1. Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. Chemistry: The Central Science, Eleventh Edition By Theodore E. Brown, H. Eugene LeMay, Bruce E. Bursten, and Catherine J. Murphy With contributions from Patrick Woodward Sample Exercise 8.1 Identifying Conjugate Acids and Bases (a) What is the conjugate base of each of the following acids: HClO 4, H 2 S, PH 4 +, HCO 3 – ? (b) What is the conjugate acid of each of the following bases: CN –, SO 4 2–, H 2 O, HCO 3 – ? Write the formula for the conjugate acid of each of the following: HSO 3 –, F –, PO 4 3–, CO. Answers: H 2 SO 3, HF, HPO 4 2–, HCO + Practice Exercise Solution Analyze: We are asked to give the conjugate base for each of a series of species and to give the conjugate acid for each of another series of species. Plan: The conjugate base of a substance is simply the parent substance minus one proton, and the conjugate acid of a substance is the parent substance plus one proton. Solve: (a) HClO 4 less one proton (H + ) is ClO 4 –. The other conjugate bases are HS –, PH 3, and CO 3 2–. (b) CN – plus one proton (H + ) is HCN. The other conjugate acids are HSO 4 –, H 3 O +, and H 2 CO 3. Notice that the hydrogen carbonate ion (HCO 3 – ) is amphiprotic. It can act as either an acid or a base.

2. Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. Chemistry: The Central Science, Eleventh Edition By Theodore E. Brown, H. Eugene LeMay, Bruce E. Bursten, and Catherine J. Murphy With contributions from Patrick Woodward Sample Exercise 8.2 Writing Equations for Proton-Transfer Reactions The hydrogen sulfite ion (HSO 3 – ) is amphiprotic. (a) Write an equation for the reaction of HSO 3 – with water, in which the ion acts as an acid. (b) Write an equation for the reaction of HSO 3 – with water, in which the ion acts as a base. In both cases identify the conjugate acid–base pairs. When lithium oxide (Li 2 O) is dissolved in water, the solution turns basic from the reaction of the oxide ion (O 2 – ) with water. Write the reaction that occurs, and identify the conjugate acid–base pairs. Answer: O 2– (aq) + H 2 O(l) → OH – (aq) + OH – (aq). OH – is the conjugate acid of the base O 2–. OH – is also the conjugate base of the acid H 2 O. Practice Exercise Solution Analyze and Plan: We are asked to write two equations representing reactions between HSO 3 – and water, one in which HSO 3 – should donate a proton to water, thereby acting as a Brønsted–Lowry acid, and one in which HSO 3 – should accept a proton from water, thereby acting as a base. We are also asked to identify the conjugate pairs in each equation. Solve: The conjugate pairs in this equation are HSO 3 – (acid) and SO 3 2– (conjugate base); and H 2 O (base) and H 3 O + (conjugate acid). The conjugate pairs in this equation are H 2 O (acid) and OH – (conjugate base), and HSO 3 – (base) and H 2 SO 3 (conjugate acid).

3. Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. Chemistry: The Central Science, Eleventh Edition By Theodore E. Brown, H. Eugene LeMay, Bruce E. Bursten, and Catherine J. Murphy With contributions from Patrick Woodward Sample Exercise 8.3 Predicting the Position of a Proton-Transfer Equilibrium For the following proton-transfer reaction, use Figure 16.4 to predict whether the equilibrium lies predominantly to the left (that is, K c 1): Solution Analyze: We are asked to predict whether the equilibrium shown lies to the right, favoring products, or to the left, favoring reactants. Plan: This is a proton-transfer reaction, and the position of the equilibrium will favor the proton going to the stronger of two bases. The two bases in the equation are CO 3 2–, the base in the forward reaction as written, and SO 4 2– the conjugate base of HSO 4 –. We can find the relative positions of these two bases in Figure 8.1 to determine which is the stronger base. Solve: CO 3 2– appears lower in the right-hand column in Figure 16.4 and is therefore a stronger base than SO 4 2–. CO 3 2–, therefore, will get the proton preferentially to become HCO 3 –, while SO 4 2– will remain mostly unprotonated. The resulting equilibrium will lie to the right, favoring products (that is, K c > 1 ). Comment: Of the two acids in the equation, HSO 4 – and HCO 3 –, the stronger one gives up a proton more readily while the weaker one tends to retain its proton. Thus, the equilibrium favors the direction in which the proton moves from the stronger acid and becomes bonded to the stronger base. For each of the following reactions, use Figure 16.4 to predict whether the equilibrium lies predominantly to the left or to the right: Answers: (a) left, (b) right Practice Exercise

4. Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. Chemistry: The Central Science, Eleventh Edition By Theodore E. Brown, H. Eugene LeMay, Bruce E. Bursten, and Catherine J. Murphy With contributions from Patrick Woodward Sample Exercise 8.4 Calculating [H + ] for Pure Water Calculate the values of [H + ] and [OH - ] in a neutral solution at 25 ºC.Indicate whether solutions with each of the following ion concentrations are neutral, acidic, or basic: (a) [H + ] = 4 × 10 –9 M ; (b) [H + ] = 4 × 10 –9 M ; (c) [OH – ] = 7 × 10 –13 M. Answers: (a) basic, (b) neutral, (c) acidic Practice Exercise Solution Analyze: We are asked to determine the concentrations of H + and OH – ions in a neutral solution at 25 ºC. Plan: We will use Equation 16.16 and the fact that, by definition, [H + ] = [OH – ] in a neutral solution. Solve: We will represent the concentration of [H + ] and [OH – ] in neutral solution with x. This gives In an acid solution [H + ] is greater than ; 1.0 × 10 –7 M in a basic solution [H + ] is less than 1.0 × 10 –7 M.

5. Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. Chemistry: The Central Science, Eleventh Edition By Theodore E. Brown, H. Eugene LeMay, Bruce E. Bursten, and Catherine J. Murphy With contributions from Patrick Woodward Sample Exercise 8.5 Calculating [H + ] from [OH - ] Calculate the concentration of H + (aq) in (a) a solution in which [OH – ] is 0.010 M, (b) a solution in which [OH – ] is 1.8 ×10 –9 M. Note: In this problem and all that follow, we assume, unless stated otherwise, that the temperature is 25 ºC. Calculate the concentration of OH – (aq) in a solution in which (a) [H + ] = 2 × 10 –6 M; (b) [H + ] = [OH – ]; (c) [H + ] = 100× [OH – ]. Answers: (a) 5 × 10 –9 M, (b) 1.0 × 10 –7 M, (c) 1.0 × 10 –8 M Practice Exercise Solution Analyze: We are asked to calculate the hydronium ion concentration in an aqueous solution where the hydroxide concentration is known. Plan: We can use the equilibrium-constant expression for the autoionization of water and the value of K w to solve for each unknown concentration. Solve: a) Using Equation 16.16, we have: This solution is basic because (b) In this instance This solution is acidic because

6. Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. Chemistry: The Central Science, Eleventh Edition By Theodore E. Brown, H. Eugene LeMay, Bruce E. Bursten, and Catherine J. Murphy With contributions from Patrick Woodward Sample Exercise 8.6 Calculating pH from [H + ] Calculate the pH values for the two solutions described in Sample Exercise 8.5. (a) In a sample of lemon juice [H + ] is 3.8 × 10 –4 M. What is the pH? (b) A commonly available window- cleaning solution has [OH – ] = 1.9 ×10 –6 M. What is the pH? Answers: (a) 3.42, (b) [H + ] = 5.3 ×10 –9 M, so pH = 8.28 Practice Exercise Solution Analyze: We are asked to determine the pH of aqueous solutions for which we have already calculated [H + ]. Plan: We can calculate pH using its defining equation, Equation 16.17. Solve: (a) In the first instance we found [H+]. to be 1.0 ×10 –12 M. Because 1.0 ×10 –12 has two significant figures, the pH has two decimal places, 12.00. (b) For the second solution, [H + ] = 5.6 × 10 –6 M. Before performing the calculation, it is helpful to estimate the pH. To do so, we note that [H + ] lies between 1 × 10 –6 and 1 × 10 –5 Thus, we expect the pH to lie between 6.0 and 5.0. We use Equation 16.17 to calculate the pH. Check: After calculating a pH, it is useful to compare it to your prior estimate. In this case the pH, as we predicted, falls between 6 and 5. Had the calculated pH and the estimate not agreed, we should have reconsidered our calculation or estimate or both.

7. Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. Chemistry: The Central Science, Eleventh Edition By Theodore E. Brown, H. Eugene LeMay, Bruce E. Bursten, and Catherine J. Murphy With contributions from Patrick Woodward Sample Exercise 8.7 Calculating [H + ] from pH A sample of freshly pressed apple juice has a pH of 3.76. Calculate [H + ]. A solution formed by dissolving an antacid tablet has a pH of 9.18. Calculate [H + ]. Answer: [H + ] = 6.6× 10 –10 M Practice Exercise Solution Analyze: We need to calculate [H + ] from pH. Plan: We will use Equation 16.17, pH = –log[H + ], for the calculation. Solve: From Equation 16.17, we have Thus, To find [H + ], we need to determine the antilog of –3.76. Scientific calculators have an antilog function (sometimes labeled INV log or 10 x ) that allows us to perform the calculation: Comment: Consult the user’s manual for your calculator to find out how to perform the antilog operation. The number of significant figures in [H + ] is two because the number of decimal places in the pH is two. Check: Because the pH is between 3.0 and 4.0, we know that [H + ] will be between 1 × 10 –3 and 1 × 10 –4 M. Our calculated [H + ] falls within this estimated range.

8. Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. Chemistry: The Central Science, Eleventh Edition By Theodore E. Brown, H. Eugene LeMay, Bruce E. Bursten, and Catherine J. Murphy With contributions from Patrick Woodward Sample Exercise 8.8 Calculating the pH of a Strong Acid What is the pH of a 0.040 M solution of HClO 4 ? An aqueous solution of HNO 3 has a pH of 2.34. What is the concentration of the acid? Answer: 0.0046 M Practice Exercise Solution Analyze and Plan: Because HClO 4 is a strong acid, it is completely ionized, giving [H + ] = [ClO 4 - ] = 0.040 M Solve: The pH of the solution is given by pH = –log(0.040) = 1.40. Check: Because [H + ] lies between 1 × 10 –2 and 1 × 10 –1, the pH will be between 2.0 and 1.0. Our calculated pH falls within the estimated range. Furthermore, because the concentration has two significant figures, the pH has two decimal places.

9. Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. Chemistry: The Central Science, Eleventh Edition By Theodore E. Brown, H. Eugene LeMay, Bruce E. Bursten, and Catherine J. Murphy With contributions from Patrick Woodward Sample Exercise 8.9 Calculating the pH of a Strong Base What is the pH of (a) a 0.028 M solution of NaOH, (b) a 0.0011 M solution of Ca(OH) 2 ? What is the concentration of a solution of (a) KOH for which the pH is 11.89; (b) Ca(OH) 2 for which the pH is 11.68? Answers: (a) 7.8 × 10 –3 M, (b) 2.4 ×10 –3 M Practice Exercise Solution Analyze: We are asked to calculate the pH of two solutions of strong bases. Plan: We can calculate each pH by either of two equivalent methods. First, we could use Equation 16.16 to calculate [H + ] and then use Equation 16.17 to calculate the pH. Alternatively, we could use [OH – ] to calculate pOH and then use Equation 16.20 to calculate the pH. Solve: (a) NaOH dissociates in water to give one OH – ion per formula unit. Therefore, the OH – concentration for the solution in (a) equals the stated concentration of NaOH, namely 0.028 M. (b) Ca(OH) 2 is a strong base that dissociates in water to give two OH – ions per formula unit. Thus, the concentration of OH – (aq) for the solution in part (b) is 2 × (0.0011 M) = 0.0022 M

10. Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. Chemistry: The Central Science, Eleventh Edition By Theodore E. Brown, H. Eugene LeMay, Bruce E. Bursten, and Catherine J. Murphy With contributions from Patrick Woodward Sample Exercise 8.10 Calculating K a from Measured pH A student prepared a 0.10 M solution of formic acid (HCOOH) and measured its pH. The pH at 25 ºC was found to be 2.38. Calculate K a for formic acid at this temperature. Solution Analyze: We are given the molar concentration of an aqueous solution of weak acid and the pH of the solution, and we are asked to determine the value of K a for the acid. Plan: Although we are dealing specifically with the ionization of a weak acid, this problem is very similar to the equilibrium problems we encountered in Chapter 15. We can solve this problem using the method first outlined in Sample Exercise 15.9, starting with the chemical reaction and a tabulation of initial and equilibrium concentrations. Solve: The first step in solving any equilibrium problem is to write the equation for the equilibrium reaction. The ionization of formic acid can be written as follows: The equilibrium-constant expression is From the measured pH, we can calculate [H + ]: We can do a little accounting to determine the concentrations of the species involved in the equilibrium. We imagine that the solution is initially 0.10 M in HCOOH molecules. We then consider the ionization of the acid into H + and HCOO –. For each HCOOH molecule that ionizes, one H + ion and one ion HCOO – are produced in solution. Because the pH measurement indicates that [H + ] = 4.2 × 10 –3 M at equilibrium, we can construct the following table:

11. Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. Chemistry: The Central Science, Eleventh Edition By Theodore E. Brown, H. Eugene LeMay, Bruce E. Bursten, and Catherine J. Murphy With contributions from Patrick Woodward Sample Exercise 8.10 Calculating K a from Measured pH Solution (Continued) Check: The magnitude of our answer is reasonable because K a for a weak acid is usually between 10 –3 and 10 –10. Notice that we have neglected the very small concentration of H + (aq) that is due to the autoionization of H 2 O. Notice also that the amount of HCOOH that ionizes is very small compared with the initial concentration of the acid. To the number of significant figures we are using, the subtraction yields 0.10 M: We can now insert the equilibrium centrations into the expression for K a : Niacin, one of the B vitamins, has the following molecular structure: A 0.020 M solution of niacin has a pH of 3.26. What is the acid-dissociation constant, K a, for niacin? Answers: 1.5 × 10 –5 Practice Exercise

12. Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. Chemistry: The Central Science, Eleventh Edition By Theodore E. Brown, H. Eugene LeMay, Bruce E. Bursten, and Catherine J. Murphy With contributions from Patrick Woodward Sample Exercise 8.11 Calculating Percent Ionization A 0.10 M solution of formic acid (HCOOH) contains 4.2 × 10 –3 M H + (aq). Calculate the percentage of the acid that is ionized. A 0.020 M solution of niacin has a pH of 3.26. Calculate the percent ionization of the niacin. Answer: 2.7% Practice Exercise Solution Analyze: We are given the molar concentration of an aqueous solution of weak acid and the equilibrium concentration of H + (aq) and asked to determine the percent ionization of the acid. Plan: The percent ionization is given by Equation 16.27. Solve:

13. Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. Chemistry: The Central Science, Eleventh Edition By Theodore E. Brown, H. Eugene LeMay, Bruce E. Bursten, and Catherine J. Murphy With contributions from Patrick Woodward Sample Exercise 8.12 Using K a to Calculate pH Calculate the pH of a 0.20 M solution of HCN. (Ka for HCN is 4.9 × 10 –10 ) Solution Analyze: We are given the molarity of a weak acid and are asked for the pH. From Table 16.2, Ka for HCN is 4.9 × 10 –10. Plan: We proceed as in the example just worked in the text, writing the chemical equation and constructing a table of initial and equilibrium concentrations in which the equilibrium concentration of H + is our unknown. Solve: Writing both the chemical equation for the ionization reaction that forms H + (aq) and the equilibrium-constant (K a ) expression for the reaction: Next, we tabulate the concentration of the species involved in the equilibrium reaction, letting x = [H + ] at equilibrium: Substituting the equilibrium concentrations from the table into the equilibrium-constant expression yields

14. Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. Chemistry: The Central Science, Eleventh Edition By Theodore E. Brown, H. Eugene LeMay, Bruce E. Bursten, and Catherine J. Murphy With contributions from Patrick Woodward Sample Exercise 8.12 Using K a to Calculate pH The Ka for niacin (Practice Exercise 16.10) is 1.5 × 10 -5. What is the pH of a 0.010 M solution of niacin? Answer: 3.41 Practice Exercise Solution (Continued) We next make the simplifying approximation that x, the amount of acid that dissociates, is small compared with the initial concentration of acid; that is, Thus, Solving for x, we have A concentration of 9.9 × 10 -6 M is much smaller than 5% of 0.20, the initial HCN concentration. Our simplifying approximation is therefore appropriate. We now calculate the pH of the solution:

15. Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. Chemistry: The Central Science, Eleventh Edition By Theodore E. Brown, H. Eugene LeMay, Bruce E. Bursten, and Catherine J. Murphy With contributions from Patrick Woodward Sample Exercise 8.13 Using K a to Calculate Percent Ionization Calculate the percentage of HF molecules ionized in (a) a 0.10 M HF solution, (b) a 0.010 M HF solution. Solution Analyze: We are asked to calculate the percent ionization of two HF solutions of different concentration. From Appendix D, we find K a = 6.8 × 10 -4. Plan: We approach this problem as we would previous equilibrium problems. We begin by writing the chemical equation for the equilibrium and tabulating the known and unknown concentrations of all species. We then substitute the equilibrium concentrations into the equilibrium-constant expression and solve for the unknown concentration, that of H +. Solve: (a) The equilibrium reaction and equilibrium concentrations are as follows: The equilibrium-constant expression is When we try solving this equation using the approximation 0.10 – x = 0.10 (that is, by neglecting the concentration of acid that ionizes in comparison with the initial concentration), we obtain

16. Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. Chemistry: The Central Science, Eleventh Edition By Theodore E. Brown, H. Eugene LeMay, Bruce E. Bursten, and Catherine J. Murphy With contributions from Patrick Woodward Sample Exercise 8.13 Using K a to Calculate Percent Ionization Solution (Continued) Because this value is greater than 5% of 0.10 M, we should work the problem without the approximation, using an equation-solving calculator or the quadratic formula. Rearranging our equation and writing it in standard quadratic form, we have This equation can be solved using the standard quadratic formula. Substituting the appropriate numbers gives Of the two solutions, only the one that gives a positive value for x is chemically reasonable. Thus, From our result, we can calculate the percent of molecules ionized: (b) Proceeding similarly for the 0.010 M solution, we have Solving the resultant quadratic expression, we obtain The percentage of molecules ionized is

17. Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. Chemistry: The Central Science, Eleventh Edition By Theodore E. Brown, H. Eugene LeMay, Bruce E. Bursten, and Catherine J. Murphy With contributions from Patrick Woodward Sample Exercise 8.13 Using K a to Calculate Percent Ionization Solution (Continued) Comment: Notice that if we do not use the quadratic formula to solve the problem properly, we calculate 8.2% ionization for (a) and 26% ionization for (b). Notice also that in diluting the solution by a factor of 10, the percentage of molecules ionized increases by a factor of 3. This result is in accord with what we see in Figure 16.9. It is also what we would expect from Le Châtelier’s principle. (Section 15.7) There are more “particles” or reaction components on the right side of the equation than on the left. Dilution causes the reaction to shift in the direction of the larger number of particles because this counters the effect of the decreasing concentration of particles. In Practice Exercise 8.11, we found that the percent ionization of niacin (K a = 1.5 × 10 -5 ) in a 0.020 M solution is 2.7%. Calculate the percentage of niacin molecules ionized in a solution that is (a) 0.010 M, (b) 1.0 × 10 -3 M. Answers: (a) 3.9%, (b) 12% Practice Exercise

18. Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. Chemistry: The Central Science, Eleventh Edition By Theodore E. Brown, H. Eugene LeMay, Bruce E. Bursten, and Catherine J. Murphy With contributions from Patrick Woodward Sample Exercise 8.14 Calculating the pH of a Polyprotic Acid Solution The solubility of CO 2 in pure water at 25 ºC and 0.1 atm pressure is 0.0037 M. The common practice is to assume that all of the dissolved CO 2 is in the form of carbonic acid (H 2 CO 3 ), which is produced by reaction between the CO 2 and H 2 O: What is the pH of a 0.0037 M solution of H 2 CO 3 ? Solution Analyze: We are asked to determine the pH of a 0.0037 M solution of a polyprotic acid. Plan: H 2 CO 3 is a diprotic acid; the two acid-dissociation constants, K a1 and K a2 (Table 16.3), differ by more than a factor of 10 3. Consequently, the pH can be determined by considering only K a1, thereby treating the acid as if it were a monoprotic acid. Solve: Proceeding as in Sample Exercises 16.12 and 16.13, we can write the equilibrium reaction and equilibrium concentrations as follows: The equilibrium-constant expression is as follows: Solving this equation using an equation-solving calculator, we get Alternatively, because K a1 is small, we can make the simplifying approximation that x is small, so that

19. Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. Chemistry: The Central Science, Eleventh Edition By Theodore E. Brown, H. Eugene LeMay, Bruce E. Bursten, and Catherine J. Murphy With contributions from Patrick Woodward Sample Exercise 8.14 Calculating the pH of a Polyprotic Acid Solution Solution (Continued) Thus, Solving for x, we have The small value of x indicates that our simplifying assumption was justified. The pH is therefore Comment: If we were asked to solve for [CO 3 2- ], we would need to use K a2. Let’s illustrate that calculation. Using the values of [HCO 3 – ] and [H + ] calculated above, and setting [CO 3 2– ] = y, we have the following initial and equilibrium concentration values: Assuming that y is small compared to 4.0 × 10 –5, we have The value calculated for y is indeed very small compared to 4.0 × 10 -5, showing that our assumption was justified. It also shows that the ionization of HCO 3– is negligible compared to that of H 2 CO 3, as far as production of H + is concerned. However, it is the only source of CO 3 2–, which has a very low concentration in the solution. Our calculations thus tell us that in a solution of carbon dioxide in water, most of the CO 2 is in the form of CO 2 or H 2 CO 3, a small fraction ionizes to form H + and HCO 3–, and an even smaller fraction ionizes to give CO 3 2–. Notice also that [CO 3 2– ] is numerically equal to K a2.

20. Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. Chemistry: The Central Science, Eleventh Edition By Theodore E. Brown, H. Eugene LeMay, Bruce E. Bursten, and Catherine J. Murphy With contributions from Patrick Woodward Sample Exercise 8.14 Calculating the pH of a Polyprotic Acid Solution (a) Calculate the pH of a 0.020 M solution of oxalic acid (H 2 C 2 O 4 ). (See Table 16.3 for K a1 and K a2.) (b) Calculate the concentration of oxalate ion [C 2 O 4 2– ], in this solution. Answers: (a) pH = 1.80, (b) [C 2 O 4 2– ] = 6.4 × 10 –5 M Practice Exercise

21. Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. Chemistry: The Central Science, Eleventh Edition By Theodore E. Brown, H. Eugene LeMay, Bruce E. Bursten, and Catherine J. Murphy With contributions from Patrick Woodward Sample Exercise 8.15 Using K b to Calculate OH¯ Calculate the concentration of OH – in a 0.15 M solution of NH 3. Solution Analyze: We are given the concentration of a weak base and are asked to determine the concentration of OH –. Plan: We will use essentially the same procedure here as used in solving problems involving the ionization of weak acids; that is, we write the chemical equation and tabulate initial and equilibrium concentrations. Solve:We first write the ionization reaction and the corresponding equilibrium-constant (K b ) expression: We then tabulate the equilibrium concentrations involved in the equilibrium: (We ignore the concentration of H 2 O because it is not involved in the equilibrium-constant expression.) Inserting these quantities into the equilibrium-constant expression gives the following: Because K b is small, we can neglect the small amount of NH 3 that reacts with water, as compared to the total NH 3 concentration; that is, we can neglect x relative to 0.15 M. Then we have

22. Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. Chemistry: The Central Science, Eleventh Edition By Theodore E. Brown, H. Eugene LeMay, Bruce E. Bursten, and Catherine J. Murphy With contributions from Patrick Woodward Sample Exercise 8.15 Using K b to Calculate OH¯ Solution (Continued) Check: The value obtained for x is only about 1% of the NH 3 concentration, 0.15 M. Therefore, neglecting x relative to 0.15 was justified. Comment: You may be asked to find the pH of a solution of a weak base. Once you have found [OH – ], you can proceed as in Sample Exercise 16.9, where we calculated the pH of a strong base. In the present sample exercise, we have seen that the 0.15 M solution of NH 3 contains [OH – ] = 1.6 × 10 -3 M. Thus, pOH = –log(1.6 × 10 -3 ) = 2.80, and pH = 14.00 – 2.80 = 11.20. The pH of the solution is above 7 because we are dealing with a solution of a base. Which of the following compounds should produce the highest pH as a 0.05 M solution: pyridine, methylamine, or nitrous acid? Answer: methylamine (because it has the largest K b value of the two amine bases in the list) Practice Exercise

23. Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. Chemistry: The Central Science, Eleventh Edition By Theodore E. Brown, H. Eugene LeMay, Bruce E. Bursten, and Catherine J. Murphy With contributions from Patrick Woodward Sample Exercise 8.16 Using pH to Determine the Concentration of a Salt A solution made by adding solid sodium hypochlorite (NaClO) to enough water to make 2.00 L of solution has a pH of 10.50. Using the information K b (ClO – ) = 3.3 × 10 –7,calculate the number of moles of NaClO that were added to the water. Solution Analyze: We are given the pH of a 2.00-L solution of NaClO and must calculate the number of moles of NaClO needed to raise the pH to 10.50. NaClO is an ionic compound consisting of Na + and ClO – ions. As such, it is a strong electrolyte that completely dissociates in solution into Na +, which is a spectator ion, and ClO – ion, which is a weak base with K b = 3.3 × 10 –7. Plan: From the pH, we can determine the equilibrium concentration of OH –. We can then construct a table of initial and equilibrium concentrations in which the initial concentration of ClO – is our unknown. We can calculate [ClO – ] using the equilibrium constant expression, K b. Solve: We can calculate [OH – ] : This concentration is high enough that we can assume that Equation 16.37 is the only source of OH – ; that is, we can neglect any OH – produced by the autoionization of H 2 O. We now assume a value of x for the initial concentration of ClO – and solve the equilibrium problem in the usual way.

24. Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. Chemistry: The Central Science, Eleventh Edition By Theodore E. Brown, H. Eugene LeMay, Bruce E. Bursten, and Catherine J. Murphy With contributions from Patrick Woodward Sample Exercise 8.16 Using pH to Determine the Concentration of a Salt A solution of NH 3 in water has a pH of 11.17. What is the molarity of the solution? Answer: 0.12 M Practice Exercise Solution (Continued) We now use the expression for the base- dissociation constant to solve for x: Thus We say that the solution is 0.31 M in NaClO even though some of the ClO – ions have reacted with water. Because the solution is 0.31 M in NaClO and the total volume of solution is 2.00 L, 0.62 mol of NaClO is the amount of the salt that was added to the water.

25. Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. Chemistry: The Central Science, Eleventh Edition By Theodore E. Brown, H. Eugene LeMay, Bruce E. Bursten, and Catherine J. Murphy With contributions from Patrick Woodward Sample Exercise 8.17 Calculating K a or K b for a Conjugate Acid-Base Pair Calculate (a) the base-dissociation constant, K b, for the fluoride ion (F – ); (b) the acid dissociation constant, K a, for the ammonium ion (NH 4 + ). K a for the acid, HF, is K a = 6.8 × 10 -4, K b for NH 3 is K b = 1.8 × 10 -5 Solution Analyze: We are asked to determine dissociation constants for F –, the conjugate base of HF, and NH 4 +, the conjugate acid of NH 3. Plan: Although neither F – nor NH 4 + appears in the tables, we can find the tabulated values for ionization constants for HF and NH 3, and use the relationship between K a and K b to calculate the ionization constants for each of the conjugates. Solve: (a) K a for the weak acid, HF, is K a = 6.8 × 10 -4. We can calculate K b for the conjugate base, F – : (b) K b for NH 3 is K b = 1.8 × 10 -5., We can calculate K a for the conjugate acid, NH 4 + :

26. Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. Chemistry: The Central Science, Eleventh Edition By Theodore E. Brown, H. Eugene LeMay, Bruce E. Bursten, and Catherine J. Murphy With contributions from Patrick Woodward Sample Exercise 8.17 Calculating K a or K b for a Conjugate Acid-Base Pair (a) Which of the following anions has the largest base-dissociation constant: NO 2 –, PO 4 3–, or N 3 – ? (b) The base quinoline has the following structure: Its conjugate acid is listed in handbooks as having a pK a of 4.90. What is the base dissociation constant for quinoline? Answers: (a) PO 4 3– (K b = 2.4 × 10 -2 ), (b) 7.9 × 10 -10 Practice Exercise

27. Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. Chemistry: The Central Science, Eleventh Edition By Theodore E. Brown, H. Eugene LeMay, Bruce E. Bursten, and Catherine J. Murphy With contributions from Patrick Woodward Sample Exercise 8.18 Determining Whether Salt Solutions Are Acidic, Basic, or Neutral Determine whether aqueous solutions of each of the following salts will be acidic, basic, or neutral: (a) Ba(CH 3 COO) 2, (b) NH 4 Cl, (c) CH 3 NH 3 Br, (d) KNO 3, (e) Al(ClO 4 ) 3. Solution Analyze: We are given the chemical formulas of five ionic compounds (salts) and asked whether their aqueous solutions will be acidic, basic, or neutral. Plan: We can determine whether a solution of a salt is acidic, basic, or neutral by identifying the ions in solution and by assessing how each ion will affect the pH. Solve: (a) This solution contains barium ions and acetate ions. The cation, Ba 2+, is an ion of one of the heavy alkaline earth metals and will therefore not affect the pH (summary point 4). The anion, CH 3 COO –, is the conjugate base of the weak acid CH 3 COOH and will hydrolyze to produce OH – ions, thereby making the solution basic (summary point 2). (b) This solution contains NH 4 + and Cl – ions. NH 4 + is the conjugate acid of a weak base (NH 3 ) and is therefore acidic (summary point 3). Cl – is the conjugate base of a strong acid (HCl) and therefore has no influence on the pH of the solution (summary point 1). Because the solution contains an ion that is acidic (NH 4 + ) and one that has no influence on pH (Cl – ), the solution of NH 4 Cl will be acidic. (c) This solution contains CH 3 NH 3 + and Br – ions. CH 3 NH 3 + is the conjugate acid of a weak base (CH 3 NH 2, an amine) and is therefore acidic (summary point 3). Br – is the conjugate base of a strong acid (HBr) and is therefore pH-neutral (summary point 1). Because the solution contains one ion that is acidic and one that is neutral, the solution of CH 3 NH 3 Br will be acidic. (d) This solution contains the K + ion, which is a cation of group 1A, and the ion NO 3 –, which is the conjugate base of the strong acid HNO 3. Neither of the ions will react with water to any appreciable extent (summary points 1 and 4), making the solution neutral.

28. Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. Chemistry: The Central Science, Eleventh Edition By Theodore E. Brown, H. Eugene LeMay, Bruce E. Bursten, and Catherine J. Murphy With contributions from Patrick Woodward Sample Exercise 8.18 Determining Whether Salt Solutions Are Acidic, Basic, or Neutral In each of the following, indicate which salt in each of the following pairs will form the more acidic (or less basic) 0.010 M solution: (a) NaNO 3, or Fe(NO 3 ) 3 ; (b) KBr, or KBrO; (c) CH 3 NH 3 Cl, or BaCl 2, (d) NH 4 NO 2, or NH 4 NO 3. Answers: (a) Fe(NO 3 ) 3, (b) KBr, (c) CH 3 NH 3 Cl, (d) NH 4 NO3 Practice Exercise Solution (Continued) (e) This solution contains Al 3+ and ClO 4 – ions. Cations, such as Al 3+, that are not in groups 1A or 2A are acidic (summary point 5). The ClO 4 – ion is the conjugate base of a strong acid (HClO 4 ) and therefore does not affect pH (summary point 1). Thus, the solution of Al(ClO 4 ) 3 will be acidic.

29. Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. Chemistry: The Central Science, Eleventh Edition By Theodore E. Brown, H. Eugene LeMay, Bruce E. Bursten, and Catherine J. Murphy With contributions from Patrick Woodward Sample Exercise 8.19 Predicting Whether the Solution of an Amphiprotic Anion is Acidic or Basic Predict whether the salt Na 2 HPO 4 will form an acidic solution or a basic solution on dissolving in water. Ka (HPO 4 2- /PO 4 3- ) = 4.2 × 10 -13, Ka (H 2 PO 4 - /HPO 4 2- ) = 6.2 × 10 -8 Solution Analyze: We are asked to predict whether a solution of Na 2 HPO 4 will be acidic or basic. This substance is an ionic compound composed of Na + and HPO 4 2– ions. Plan: We need to evaluate each ion, predicting whether each is acidic or basic. Because Na + is a cation of group 1A, we know that it has no influence on pH. It is merely a spectator ion in acid–base chemistry. Thus, our analysis of whether the solution is acidic or basic must focus on the behavior of the HPO 4 2– ion. We need to consider the fact that HPO 4 2– can act as either an acid or a base. The reaction with the larger equilibrium constant will determine whether the solution is acidic or basic Solve: The value of K a for Equation 16.45,, is 4.2 × 10 -13. We must calculate the value of K b for Equation 16.46 from the value of K a for its conjugate acid, H 2 PO 4 –. We make use of the relationship shown in Equation : We want to know K b for the base HPO 4 2–, knowing the value of K a for the conjugate acid HPO 4 2– : Because K a for H 2 PO 4 – is 6.2 × 10 -8, we calculate K b for H 2 PO 4 2– to be 1.6 × 10 -7. This is more than 10 5 times larger than K a for H 2 PO 4 2– ; thus, the reaction shown in Equation 16.46 predominates over that in Equation 16.45, and the solution will be basic.

30. Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. Chemistry: The Central Science, Eleventh Edition By Theodore E. Brown, H. Eugene LeMay, Bruce E. Bursten, and Catherine J. Murphy With contributions from Patrick Woodward Sample Exercise 8.19 Predicting Whether the Solution of an Amphiprotic Anion is Acidic or Basic Predict whether the dipotassium salt of citric acid (K 2 HC 6 H 5 O 7 ) will form an acidic or basic solution in water (see Table 16.3 for data). Answer: acidic Practice Exercise

31. Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. Chemistry: The Central Science, Eleventh Edition By Theodore E. Brown, H. Eugene LeMay, Bruce E. Bursten, and Catherine J. Murphy With contributions from Patrick Woodward Sample Exercise 8.20 Calculating the pH When a Common Ion is Involved What is the pH of a solution made by adding 0.30 mol of acetic acid and 0.30 mol of sodium acetate to enough water to make 1.0 L of solution? K a (CH 3 COOH/CH 3 COO – ) = 1.8 10 -5 Solution Analyze: We are asked to determine the pH of a solution of a weak electrolyte (CH 3 COOH) and a strong electrolyte (CH 3 COONa) that share a common ion, CH 3 COO –. Plan: In any problem in which we must determine the pH of a solution containing a mixture of solutes, it is helpful to proceed by a series of logical steps: 1. Consider which solutes are strong electrolytes and which are weak electrolytes, and identify the major species in solution. 2. Identify the important equilibrium that is the source of H + and therefore determines pH. 3. Tabulate the concentrations of ions involved in the equilibrium. 4. Use the equilibrium-constant expression to calculate [H + ] and then pH. Solve: First, because CH 3 COOH is a weak electrolyte and CH 3 COONa is a strong electrolyte, the major species in the solution are CH 3 COOH (a weak acid), Na + (which is neither acidic nor basic and is therefore a spectator in the acid–base chemistry), and CH 3 COO – (which is the conjugate base of CH 3 COOH). Second, [H + ] and, therefore, the pH are controlled by the dissociation equilibrium of CH 3 COOH: (We have written the equilibrium Using H + (aq) rather than H 3 O + (aq) but both representations of the hydrated hydrogen ion are equally valid.)

32. Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. Chemistry: The Central Science, Eleventh Edition By Theodore E. Brown, H. Eugene LeMay, Bruce E. Bursten, and Catherine J. Murphy With contributions from Patrick Woodward Sample Exercise 8.20 Calculating the pH When a Common Ion is Involved Solution (Continued) Third, we tabulate the initial and equilibrium concentrations as we did in solving other equilibrium problems in Chapters 15 and 16: The equilibrium concentration of CH 3 COO – (the common ion) is the initial concentration that is due to CH 3 COONa (0.30 M) plus the change in concentration (x) that is due to the ionization of CH 3 COOH. Now we can use the equilibrium- constant expression: (The dissociation constant for CH 3 COOH at 25 ºC is from Appendix D; addition of CH 3 COONa does not change the value of this constant.) Substituting the equilibrium-constant concentrations from our table into the equilibrium expression gives

33. Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. Chemistry: The Central Science, Eleventh Edition By Theodore E. Brown, H. Eugene LeMay, Bruce E. Bursten, and Catherine J. Murphy With contributions from Patrick Woodward Sample Exercise 8.20 Calculating the pH When a Common Ion is Involved Calculate the pH of a solution containing 0.085 M nitrous acid (HNO 2 ; K a = 4.5 × 10 -4 ) and 0.10 M potassium nitrite (KNO 2 ). Answer: 3.42 Practice Exercise Solution (Continued) Because K a is small, we assume that x is small compared to the original concentrations of CH 3 COOH and CH 3 COO – (0.30 M each). Thus, we can ignore the very small x relative to 0.30 M, giving The resulting value of x is indeed small relative to 0.30, justifying the approximation made in simplifying the problem. The resulting value of x is indeed small relative to 0.30, justifying the approximation made in simplifying the problem. Finally, we calculate the pH from the equilibrium concentration of H + (aq): Comment: In Section 16.6 we calculated that a 0.30 M solution of CH 3 COOH has a pH of 2.64, corresponding to H + ] = 2.3 × 10 -3 M. Thus, the addition of CH 3 COONa has substantially decreased, [H + ] as we would expect from Le Châtelier’s principle.

34. Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. Chemistry: The Central Science, Eleventh Edition By Theodore E. Brown, H. Eugene LeMay, Bruce E. Bursten, and Catherine J. Murphy With contributions from Patrick Woodward Sample Exercise 8.21 Calculating Ion Concentrations When a Common is Involved Calculate the fluoride ion concentration and pH of a solution that is 0.20 M in HF and 0.10 M in HCl. K a (HF/F – ) = 6.8 10 -4 Solution Plan: We can again use the four steps outlined in Sample Exercise 8.20. Solve: Because HF is a weak acid and HCl is a strong acid, the major species in solution are HF, H+, and Cl –. The Cl –, which is the conjugate base of a strong acid, is merely a spectator ion in any acid–base chemistry. The problem asks for [F – ], which is formed by ionization of HF. Thus, the important equilibrium is The common ion in this problem is the hydrogen (or hydronium) ion. Now we can tabulate the initial and equilibrium concentrations of each species involved in this equilibrium: The equilibrium constant for the ionization of HF, from Appendix D, is 6.8 × 10 -4. Substituting the equilibrium-constant concentrations into the equilibrium expression gives

35. Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. Chemistry: The Central Science, Eleventh Edition By Theodore E. Brown, H. Eugene LeMay, Bruce E. Bursten, and Catherine J. Murphy With contributions from Patrick Woodward Sample Exercise 8.21 Calculating Ion Concentrations When a Common is Involved Calculate the formate ion concentration and pH of a solution that is 0.050 M in formic acid (HCOOH; K a = 1.8 × 10 -4 ) and 0.10 M in HNO 3. Answer: [HCOO – ] = 9.0 × 10 -5 ; pH = 1.00 Practice Exercise Comment: Notice that for all practical purposes, [H + ] is due entirely to the HCl; the HF makes a negligible contribution by comparison. Solution (Continued) If we assume that x is small relative to 0.10 or 0.20 M, this expression simplifies to This F – concentration is substantially smaller than it would be in a 0.20 M solution of HF with no added HCl. The common ion, H +, suppresses the ionization of HF. The concentration of H + (aq) is Thus,

36. Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. Chemistry: The Central Science, Eleventh Edition By Theodore E. Brown, H. Eugene LeMay, Bruce E. Bursten, and Catherine J. Murphy With contributions from Patrick Woodward Sample Exercise 8.22 Calculating the pH of a Buffer What is the pH of a buffer that is 0.12 M in lactic acid [CH 3 CH(OH)COOH, or HC 3 H 5 O 3 ] and 0.10 M in sodium lactate [CH 3 CH(OH)COONa or NaC 3 H 5 O 3 ]? For lactic acid, K a = 1.4 × 10 -4. Solution Analyze: We are asked to calculate the pH of a buffer containing lactic acid HC 3 H 5 O 3 and its conjugate base, the lactate ion (C 3 H 5 O 3 – ). Plan: We will first determine the pH using the method described in Section 17.1. Because HC 3 H 5 O 3 is a weak electrolyte and NaC 3 H 5 O 3 is a strong electrolyte, the major species in solution are HC 3 H 5 O 3, Na +, and C 3 H 5 O 3 –. The Na + ion is a spectator ion. The HC 3 H 5 O 3 –C 3 H 5 O 3 – conjugate acid–base pair determines [H + ] and thus pH; [H + ] can be determined using the aciddissociation equilibrium of lactic acid. Solve: The initial and equilibrium concentrations of the species involved in this equilibrium are The equilibrium concentrations are governed by the equilibrium expression:

37. Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. Chemistry: The Central Science, Eleventh Edition By Theodore E. Brown, H. Eugene LeMay, Bruce E. Bursten, and Catherine J. Murphy With contributions from Patrick Woodward Sample Exercise 8.22 Calculating the pH of a Buffer Calculate the pH of a buffer composed of 0.12 M benzoic acid and 0.20 M sodium benzoate. (Refer to Appendix D.) Answer: 4.42 Practice Exercise Solution (Continued) Because K a is small and a common ion is present, we expect x to be small relative to either 0.12 or 0.10 M. Thus, our equation can be simplified to give Solving for x gives a value that justifies our approximation: Alternatively, we could have used the Henderson–Hasselbalch equation to calculate pH directly:

38. Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. Chemistry: The Central Science, Eleventh Edition By Theodore E. Brown, H. Eugene LeMay, Bruce E. Bursten, and Catherine J. Murphy With contributions from Patrick Woodward Sample Exercise 8.23 Preparing a Buffer How many moles of NH 4 Cl must be added to 2.0 L of 0.10 M NH 3 to form a buffer whose pH is 9.00? (Assume that the addition of NH 4 Cl does not change the volume of the solution.) K b (NH 3 /NH 4 + ) = 1.8 10 -5 Solution Analyze: Here we are asked to determine the amount of NH4 + ion required to prepare a buffer of a specific pH. pH = pK a + log [NH 3 ] [NH 4 + ] pK b = 4.74, pK a = pK w – pK b = 14 – 4.74 = 9.255 [NH 3 ] = 0.1 M 9 = 9.255 + log 0.1 [NH 4 + ] [NH 4 + ] = 0.18 M

39. Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. Chemistry: The Central Science, Eleventh Edition By Theodore E. Brown, H. Eugene LeMay, Bruce E. Bursten, and Catherine J. Murphy With contributions from Patrick Woodward Sample Exercise 8.23 Preparing a Buffer Calculate the concentration of sodium benzoate that must be present in a 0.20 M solution of benzoic acid (C 6 H 5 COOH) to produce a pH of 4.00. Answer: 0.13 M Practice Exercise Solution (Continued) Comment: Because NH 4 + and NH 3 are a conjugate acid–base pair, we could use the Henderson– Hasselbalch equation (Equation 17.9) to solve this problem. To do so requires first using Equation 16.41 to calculate pK a for NH 4 + from the value of pK b for NH 3. We suggest you try this approach to convince yourself that you can use the Henderson–Hasselbalch equation for buffers for which you are given K b for the conjugate base rather than K a for the conjugate acid. [NH 4 + ] must equal 0.18 M. The number of moles of NH 4 Cl needed to produce this concentration is given by the product of the volume of the solution and its molarity:

40. Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. Chemistry: The Central Science, Eleventh Edition By Theodore E. Brown, H. Eugene LeMay, Bruce E. Bursten, and Catherine J. Murphy With contributions from Patrick Woodward Sample Exercise 8.24 Calculating pH Changes in Buffers A buffer is made by adding 0.300 mol CH 3 COOH and 0.300 mol CH 3 COONa to enough water to make 1.00 L of solution. The pH of the buffer is 4.74 (Sample Exercise 17.1). (a) Calculate the pH of this solution after 0.020 mol of NaOH is added. (b) For comparison, calculate the pH that would result if 0.020 mol of NaOH were added to 1.00 L of pure water (neglect any volume changes). Solution Analyze: We are asked to determine the pH of a buffer after addition of a small amount of strong base and to compare the pH change to the pH that would result if we were to add the same amount of strong base to pure water. Plan: (a) Solving this problem involves the two steps outlined in Figure 17.3. Thus, we must first do a stoichiometry calculation to determine how the added OH – reacts with the buffer and affects its composition. Then we can use the resultant composition of the buffer and either the Henderson–Hasselbalch equation or the equilibriumconstant expression for the buffer to determine the pH. Solve: Stoichiometry Calculation: The OH – provided by NaOH reacts with CH 3 COOH, the weak acid component of the buffer. Prior to this neutralization reaction, there are 0.300 mol each of CH 3 COOH and CH 3 COO –. Neutralizing the 0.020 mol OH – requires 0.020 mol of CH 3 COOH. Consequently, the amount of CH 3 COOH decreases by 0.020 mol, and the amount of the product of the neutralization, CH 3 COO –, increases by 0.020 mol. We can create a table to see how the composition of the buffer changes as a result of its reaction with OH – :

41. Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. Chemistry: The Central Science, Eleventh Edition By Theodore E. Brown, H. Eugene LeMay, Bruce E. Bursten, and Catherine J. Murphy With contributions from Patrick Woodward Sample Exercise 8.24 Calculating pH Changes in Buffers Determine (a) the pH of the original buffer described in Sample Exercise 17.5 after the addition of 0.020 mol HCl and (b) the pH of the solution that would result from the addition of 0.020 mol HCl to 1.00 L of pure water Answers: (a) 4.68, (b) 1.70 Practice Exercise Solution (Continued) Equilibrium Calculation: We now turn our attention to the equilibrium that will determine the pH of the buffer, namely the ionization of acetic acid. Using the quantities of CH 3 COOH and CH 3 COO – remaining in the buffer, we can determine the pH using the Henderson–Hasselbalch equation. Comment Notice that we could have used mole amounts in place of concentrations in the Henderson– Hasselbalch equation and gotten the same result. The volumes of the acid and base are equal and cancel. If 0.020 mol of H + was added to the buffer, we would proceed in a similar way to calculate the resulting pH of the buffer. In this case the pH decreases by 0.06 units, giving pH = 4.68, as shown in the figure in the margin. (b) To determine the pH of a solution made by adding 0.020 mol of NaOH to 1.00 L of pure water, we can first determine pOH using Equation 16.18 and subtracting from 14. Note that although the small amount of NaOH changes the pH of water significantly, the pH of the buffer changes very little.

42. Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. Chemistry: The Central Science, Eleventh Edition By Theodore E. Brown, H. Eugene LeMay, Bruce E. Bursten, and Catherine J. Murphy With contributions from Patrick Woodward Sample Exercise 8.25 Writing Solubility-Product (K sp ) Expressions Write the expression for the solubility-product constant for CaF 2, and look up the corresponding K sp value in Appendix D. Solution Analyze: We are asked to write an equilibrium-constant expression for the process by which CaF 2 dissolves in water. Plan: We apply the same rules for writing any equilibrium-constant expression, excluding the solid reactant from the expression. We assume that the compound dissociates completely into its component ions. Solve: Following the italicized rule stated previously, the expression for is In Appendix D we see that this K sp has a value of 3.9 × 10 -11. Give the solubility-product-constant expressions and the values of the solubility-product constants (from Appendix D) for the following compounds: (a) barium carbonate, (b) silver sulfate. Practice Exercise

43. Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. Chemistry: The Central Science, Eleventh Edition By Theodore E. Brown, H. Eugene LeMay, Bruce E. Bursten, and Catherine J. Murphy With contributions from Patrick Woodward Sample Exercise 8.26 Calculating K sp from Solubility Solid silver chromate is added to pure water at 25 ºC. Some of the solid remains undissolved at the bottom of the flask. The mixture is stirred for several days to ensure that equilibrium is achieved between the undissolved Ag 2 CrO 4 (s) and the solution. Analysis of the equilibrated solution shows that its silver ion concentration is 1.3 × 10 -4 M. Assuming that Ag 2 CrO 4 dissociates completely in water and that there are no other important equilibria involving the Ag + or CrO 4 2– ions in the solution, calculate K sp for this compound. Solution Analyze: We are given the equilibrium concentration of Ag + in a saturated solution of Ag 2 CrO 4. From this information, we are asked to determine the value of the solubility product constant, K sp, for Ag 2 CrO 4. Plan: The equilibrium equation and the expression for K sp are To calculate K sp, we need the equilibrium concentrations of Ag + and CrO 4 2–. We know that at equilibrium [Ag + ] = 1.3 × 10 -4 M. All the Ag + and CrO 4 2– ions in the solution come from the Ag 2 CrO 4 that dissolves. Thus, we can use [Ag + ] to calculate [CrO 4 2– ]. Solve: From the chemical formula of silver chromate, we know that there must be 2 Ag + ions in solution for each CrO 4 2– ion in solution. Consequently, the concentration Of CrO 4 2– is half the concentration of Ag + : We can now calculate the value of K sp.

44. Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. Chemistry: The Central Science, Eleventh Edition By Theodore E. Brown, H. Eugene LeMay, Bruce E. Bursten, and Catherine J. Murphy With contributions from Patrick Woodward Sample Exercise 8.26 Calculating K sp from Solubility A saturated solution of Mg(OH) 2 in contact with undissolved solid is prepared at 25 ºC. The pH of the solution is found to be 10.17. Assuming that Mg(OH) 2 dissociates completely in water and that there are no other simultaneous equilibria involving the Mg 2+ or OH – ions in the solution, calculate K sp for this compound. Answer: 1.6 ×10 -12 Practice Exercise Solution (Continued) Check: We obtain a small value, as expected for a slightly soluble salt. Furthermore, the calculated value agrees well with the one given in Appendix D, 1.2 × 10 -12.

45. Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. Chemistry: The Central Science, Eleventh Edition By Theodore E. Brown, H. Eugene LeMay, Bruce E. Bursten, and Catherine J. Murphy With contributions from Patrick Woodward Sample Exercise 8.27 Calculating Solubility from K sp The K sp for CaF 2 is 3.9 ×10 -11 at 25 ºC. Assuming that CaF 2 dissociates completely upon dissolving and that there are no other important equilibria affecting its solubility, calculate the solubility of CaF 2 in grams per liter. Solution Analyze: We are given K sp for CaF 2 and are asked to determine solubility. Recall that the solubility of a substance is the quantity that can dissolve in solvent, whereas the solubility-product constant, K sp, is an equilibrium constant. Plan: We can approach this problem by using our standard techniques for solving equilibrium problems. We write the chemical equation for the dissolution process and set up a table of the initial and equilibrium concentrations. We then use the equilibrium constant expression. In this case we know K sp, and so we solve for the concentrations of the ions in solution. Solve: Assume initially that none of the salt has dissolved, and then allow x moles/liter of CaF 2 to dissociate completely when equilibrium is achieved. The stoichiometry of the equilibrium dictates that 2x moles/liter of F – are produced for each x moles/liter of CaF 2 that dissolve. We now use the expression for K sp and substitute the equilibrium concentrations to solve for the value of x:

46. Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. Chemistry: The Central Science, Eleventh Edition By Theodore E. Brown, H. Eugene LeMay, Bruce E. Bursten, and Catherine J. Murphy With contributions from Patrick Woodward Sample Exercise 8.27 Calculating Solubility from K sp The K sp for LaF 3 is 2 × 10 -19. What is the solubility of LaF 3 in water in moles per liter? Answer: 9 × 10 -6 mol/L Practice Exercise Check: We expect a small number for the solubility of a slightly soluble salt. If we reverse the calculation, we should be able to recalculate the solubility product: K sp = (2.1 × 10 -4 )(4.2 × 10 -4 ) 2 = 3.7 × 10 -11, close to the starting value for K sp, 3.9 × 10 -11, Comment: Because F - is the anion of a weak acid, you might expect that the hydrolysis of the ion would affect the solubility of CaF 2. The basicity of F – is so small (K b = 1.5 × 10 -11 ), however, that the hydrolysis occurs to only a slight extent and does not significantly influence the solubility. The reported solubility is 0.017 g/L at 25 ºC, in good agreement with our calculation Solution (Remember that to calculate the cube root of a number, you can use the yx function on your calculator,with x =.) Thus, the molar solubility of CaF 2 is 2.1 × 10 -4 mol/L. The mass of CaF 2 that dissolves in water to form a liter of solution is

47. Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. Chemistry: The Central Science, Eleventh Edition By Theodore E. Brown, H. Eugene LeMay, Bruce E. Bursten, and Catherine J. Murphy With contributions from Patrick Woodward Sample Exercise 8.28 Calculating the Effect of a Common Ion on Solubility Calculate the molar solubility of CaF 2 at 25 °C in a solution that is (a) 0.010 M in Ca(NO 3 ) 2, (b) 0.010 M in NaF. Solve: (a) In this instance the initial concentration of Ca 2+ is 0.010 M because of the dissolved Ca(NO 3 ) 2 : Substituting into the solubility-product expression gives Solution Analyze: We are asked to determine the solubility of CaF 2 in the presence of two strong electrolytes, each of which contains an ion common to CaF 2. In (a) the common ion is Ca 2+, and NO 3 – is a spectator ion. In (b) the common ion is F –, and Na + is a spectator ion. Plan: Because the slightly soluble compound is CaF 2, we need to use the K sp for this compound, which is available in Appendix D: The value of K sp is unchanged by the presence of additional solutes. Because of the common-ion effect, however, the solubility of the salt will decrease in the presence of common ions. We can again use our standard equilibrium techniques of starting with the equation for CaF 2 dissolution, setting up a table of initial and equilibrium concentrations, and using the K sp expression to determine the concentration of the ion that comes only from CaF 2.

48. Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. Chemistry: The Central Science, Eleventh Edition By Theodore E. Brown, H. Eugene LeMay, Bruce E. Bursten, and Catherine J. Murphy With contributions from Patrick Woodward Sample Exercise 8.28 Calculating the Effect of a Common Ion on Solubility Solution (Continued) This would be a messy problem to solve exactly, but fortunately it is possible to simplify matters greatly. Even without the common-ion effect, the solubility of CaF 2 is very small (2.1 × 10 -4 M). Thus, we assume that the 0.010 M concentration of Ca 2+ fromCa(NO 3 ) 2 is very much greater than the small additional concentration resulting from the solubility of CaF 2 ; that is, x is small compared to 0.010 M, and 0.010 + x 0.010. We then have The very small value for x validates the simplifying assumption we have made. Our calculation indicates that 3.1 × 10 -5 mol of solid CaF 2 dissolves per liter of the 0.010 M Ca(NO 3 ) 2 solution. (b) In this case the common ion is F –, and at equilibrium we have Assuming that 2x is small compared to 0.010 M (that is, 0.010 + 2x 0.010), we have Thus, 3.9 × 10 -7 mol of solid CaF 2 should dissolve per liter of 0.010 M NaF solution.

49. Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. Chemistry: The Central Science, Eleventh Edition By Theodore E. Brown, H. Eugene LeMay, Bruce E. Bursten, and Catherine J. Murphy With contributions from Patrick Woodward Sample Exercise 8.28 Calculating the Effect of a Common Ion on Solubility The value for K sp for manganese(II) hydroxide, Mn(OH) 2, is 1.6 ×10 -13. Calculate the molar solubility of Mn(OH) 2 in a solution that contains 0.020 M NaOH. Answer: 4.0 × 10 -10 M Practice Exercise Solution (Continued) Comment: The molar solubility of CaF 2 in pure water is 2.1 × 10 -4 M (Sample Exercise 17.11). By comparison, our calculations above show that the solubility of CaF 2 in the presence of 0.010 M Ca 2+ is 3.1 × 10 -5 M, and in the presence of 0.010 M F – ion it is 3.9 × 10 -7 M. Thus, the addition of either Ca 2+ or F – to a solution of CaF 2 decreases the solubility. However, the effect of F - on the solubility is more pronounced than that of Ca 2+ because [F – ] appears to the second power in the K sp expression for CaF 2, whereas Ca 2+ appears to the first power.

50. Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. Chemistry: The Central Science, Eleventh Edition By Theodore E. Brown, H. Eugene LeMay, Bruce E. Bursten, and Catherine J. Murphy With contributions from Patrick Woodward Sample Exercise 8.29 Predicting the Effect of Acid on Solubility Which of the following substances will be more soluble in acidic solution than inbasic solution: (a) Ni(OH) 2 (s), (b) CaCO 3 (s), (c) BaF2(s), (d) AgCl(s)? Solution Analyze: The problem lists four sparingly soluble salts, and we are asked to determine which will be more soluble at low pH than at high pH. Plan: Ionic compounds that dissociate to produce a basic anion will be more soluble in acid solution. The reaction between CO 3 2– and H + occurs in a stepwise fashion, first forming HCO 3 –. H 2 CO 3 forms in appreciable amounts only when the concentration of H + is sufficiently high.

51. Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. Chemistry: The Central Science, Eleventh Edition By Theodore E. Brown, H. Eugene LeMay, Bruce E. Bursten, and Catherine J. Murphy With contributions from Patrick Woodward Sample Exercise 8.29 Predicting the Effect of Acid on Solubility Solution (d) The solubility of AgCl is unaffected by changes in pH because Cl – is the anion of a strong acid and therefore has negligible basicity. Write the net ionic equation for the reaction of the following copper(II) compounds with acid: (a) CuS, (b) Cu(N 3 ) 2. Practice Exercise

52. Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. Chemistry: The Central Science, Eleventh Edition By Theodore E. Brown, H. Eugene LeMay, Bruce E. Bursten, and Catherine J. Murphy With contributions from Patrick Woodward Sample Exercise 8.30 Evaluating an Equilibrium Involving a Complex Ion Calculate the concentration of Ag + present in solution at equilibrium when concentrated ammonia is added to a 0.010 M solution of AgNO 3 to give an equilibrium concentration of [NH 3 ] = 0.20 M. Neglect the small volume change that occurs when NH 3 is added. Solution Analyze: When NH 3 (aq) is added to Ag + (aq), a reaction occurs forming Ag(NH 3 ) 2 + as shown in Equation 17.22. We are asked to determine what concentration of Ag + (aq) will remain uncombined when the NH 3 concentration is brought to 0.20 M in a solution originally 0.010 M in AgNO 3. Plan: We first assume that the AgNO 3 is completely dissociated, giving 0.10 M Ag +. Because K f for the formation of Ag(NH 3 ) 2 + is quite large, we assume that essentially all the Ag + is then converted to Ag(NH 3 ) 2 + and approach the problem as though we are concerned with the dissociation of Ag(NH 3 ) 2 + rather than its formation. To facilitate this approach, we will need to reverse the equation to represent the formation of Ag + and NH 3 from Ag(NH 3 ) 2 + and also make the corresponding change to the equilibrium constant. Solve: If [Ag + ] is 0.010 M initially, then [Ag(NH 3 ) 2 + will be 0.010 M following addition of the NH 3. We now construct a table to solve this equilibrium problem. Note that the NH 3 concentration given in the problem is an equilibrium concentration rather than an initial concentration.

53. Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. Chemistry: The Central Science, Eleventh Edition By Theodore E. Brown, H. Eugene LeMay, Bruce E. Bursten, and Catherine J. Murphy With contributions from Patrick Woodward Sample Exercise 8.30 Evaluating an Equilibrium Involving a Complex Ion Calculate [Cr 3+ ] in equilibrium with Cr(OH) 4 - when 0.010 mol of Cr(NO 3 ) 3 is dissolved in a liter of solution buffered at pH 10.0. Answer: 1× 10 -16 M Practice Exercise Solution (Continued) Because the concentration of Ag + is very small, we can ignore x in comparison with 0.010. Thus, 0.010 – x 0.010 M. Substituting these values into the equilibriumconstantexpression for the dissociation of Ag(NH 3 ) 2 +, we obtain Solving for x, we obtain x = 1.5 × 10 -8 M = [Ag + ]. Thus, formation of the Ag(NH 3 ) 2 + complex drastically reduces the concentration of free Ag + ion in solution.

54. Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. Chemistry: The Central Science, Eleventh Edition By Theodore E. Brown, H. Eugene LeMay, Bruce E. Bursten, and Catherine J. Murphy With contributions from Patrick Woodward Sample Exercise 8.31 Predicting Whether a Precipitate Will Form Will a precipitate of PbSO 4 form when 0.10 L of 8.0 × 10 -3 M Pb(NO 3 ) 2 is added to 0.40 L of 5.0 ×10 -3 M Na 2 SO 4 ? PbSO 4 has a K sp of 6.3 × 10 -7 Solution Analyze: The problem asks us to determine whether a precipitate will form when two salt solutions are combined. Plan: We should determine the concentrations of all ions immediately upon mixing of the solutions and compare the value of the reaction quotient, Q, to the solubility-product constant, K sp, for any potentially insoluble product. The possible metathesis products are PbSO 4 and NaNO 3. Sodium salts are quite soluble; PbSO 4 has a K sp of 6.3 × 10 -7 (Appendix D), however, and will precipitate if the Pb 2+ and SO 4 2– concentrations are high enough for Q to exceed K sp for the salt. Solve: When the two solutions are mixed, the total volume becomes 0.10 L + 0.40 L = 0.50 L. The number of moles of Pb 2+ in 0.10 L of 8.0 × 10 -3 M Pb(NO 3 ) 2 is The concentration of Pb 2+ in the 0.50-L mixture is therefore The number of moles of SO 4 2– in 0.40 L of 5.0 × 10 -3 MNa 2 SO 4 is Therefore, [SO 4 2– ] in the 0.50-L mixture is We then have Because Q > K sp, PbSO 4 will precipitate.

55. Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. Chemistry: The Central Science, Eleventh Edition By Theodore E. Brown, H. Eugene LeMay, Bruce E. Bursten, and Catherine J. Murphy With contributions from Patrick Woodward Sample Exercise 8.31 Predicting Whether a Precipitate Will Form Will a precipitate form when 0.050 L of 2.0 ×10 -2 M NaF is mixed with 0.010 L of 1.0 × 10 -2 M Ca(NO 3 ) 2 ? Answer: Yes, CaF 2 precipitates because Q = 4.6 × 10 -8 is larger than K sp = 3.9 ×10 -11 Practice Exercise

56. Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. Chemistry: The Central Science, Eleventh Edition By Theodore E. Brown, H. Eugene LeMay, Bruce E. Bursten, and Catherine J. Murphy With contributions from Patrick Woodward Sample Exercise 8.32 Calculating Ion Concentrations for Precipitation A solution contains 1.0 × 10 -2 M Ag + and 2.0 ×10 -2 M Pb 2+. When Cl – is added to the solution, both AgCl (K sp = 1.8× 10 -10 ) and PbCl 2 (K sp = 1.7×10 -5 ) precipitate from the solution. What concentration of Cl – is necessary to begin the precipitation of each salt? Which salt precipitates first? Solution Analyze: We are asked to determine the concentration of Cl – necessary to begin the precipitation from a solution containing Ag + and Pb 2+, and to predict which metal chloride will begin to precipitate first. Plan: We are given K sp values for the two possible precipitates. Using these and the metal ion concentrations, we can calculate what concentration of Cl – ion would be necessary to begin precipitation of each. The salt requiring the lower Cl – ion concentration will precipitate first. Solve: For AgCl we have Because [Ag + ] = 1.0 × 10 -2 M, the greatest concentration of Cl – that can be present without causing precipitation of AgCl can be calculated from the K sp expression Any Cl – in excess of this very small concentration will cause AgCl to precipitate from solution. Proceeding similarly for PbCl 2, we have

57. Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. Chemistry: The Central Science, Eleventh Edition By Theodore E. Brown, H. Eugene LeMay, Bruce E. Bursten, and Catherine J. Murphy With contributions from Patrick Woodward Sample Exercise 8.32 Calculating Ion Concentrations for Precipitation Asolution consists of 0.050 M Mg 2+ and 0.020 M Cu 2+. Which ion will precipitate first as OH – is added to the solution? What concentration of OH – is necessary to begin the precipitation of each cation? [K sp = 1.8 × 10 -11 for Mg(OH) 2, and K sp = 4.8 ×10 -20 for Cu(OH) 2.] Answer: Cu(OH) 2 precipitates first. Cu(OH) 2 begins to precipitate when [OH – ] exceeds 1.5 × 10 -9 M; Mg(OH) 2 begins to precipitate when [OH – ] exceeds 1.9 × 10 -5 M. Practice Exercise Solution (Continued) Thus, a concentration of Cl – in excess of 2.9 × 10 -2 M will cause PbCl 2 to precipitate. Comparing the concentrations of Cl – required to precipitate each salt, we see that as Cl – is added to the solution, AgCl will precipitate first because it requires a much smaller concentration of Cl –. Thus, Ag + can be separated from by slowly adding Cl – so [Cl – ] is between 1.8 × 10 -8 M and 2.9 × 10 -2 M.