Sample Exercise 8.1 Identifying Conjugate Acids and Bases

1. 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) HClO4 less one proton (H+) is ClO4–. The other conjugate bases are HS–, PH3, and CO32–. (b) CN– plus one proton (H+) is HCN. The other conjugate acids are HSO4–, H3O+, and H2CO3. Notice that the hydrogen carbonate ion (HCO3–) is amphiprotic. It can act as either an acid or a base. Practice Exercise Write the formula for the conjugate acid of each of the following: HSO3–, F– , PO43–, CO. Answers: H2SO3, HF, HPO42–, HCO+ Sample Exercise 8.1 Identifying Conjugate Acids and Bases (a) What is the conjugate base of each of the following acids: HClO4, H2S, PH4+, HCO3–? (b) What is the conjugate acid of each of the following bases: CN–, SO42–, H2O, HCO3– ?

2. Solution Analyze and Plan: We are asked to write two equations representing reactions between HSO3– and water, one in which HSO3– should donate a proton to water, thereby acting as a Brønsted–Lowry acid, and one in which HSO3– 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 HSO3– (acid) and SO32– (conjugate base); and H2O (base) and H3O+ (conjugate acid). The conjugate pairs in this equation are H2O (acid) and OH– (conjugate base), and HSO3– (base) and H2SO3 (conjugate acid). Practice Exercise When lithium oxide (Li2O) is dissolved in water, the solution turns basic from the reaction of the oxide ion (O2–) with water. Write the reaction that occurs, and identify the conjugate acid–base pairs. Answer: O2–(aq) + H2O(l) → OH–(aq) + OH–(aq). OH– is the conjugate acid of the base O2–. OH– is also the conjugate base of the acid H2O. Sample Exercise 8.2 Writing Equations for Proton-Transfer Reactions The hydrogen sulfite ion (HSO3–) is amphiprotic. (a) Write an equation for the reaction of HSO3– with water, in which the ion acts as an acid. (b) Write an equation for the reaction of HSO3– with water, in which the ion acts as a base. In both cases identify the conjugate acid–base pairs.

3. For the following proton-transfer reaction, use Figure 16.4 to predict whether the equilibrium lies predominantly to the left (that is, Kc < 1 ) or to the right (Kc > 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 CO32–, the base in the forward reaction as written, and SO42– the conjugate base of HSO4–. We can find the relative positions of these two bases in Figure 8.1 to determine which is the stronger base. Solve: CO32–appears lower in the right-hand column in Figure 16.4 and is therefore a stronger base than SO42–. CO32–, therefore, will get the proton preferentially to become HCO3–, while SO42– will remain mostly unprotonated. The resulting equilibrium will lie to the right, favoring products (that is, Kc > 1 ). Comment: Of the two acids in the equation, HSO4– and HCO3–, 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. Practice Exercise 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 Sample Exercise 8.3 Predicting the Position of a Proton-Transfer Equilibrium

4. 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 × 10–7M. Practice Exercise Indicate whether solutions with each of the following ion concentrations are neutral, acidic, or basic: (a) [H+] = 4 × 10–9 M ; (b) [H+] = 4 × 10–9M ; (c) [OH–] = 7 × 10–13M. Answers: (a) basic, (b) neutral, (c) acidic Sample Exercise 8.4 Calculating [H+] for Pure Water Calculate the values of [H+] and [OH-] in a neutral solution at 25 ºC.

5. 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 Kwto solve for each unknown concentration. Practice Exercise Solve: a) Using Equation 16.16, we have: This solution is basic because (b) In this instance This solution is acidic because Calculate the concentration of OH–(aq) in a solution in which (a) [H+] = 2 × 10–6M; (b) [H+] = [OH–]; (c) [H+] = 100× [OH–]. Answers: (a) 5 × 10–9M, (b) 1.0 × 10–7M, (c) 1.0 × 10–8 M 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.

6. 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 × 10–6M. 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. Practice Exercise • (a) In a sample of lemon juice [H+] is 3.8 × 10–4M. 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 Sample Exercise 8.6 Calculating pH from [H+] Calculate the pH values for the two solutions described in Sample Exercise 8.5.

7. 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 10x) that allows us to perform the calculation: Solution Analyze: We need to calculate [H+] from pH. Plan: We will use Equation 16.17, pH = –log[H+], for the calculation. Practice Exercise A solution formed by dissolving an antacid tablet has a pH of 9.18. Calculate [H+] . Answer: [H+] = 6.6× 10–10M 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 × 10–4M. Our calculated [H+] falls within this estimated range. Sample Exercise 8.7 Calculating [H+] from pH A sample of freshly pressed apple juice has a pH of 3.76. Calculate [H+].

8. Solution Analyze and Plan: Because HClO4 is a strong acid, it is completely ionized, giving [H+] = [ClO4-] = 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. Practice Exercise An aqueous solution of HNO3 has a pH of 2.34. What is the concentration of the acid? Answer: 0.0046 M Sample Exercise 8.8 Calculating the pH of a Strong Acid What is the pH of a 0.040 M solution of HClO4?

9. 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 Practice Exercise 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 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?

10. 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 Kafor 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 × 10–3M at equilibrium, we can construct the following table: Sample Exercise 8.10 Calculating Kafrom 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 Ka for formic acid at this temperature.

11. Notice that we have neglected the very small concentration of H+(aq) that is due to the autoionization of H2O. 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 Ka: Solution (Continued) Practice Exercise 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, Ka, for niacin? Answers: 1.5 × 10–5 Check: The magnitude of our answer is reasonable because Kafor a weak acid is usually between 10–3 and 10–10 . Sample Exercise 8.10 Calculating Kafrom Measured pH

12. 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: Practice Exercise A 0.020 M solution of niacin has a pH of 3.26. Calculate the percent ionization of the niacin. Answer: 2.7% 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.

13. 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 (Ka) 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 Sample Exercise 8.12 Using Ka to Calculate pH Calculate the pH of a 0.20 M solution of HCN. (Ka for HCN is 4.9 × 10–10)

14. 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: Practice Exercise 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 Sample Exercise 8.12 Using Ka to Calculate pH

15. Solution Analyze: We are asked to calculate the percent ionization of two HF solutions of different concentration. From Appendix D, we find Ka = 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 Sample Exercise 8.13 Using Ka 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.

16. 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 Sample Exercise 8.13 Using Ka to Calculate Percent Ionization

17. 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. Solution (Continued) Practice Exercise In Practice Exercise 8.11, we found that the percent ionization of niacin (Ka = 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 × 10-3M. Answers: (a) 3.9%, (b) 12% Sample Exercise 8.13 Using Ka to Calculate Percent Ionization

18. The solubility of CO2 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 CO2 is in the form of carbonic acid (H2CO3), which is produced by reaction between the CO2 and H2O: What is the pH of a 0.0037 M solution of H2CO3? Solution Analyze: We are asked to determine the pH of a 0.0037 M solution of a polyprotic acid. Plan: H2CO3 is a diprotic acid; the two acid-dissociation constants, Ka1 and Ka2 (Table 16.3), differ by more than a factor of 103. Consequently, the pH can be determined by considering only Ka1, 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 Ka1 is small, we can make the simplifying approximation that x is small, so that Sample Exercise 8.14 Calculating the pH of a Polyprotic Acid Solution

19. 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 [CO32-], we would need to use Ka2. Let’s illustrate that calculation. Using the values of [HCO3–] and [H+] calculated above, and setting [CO32–] = y, we have the following initial and equilibrium concentration values: Assuming that y is small compared to 4.0 × 10–5, we have Solution (Continued) 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 HCO3– is negligible compared to that of H2CO3, as far as production of H+ is concerned. However, it is the only source of CO32–, 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 CO2 is in the form of CO2 or H2CO3, a small fraction ionizes to form H+ and HCO3–, and an even smaller fraction ionizes to give CO32–. Notice also that [CO32–] is numerically equal to Ka2. Sample Exercise 8.14 Calculating the pH of a Polyprotic Acid Solution

20. Practice Exercise (a) Calculate the pH of a 0.020 M solution of oxalic acid (H2C2O4). (See Table 16.3 for Ka1 and Ka2.) (b) Calculate the concentration of oxalate ion [C2O42–] , in this solution. Answers: (a) pH = 1.80, (b) [C2O42–] = 6.4 × 10–5M Sample Exercise 8.14 Calculating the pH of a Polyprotic Acid Solution

21. 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 (Kb) expression: We then tabulate the equilibrium concentrations involved in the equilibrium: (We ignore the concentration of H2O because it is not involved in the equilibrium-constant expression.) Inserting these quantities into the equilibrium-constant expression gives the following: Because Kbis small, we can neglect the small amount of NH3 that reacts with water, as compared to the total NH3 concentration; that is, we can neglect x relative to 0.15 M. Then we have Sample Exercise 8.15 Using Kb to Calculate OH¯ Calculate the concentration of OH– in a 0.15 M solution of NH3.

22. Solution (Continued) Check: The value obtained for x is only about 1% of the NH3 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 NH3 contains [OH–] = 1.6 × 10-3M. 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. Practice Exercise 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 Kbvalue of the two amine bases in the list) Sample Exercise 8.15 Using Kb to Calculate OH¯

23. 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 Kb = 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, Kb. 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 H2O. We now assume a value of x for the initial concentration of ClO– and solve the equilibrium problem in the usual way. 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 Kb (ClO–) = 3.3 × 10–7 ,calculate the number of moles of NaClO that were added to the water.

24. We now use the expression for the base-dissociation constant to solve for x: Thus Solution (Continued) 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. Practice Exercise A solution of NH3 in water has a pH of 11.17. What is the molarity of the solution? Answer: 0.12 M Sample Exercise 8.16 Using pH to Determine the Concentration of a Salt

25. Solution Analyze: We are asked to determine dissociation constants for F–, the conjugate base of HF, and NH4+ , the conjugate acid of NH3. Plan: Although neither F– nor NH4+ appears in the tables, we can find the tabulated values for ionization constants for HF and NH3, and use the relationship between Ka and Kbto calculate the ionization constants for each of the conjugates. Solve: (a) Kafor the weak acid, HF, is Ka = 6.8 × 10-4. We can calculate Kbfor the conjugate base, F–: (b) Kbfor NH3 is Kb = 1.8 × 10-5. , We can calculate Kafor the conjugate acid, NH4+ : Sample Exercise 8.17 Calculating Kaor Kb for a Conjugate Acid-Base Pair Calculate (a) the base-dissociation constant, Kb, for the fluoride ion (F–); (b) the acid dissociation constant, Ka, for the ammonium ion (NH4+). Kafor the acid, HF, is Ka= 6.8 × 10-4,Kbfor NH3 is Kb = 1.8 × 10-5

26. Practice Exercise • (a) Which of the following anions has the largest base-dissociation constant: NO2–, PO43–, or N3–? (b) The base quinoline has the following structure: • Its conjugate acid is listed in handbooks as having a pKaof 4.90. What is the base dissociation • constant for quinoline? • Answers: (a) PO43– (Kb = 2.4 × 10-2), (b) 7.9 × 10-10 Sample Exercise 8.17 Calculating Kaor Kb for a Conjugate Acid-Base Pair

27. 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, Ba2+, is an ion of one of the heavy alkaline earth metals and will therefore not affect the pH (summary point 4). The anion, CH3COO–, is the conjugate base of the weak acid CH3COOH and will hydrolyze to produce OH– ions, thereby making the solution basic (summary point 2). (b) This solution contains NH4+ and Cl– ions. NH4+ is the conjugate acid of a weak base (NH3) 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 (NH4+) and one that has no influence on pH (Cl–), the solution of NH4Cl will be acidic. (c) This solution contains CH3NH3+ and Br– ions. CH3NH3+ is the conjugate acid of a weak base (CH3NH2, 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 CH3NH3Br will be acidic. (d) This solution contains the K+ ion, which is a cation of group 1A, and the ion NO3–, which is the conjugate base of the strong acid HNO3. Neither of the ions will react with water to any appreciable extent (summary points 1 and 4), making the solution neutral. 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(CH3COO)2, (b) NH4Cl, (c) CH3NH3Br, (d) KNO3, (e) Al(ClO4)3.

28. Solution (Continued) (e) This solution contains Al3+ and ClO4– ions. Cations, such as Al3+, that are not in groups 1A or 2A are acidic (summary point 5). The ClO4– ion is the conjugate base of a strong acid (HClO4) and therefore does not affect pH (summary point 1). Thus, the solution of Al(ClO4)3 will be acidic. Practice Exercise 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) NaNO3, or Fe(NO3)3; (b) KBr, or KBrO; (c) CH3NH3Cl, or BaCl2, (d) NH4NO2, or NH4NO3. Answers: (a) Fe(NO3)3, (b) KBr, (c) CH3NH3Cl, (d) NH4NO3 Sample Exercise 8.18 Determining Whether Salt Solutions Are Acidic, Basic, or Neutral

29. Solution Analyze: We are asked to predict whether a solution of Na2HPO4 will be acidic or basic. This substance is an ionic compound composed of Na+ and HPO42– 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 HPO42– ion. We need to consider the fact that HPO42– 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 Kafor Equation 16.45, , is 4.2 × 10-13 . We must calculate the value of Kbfor Equation 16.46 from the value of Kafor its conjugate acid, H2PO4–. We make use of the relationship shown in Equation : We want to know Kbfor the base HPO42–, knowing the value of Kafor the conjugate acid HPO42–: Because Kafor H2PO4– is 6.2 × 10-8 , we calculate Kbfor H2PO42– to be 1.6 × 10-7. This is more than 105 times larger than Kafor H2PO42–; thus, the reaction shown in Equation 16.46 predominates over that in Equation 16.45, and the solution will be basic. Sample Exercise 8.19 Predicting Whether the Solution of an Amphiprotic Anion is Acidic or Basic Predict whether the salt Na2HPO4 will form an acidic solution or a basic solution on dissolving in water. Ka (HPO42-/PO43-) = 4.2 × 10-13 , Ka (H2PO4-/HPO42-) = 6.2 × 10-8

30. Practice Exercise Predict whether the dipotassium salt of citric acid (K2HC6H5O7) will form an acidic or basic solution in water (see Table 16.3 for data). Answer: acidic Sample Exercise 8.19 Predicting Whether the Solution of an Amphiprotic Anion is Acidic or Basic

31. Solution Analyze: We are asked to determine the pH of a solution of a weak electrolyte (CH3COOH) and a strong electrolyte (CH3COONa) that share a common ion, CH3COO–. 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. 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: Solve: First, because CH3COOH is a weak electrolyte and CH3COONa is a strong electrolyte, the major species in the solution are CH3COOH (a weak acid), Na+ (which is neither acidic nor basic and is therefore a spectator in the acid–base chemistry), and CH3COO– (which is the conjugate base of CH3COOH). Second, [H+] and, therefore, the pH are controlled by the dissociation equilibrium of CH3COOH: (We have written the equilibrium Using H+(aq) rather than H3O+(aq) but both representations of the hydrated hydrogen ion are equally valid.) 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? Ka (CH3COOH/CH3COO– ) = 1.8 10-5

32. 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 CH3COO– (the common ion) is the initial concentration that is due to CH3COONa (0.30 M) plus the change in concentration (x) that is due to the ionization of CH3COOH. Now we can use the equilibrium-constant expression: (The dissociation constant for CH3COOH at 25 ºC is from Appendix D; addition of CH3COONa does not change the value of this constant.) Substituting the equilibrium-constant concentrations from our table into the equilibrium expression gives Solution (Continued) Sample Exercise 8.20 Calculating the pH When a Common Ion is Involved

33. Because Kais small, we assume that x is small compared to the original concentrations of CH3COOH and CH3COO– (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): Solution (Continued) Practice Exercise Calculate the pH of a solution containing 0.085 M nitrous acid (HNO2; Ka = 4.5 × 10-4) and 0.10 M potassium nitrite (KNO2). Answer: 3.42 Comment: In Section 16.6 we calculated that a 0.30 M solution of CH3COOH has a pH of 2.64, corresponding to H+] = 2.3 × 10-3M. Thus, the addition of CH3COONa has substantially decreased , [H+] as we would expect from Le Châtelier’s principle. Sample Exercise 8.20 Calculating the pH When a Common Ion is Involved

34. 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 Solution Plan: We can again use the four steps outlined in Sample Exercise 8.20. 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. Ka (HF/F– ) = 6.8 10-4

35. 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, Practice Exercise Calculate the formate ion concentration and pH of a solution that is 0.050 M in formic acid (HCOOH; Ka= 1.8 × 10-4) and 0.10 M in HNO3. Answer: [HCOO–] = 9.0 × 10-5; pH = 1.00 Comment: Notice that for all practical purposes, [H+] is due entirely to the HCl; the HF makes a negligible contribution by comparison. Sample Exercise 8.21 Calculating Ion Concentrations When a Common is Involved

36. Solution Analyze: We are asked to calculate the pH of a buffer containing lactic acid HC3H5O3 and its conjugate base, the lactate ion (C3H5O3–). Plan: We will first determine the pH using the method described in Section 17.1. Because HC3H5O3 is a weak electrolyte and NaC3H5O3 is a strong electrolyte, the major species in solution are HC3H5O3, Na+, and C3H5O3–. The Na+ ion is a spectator ion. The HC3H5O3–C3H5O3– 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: 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 [CH3CH(OH)COOH, or HC3H5O3] and 0.10 M in sodium lactate [CH3CH(OH)COONa or NaC3H5O3]? For lactic acid, Ka = 1.4 × 10-4.

37. Solution (Continued) Because Kais 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: Practice Exercise 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 Sample Exercise 8.22 Calculating the pH of a Buffer

38. Solution Analyze: Here we are asked to determine the amount of NH4+ ion required to prepare a buffer of a specific pH. [NH3] [NH4+] 0.1 [NH4+] pH = pKa + log 9 = 9.255 + log Sample Exercise 8.23 Preparing a Buffer How many moles of NH4Cl must be added to 2.0 L of 0.10 M NH3 to form a buffer whose pH is 9.00? (Assume that the addition of NH4Cl does not change the volume of the solution.) Kb (NH3/NH4+) = 1.8 10-5 pKb = 4.74, pKa = pKw – pKb = 14 – 4.74 = 9.255 [NH3] = 0.1 M [NH4+] = 0.18 M

39. [NH4+] must equal 0.18 M. The number of moles of NH4Cl needed to produce this concentration is given by the product of the volume of the solution and its molarity: Solution (Continued) Comment: Because NH4+ and NH3 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 pKafor NH4+ from the value of pKbfor NH3. We suggest you try this approach to convince yourself that you can use the Henderson–Hasselbalch equation for buffers for which you are given Kbfor the conjugate base rather than Kafor the conjugate acid. Practice Exercise Calculate the concentration of sodium benzoate that must be present in a 0.20 M solution of benzoic acid (C6H5COOH) to produce a pH of 4.00. Answer: 0.13 M Sample Exercise 8.23 Preparing a Buffer

40. 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 CH3COOH, the weak acid component of the buffer. Prior to this neutralization reaction, there are 0.300 mol each of CH3COOH and CH3COO–. Neutralizing the 0.020 mol OH– requires 0.020 mol of CH3COOH. Consequently, the amount of CH3COOH decreases by 0.020 mol, and the amount of the product of the neutralization, CH3COO–, 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–: Sample Exercise 8.24 Calculating pH Changes in Buffers A buffer is made by adding 0.300 mol CH3COOH and 0.300 mol CH3COONa 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).

41. 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 CH3COOH and CH3COO– 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. Practice Exercise 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 Sample Exercise 8.24 Calculating pH Changes in Buffers

42. Solution Analyze: We are asked to write an equilibrium-constant expression for the process by which CaF2 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 Ksp has a value of 3.9 × 10-11. Practice Exercise 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. Sample Exercise 8.25 Writing Solubility-Product (Ksp) Expressions Write the expression for the solubility-product constant for CaF2, and look up the corresponding Kspvalue in Appendix D.

43. Solution Analyze: We are given the equilibrium concentration of Ag+ in a saturated solution of Ag2CrO4. From this information, we are asked to determine the value of the solubility product constant, Ksp, for Ag2CrO4. Plan: The equilibrium equation and the expression for Kspare To calculate Ksp, we need the equilibrium concentrations of Ag+ and CrO42–. We know that at equilibrium [Ag+] = 1.3 × 10-4M. All the Ag+ and CrO42– ions in the solution come from the Ag2CrO4 that dissolves. Thus, we can use [Ag+] to calculate [CrO42–]. Solve: From the chemical formula of silver chromate, we know that there must be 2 Ag+ ions in solution for each CrO42– ion in solution. Consequently, the concentration Of CrO42– is half the concentration of Ag+: We can now calculate the value of Ksp. Sample Exercise 8.26 Calculating Ksp 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 Ag2CrO4(s) and the solution. Analysis of the equilibrated solution shows that its silver ion concentration is 1.3 × 10-4M. Assuming that Ag2CrO4 dissociates completely in water and that there are no other important equilibria involving the Ag+ or CrO42– ions in the solution, calculate Ksp for this compound.

44. 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. Practice Exercise 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 Mg2+ or OH– ions in the solution, calculate Ksp for this compound. Answer: 1.6 ×10-12 Sample Exercise 8.26 Calculating Kspfrom Solubility

45. Solution Analyze: We are given Kspfor CaF2 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, Ksp, 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 Ksp, 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 CaF2 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 CaF2 that dissolve. We now use the expression for Kspand substitute the equilibrium concentrations to solve for the value of x: Sample Exercise 8.27 Calculating Solubility from Ksp The Kspfor CaF2 is 3.9 ×10-11 at 25 ºC. Assuming that CaF2 dissociates completely upon dissolving and that there are no other important equilibria affecting its solubility, calculate the solubility of CaF2 in grams per liter.

46. 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 CaF2 is 2.1 × 10-4 mol/L. The mass of CaF2 that dissolves in water to form a liter of solution is Practice Exercise The Kspfor LaF3 is 2 × 10-19. What is the solubility of LaF3 in water in moles per liter? Answer: 9 × 10-6 mol/L Sample Exercise 8.27 Calculating Solubility from Ksp 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: Ksp = (2.1 × 10-4)(4.2 × 10-4)2 = 3.7 × 10-11 , close to the starting value for Ksp,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 CaF2. The basicity of F– is so small (Kb = 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

47. Solution Analyze: We are asked to determine the solubility of CaF2 in the presence of two strong electrolytes, each of which contains an ion common to CaF2. In (a) the common ion is Ca2+, and NO3– is a spectator ion. In (b) the common ion is F–, and Na+ is a spectator ion. Plan: Because the slightly soluble compound is CaF2, we need to use the Kspfor this compound, which is available in Appendix D: The value of Ksp 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 CaF2 dissolution, setting up a table of initial and equilibrium concentrations, and using the Ksp expression to determine the concentration of the ion that comes only from CaF2. Solve: (a) In this instance the initial concentration of Ca2+ is 0.010 M because of the dissolved Ca(NO3)2: Substituting into the solubility-product expression gives Sample Exercise 8.28 Calculating the Effect of a Common Ion on Solubility Calculate the molar solubility of CaF2 at 25 °C in a solution that is (a) 0.010 M in Ca(NO3)2, (b) 0.010 M in NaF.

48. 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 CaF2 is very small (2.1 × 10-4M). Thus, we assume that the 0.010 M concentration of Ca2+ fromCa(NO3)2 is very much greater than the small additional concentration resulting from the solubility of CaF2; 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 CaF2 dissolves per liter of the 0.010 M Ca(NO3)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 CaF2 should dissolve per liter of 0.010 M NaF solution. Sample Exercise 8.28 Calculating the Effect of a Common Ion on Solubility

49. Solution (Continued) Comment: The molar solubility of CaF2 in pure water is 2.1 × 10-4M (Sample Exercise 17.11). By comparison, our calculations above show that the solubility of CaF2 in the presence of 0.010 M Ca2+is 3.1 × 10-5M, and in the presence of 0.010 M F– ion it is 3.9 × 10-7M. Thus, the addition of either Ca2+ or F– to a solution of CaF2 decreases the solubility. However, the effect of F- on the solubility is more pronounced than that of Ca2+ because [F–] appears to the second power in the Ksp expression for CaF2, whereas Ca2+ appears to the first power. Practice Exercise The value for Kspfor 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 × 10-10M Sample Exercise 8.28 Calculating the Effect of a Common Ion on Solubility

50. 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 CO32– and H+ occurs in a stepwise fashion, first forming HCO3–. H2CO3 forms in appreciable amounts only when the concentration of H+ is sufficiently high. 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) CaCO3(s), (c) BaF2(s), (d) AgCl(s)?