_i = KE_f + PE_f\] or
\[\dfrac{1}{2} mv_i^2 + mgh_i + \dfrac{1}{2}kx_i^2 = \dfrac{1}{2}mv_f^2 + mgh_f + \dfrac{1}{2}kx_f^2 \nonumber\]
where \(h\) is the height (vertical position) and \(x\) is the compression of the spring. This general statement looks complex but becomes much simpler when we start considering specific situations. First, we must identify the initial and final conditions in a problem; then, we enter them into the last equation to solve for an unknown.
Solution for (a)
This part of the problem is limited to conditions just before the car is released and just after it leaves the spring. Take the initial height to be zero, so that both \(h_i\) and \(h_f\) are zero. Furthermore, the initial speed \(v_i\) is zero and the final compression of the spring \(x_f\) is zero, and so several terms in the conservation of mechanical energy equation are zero and it simplifies to
\[\dfrac{1}{2}kx_i^2 = \dfrac{1}{2}mv_f^2. \nonumber\]
In other words, the initial potential energy in the spring is converted completely to kinetic energy in the absence of friction. Solving for the final speed and entering known values yields
\[ \begin{align*} v_f &= \sqrt{\dfrac{k}{m}}x_i \\[5pt] &= \sqrt{\dfrac{250N/m}{0.100kg}}(0.0400\space m) \\[5pt] &= 2.00 m/s \end{align*}\]
Solution for (b)
One method of finding the speed at the top of the slope is to consider conditions just before the car is released and just after it reaches the top of the slope, completely ignoring everything in between. Doing the same type of analysis to find which terms are zero, the conservation of mechanical energy becomes
\[\dfrac{1}{2}kx_i^2 = \dfrac{1}{2}mv_f^2 + mgh_f. \nonumber\]
This form of the equation means that the spring’s initial potential energy is converted partly to gravitational potential energy and partly to kinetic energy. The final speed at the top of the slope will be less than at the bottom. Solving for \(v_f\) and substituting known values gives
\[ \begin{align*} v_f = \sqrt{\dfrac{kx^2}{m} - 2gh_f} \\[5pt] &= \sqrt{\left(\dfrac{250N/m}{0.100kg}\right)(0.0400m)^2 - 2(9.80 m/s^2)(0.180m)} \\[5pt] &= 0.687 \, m/s \end{align*}\]
Discussion
Another way to solve this problem is to realize that the car’s kinetic energy before it goes up the slope is converted partly to potential energy—that is, to take the final conditions in part (a) to be the initial conditions in part (b).
Note that, for conservative forces, we do not directly calculate the work they do; rather, we consider their effects through their corresponding potential energies, just as we did in Example. Note also that we do not consider details of the path taken—only the starting and ending points are important (as long as the path is not impossible). This assumption is usually a tremendous simplification, because the path may be complicated and forces may vary along the way.
PHET Explorations: Energy Skate Park
Learn about conservation of energy with a skater dude! Build tracks, ramps and jumps for the skater and view the kinetic energy, potential energy and friction as he moves. You can also take the skater to different planets or even space!
Summary
- A conservative force is one for which work depends only on the starting and ending points of a motion, not on the path taken.
- We can define potential energy \((PE\) for any conservative force, just as we defined \(PE_g\) for the gravitational force.
- The potential energy of a spring is \(PE_s = \frac{1}{2}kx^2\), where \(k\) is the spring’s force constant and |(x\) is the displacement from its undeformed position.
- Mechanical energy is defined to be \(KE = PE\) for conservative force.
- When only conservative forces act on and within a system, the total mechanical energy is constant. In equation form, \[ KE + PE = \, constant \] or \[KE_i +PE_i = KE_f + PE_f \] where i and f denote initial and final values. This is known as the conservation of mechanical energy.
Glossary
- conservative force
- a force that does the same work for any given initial and final configuration, regardless of the path followed
- potential energy
- energy due to position, shape, or configuration
- potential energy of a spring
- the stored energy of a spring as a function of its displacement; when Hooke’s law applies, it is given by the expression \(\frac{1}{2}kx^2\) where \(x\) is the distance the spring is compressed or extended and \(k\) is the spring constant
- conservation of mechanical energy
- the rule that the sum of the kinetic energies and potential energies remains constant if only conservative forces act on and within a system
- mechanical energy
- the sum of kinetic energy and potential energy