CHAPTER 16 ACID-BASE EQUILIBRIA AND SOLUBILITY EQUILIBRIA

CHAPTER 16 ACID-BASE EQUILIBRIA AND SOLUBILITY EQUILIBRIA 16.3 (a) This is a weak acid problem. Setting up the standard equilibrium table: CH 3 COOH(aq) H + (aq) + CH 3 COO (aq) Initial (M): 0.40 0.00 0.00 Change (M): x +x +x Equilibrium (M): (0.40 x) +x +x K a = 1.8 10 5 = + 2 2 [H ][CH3COO ] x x = [CH3COOH] (0.40 x) 0.40 x = [H + ] = 2.7 10 3 M ph = 2.57 In addition to the acetate ion formed from the ionization of acetic acid, we also have acetate ion formed from the sodium acetate dissolving. CH 3 COONa(aq) CH 3 COO (aq) + Na + (aq) Dissolving 0.20 M sodium acetate initially produces 0.20 M CH 3 COO and 0.20 M Na +. The sodium ions are not involved in any further equilibrium (why?), but the acetate ions must be added to the equilibrium in part (a). CH 3 COOH(aq) H + (aq) + CH 3 COO (aq) Initial (M): 0.40 0.00 0.20 Change (M): x +x +x Equilibrium (M): (0.40 x) +x (0.20 + x) K a = 1.8 10 5 = + [H ][CH3COO ] ( x)(0.20 + x) x(0.20) = [CH3COOH] (0.40 x) 0.40 x = [H + ] = 3.6 10 5 M ph = 4.44 Could you have predicted whether the ph should have increased or decreased after the addition of the sodium acetate to the pure 0.40 M acetic acid in part (a)? An alternate way to work part of this problem is to use the Henderson-Hasselbalch equation. [conjugate base] ph = pk a + log [acid] ph 5 0.20 M = log(1.8 10 ) + log 0.40 M = 4.74 0.30 = 4.44

356 CHAPTER 16: ACID-BASE EQUILIBRIA AND SOLUBILITY EQUILIBRIA 16.4 (a) This is a weak base calculation. K b = 1.8 10 5 = + [NH 4 ][OH ] ( x)( x) = [NH 3] 0.20 x x = 1.90 10 3 M = [OH ] poh = 2.72 ph = 11.28 Table 15.4 gives the value of K a for the ammonium ion. Using this and the given concentrations with the Henderson-Hasselbalch equation gives: [ ] 10 ph = pk a + log ( ) ( ) ( ) conjugate base = log 5.6 10 + log 0.20 acid 0.30 ph = 9.25 0.18 = 9.07 Is there any difference in the Henderson-Hasselbalch equation in the cases of a weak acid and its conjugate base and a weak base and its conjugate acid? 16.8 (a) HCl (hydrochloric acid) is a strong acid. A buffer is a solution containing both a weak acid and a weak base. Therefore, this is not a buffer system. (c) (d) (e) (f) NH 3 (ammonia) is a weak base, and its conjugate acid, NH 4 + is a weak acid. Therefore, this is a buffer system. This solution contains both a weak acid, H 2 PO 4 and its conjugate base, HPO4 2. Therefore, this is a buffer system. HNO 2 (nitrous acid) is a weak acid, and its conjugate base, NO 2 (nitrite ion, the anion of the salt KNO 2 ), is a weak base. Therefore, this is a buffer system. H 2 SO 4 (sulfuric acid) is a strong acid. A buffer is a solution containing both a weak acid and a weak base. Therefore, this is not a buffer system. HCOOH (formic acid) is a weak acid, and its conjugate base, HCOO (formate ion, the anion of the salt HCOOK), is a weak base. Therefore, this is a buffer system. + 16.9 NH 4 (aq) NH3 (aq) + H + (aq) K a = 5.6 10 10 pk a = 9.25 [NH 3] 0.15 ph = p Ka + log = + [NH 9.25 + log M = 8.88 0.35 4 ] M 16.10 (a) This problem is greatly simplified because the concentration of the weak acid (acetic acid) is equal to the concentration of its conjugate base (acetate ion). Let s set up a table of initial concentrations, change in concentrations, and equilibrium concentrations.

CHAPTER 16: ACID-BASE EQUILIBRIA AND SOLUBILITY EQUILIBRIA 357 CH 3 COOH (aq) H + (aq) + CH 3 COO (aq) Initial 2.0 M 0 2.0 M Change x +x +x Equilibrium 2.0 M x x 2.0 M + x K a = K a = + [H ][CH3COO ] [CH3COOH] + + [H ](2.0 + x) [H ](2.0) (2.0 x) 2.0 K a = [H + ] Taking the log of both sides, pk a = ph Thus, for a buffer system in which the [weak acid] = [weak base], ph = pk a ph = log(1.8 10 5 ) = 4.74 Similar to part (a), ph = pk a = 4.74 Buffer (a) will be a more effective buffer because the concentrations of acid and base components are ten times higher than those in. Thus, buffer (a) can neutralize 10 times more added acid or base compared to buffer. 16.11 H 2 CO 3 (aq) HCO 3 (aq) + H + (aq) K a1 = 4.2 10 7 pk a1 = 6.38 ph = pk a + [HCO 3 ] log [H 2 CO 3 ] 8.00 = 6.38 + log [ HCO3 ] [ HCO 2 3] log [ HCO3 ] = 162. [ HCO 2 3] [ HCO3 ] = 41.7 [ HCO 2 3] [ HCO 2 3] [ HCO3 = 0.024 ]

358 CHAPTER 16: ACID-BASE EQUILIBRIA AND SOLUBILITY EQUILIBRIA 16.12 Step 1: Write the equilibrium that occurs between H 2 PO 4 and HPO4 2. Set up a table relating the initial concentrations, the change in concentration to reach equilibrium, and the equilibrium concentrations. H 2 PO 4 (aq) + 2 H (aq) + HPO4 (aq) Initial (M): 0.15 0 0.10 Change (M): x +x +x Equilibrium (M): 0.15 x x 0.10 + x Step 2: Write the ionization constant expression in terms of the equilibrium concentrations. Knowing the value of the equilibrium constant (K a ), solve for x. K a = + 2 [H ][HPO 4 ] [H2PO 4] You can look up the K a value for dihydrogen phosphate in Table 15.5 of your text. 6.2 10 8 = 6.2 10 8 ( x)(0.10 + x) (0.15 x) ( x)(0.10) (0.15) x = [H + ] = 9.3 10 8 M Step 3: Having solved for the [H + ], calculate the ph of the solution. ph = log[h + ] = log(9.3 10 8 ) = 7.03 16.13 Using the HendersonHasselbalch equation: [CH3COO ] ph = pka + log [CH 3 COOH] Thus, [CH3COO ] 4.50 = 4.74 + log [CH 3 COOH] [CH3COO ] = 0.58 [CH3COOH] 16.14 We can use the Henderson-Hasselbalch equation to calculate the ratio [HCO 3 ]/[H2 CO 3 ]. The Henderson- Hasselbalch equation is: [conjugate base] ph = pk a + log [acid] For the buffer system of interest, HCO 3 is the conjugate base of the acid, H2 CO 3. We can write: ph = 7.40 = 7 [HCO 3 ] log(4.2 10 ) + log [H 2 CO 3 ]

CHAPTER 16: ACID-BASE EQUILIBRIA AND SOLUBILITY EQUILIBRIA 359 7.40 = [HCO 3 ] 6.38 + log [H 2 CO 3 ] The [conjugate base]/[acid] ratio is: [HCO 3 ] log [H2CO 3] = 7.40 6.38 = 1.02 [HCO 3 ] [H2CO 3] = 10 1.02 = 1.0 10 1 The buffer should be more effective against an added acid because ten times more base is present compared to acid. Note that a ph of 7.40 is only a two significant figure number (Why?); the final result should only have two significant figures. 16.15 For the first part we use K a for ammonium ion. (Why?) The HendersonHasselbalch equation is 10 (0.20 M ) ph = log(5.6 10 ) + log (0.20 M = 9.25 ) For the second part, the acidbase reaction is NH 3 (g) + H + + (aq) NH 4 (aq) We find the number of moles of HCl added 1 L 0.10 mol HCl 10.0 ml = 0.0010 mol HCl 1000 ml 1 L + The number of moles of NH 3 and NH 4 originally present are 1L 0.20mol 65.0 ml = 0.013 mol 1000 ml 1 L + Using the acidbase reaction, we find the number of moles of NH 3 and NH 4 after addition of the HCl. moles NH 3 = (0.013 0.0010) mol = 0.012 mol NH 3 + + moles NH 4 = (0.013 + 0.0010) mol = 0.014 mol NH4 We find the new ph: (0.012) ph = 9.25 + log = 9.18 (0.014) 16.16 As calculated in Problem 16.10, the ph of this buffer system is equal to pk a. ph = pk a = log(1.8 10 5 ) = 4.74 (a) The added NaOH will react completely with the acid component of the buffer, CH 3 COOH. NaOH ionizes completely; therefore, 0.080 mol of OH are added to the buffer.

360 CHAPTER 16: ACID-BASE EQUILIBRIA AND SOLUBILITY EQUILIBRIA Step 1: The neutralization reaction is: CH 3 COOH (aq) + OH (aq) CH 3 COO (aq) + H 2 O (l) Initial (mol) 1.00 0.080 1.00 After reaction (mol) 0.92 0 1.08 Step 2: Now, the acetic acid equilibrium is reestablished. Since the volume of the solution is 1.00 L, we can convert directly from moles to molar concentration. CH 3 COOH (aq) H + (aq) + CH 3 COO (aq) Initial (M) 0.92 0 1.08 Change (M) x +x +x Equilibrium (M) 0.92 x x 1.08 + x Write the K a expression, then solve for x. K a = + [H ][CH3COO ] [CH3COOH] 1.8 10 5 = ( x )(1.08 + x ) (1.08) x (0.92 x) 0.92 x = [H + ] = 1.5 10 5 M Step 3: Having solved for the [H + ], calculate the ph of the solution. ph = log[h + ] = log(1.5 10 5 ) = 4.82 The ph of the buffer increased from 4.74 to 4.82 upon addition of 0.080 mol of strong base. The added acid will react completely with the base component of the buffer, CH 3 COO. HCl ionizes completely; therefore, 0.12 mol of H + ion are added to the buffer Step 1: The neutralization reaction is: CH 3 COO (aq) + H + (aq) CH 3 COOH (aq) Initial (mol) 1.00 0.12 1.00 After reaction (mol) 0.88 0 1.12 Step 2: Now, the acetic acid equilibrium is reestablished. Since the volume of the solution is 1.00 L, we can convert directly from moles to molar concentration. CH 3 COOH (aq) H + (aq) + CH 3 COO (aq) Initial (M) 1.12 0 0.88 Change (M) x +x +x Equilibrium (M) 1.12 x x 0.88 + x Write the K a expression, then solve for x. K a = + [H ][CH3COO ] [CH3COOH]

CHAPTER 16: ACID-BASE EQUILIBRIA AND SOLUBILITY EQUILIBRIA 361 1.8 10 5 = ( x )(0.88 + x ) (0.88) x (1.12 x) 1.12 x = [H + ] = 2.3 10 5 M Step 3: Having solved for the [H + ], calculate the ph of the solution. ph = log[h + ] = log(2.3 10 5 ) = 4.64 The ph of the buffer decreased from 4.74 to 4.64 upon addition of 0.12 mol of strong acid. 16.17 We write K a1 = 1.1 10 3 pk a1 = 2.96 K a2 = 2.5 10 6 pk a2 = 5.60 In order for the buffer solution to behave effectively, the pk a of the acid component must be close to the desired ph. Therefore, the proper buffer system is Na 2 A/NaHA. 16.18 Recall that to prepare a solution of a desired ph,, we should choose a weak acid with a pk a value close to the desired ph. Calculating the pk a for each acid: For HA, pk a = log(2.7 10 3 ) = 2.57 For HB, pk a = log(4.4 10 6 ) = 5.36 For HC, pk a = log(2.6 10 9 ) = 8.59 The buffer solution with a pk a closest to the desired ph is HC. Thus, HC is the best choice to prepare a buffer solution with ph = 8.60. 16.21 Since the acid is monoprotic, the number of moles of KOH is equal to the number of moles of acid. 1 L 0.08133 mol Moles acid = 16.4 ml 1000 ml 1 L = 0.00133 mol Molar mass = 0.2688 g 0.00133 mol = 202 g/mol 16.22 The neutralization reaction is: 2 KOH (aq) + H 2 A (aq) K 2 A (aq) + 2 H 2 O (l) The number of moles of H 2 A reacted is: 1.00 mol KOH 1molH 3 11.1 ml KOH 2A = 5.55 10 mol H2A 1000 ml 2 mol KOH

362 CHAPTER 16: ACID-BASE EQUILIBRIA AND SOLUBILITY EQUILIBRIA We know that 0.500 g of the diprotic acid were reacted (1/10 of the 250 ml was tested). Divide the number of grams by the number of moles to calculate the molar mass. 0.500 g H M (H 2 A) = 2A 3 5.55 10 mol H2A = 90.1 g/mol 16.23 The neutralization reaction is: H 2 SO 4 (aq) + 2NaOH(aq) Na 2 SO 4 (aq) + 2H 2 O(l) Since one mole of sulfuric acid combines with two moles of sodium hydroxide, we write: 0.500 mol H mol NaOH = (0.0125 L H 2 SO 4 ) 2SO4 2 mol NaOH 1Lsoln 1molH2SO4 = 0.0125 mol NaOH concentration of NaOH = 0.0125 mol NaOH 1000 ml = 0.25 M 50.0 ml soln 1 L 16.24 2HCOOH + Ba(OH) 2 (HCOO) 2 Ba + 2H 2 O Number of moles of HCOOH reacted = 0.883 mol 1L 3 (20.4 10 L) = 0.0180 mol HCOOH The mole ratio between Ba(OH) 2 and HCOOH is 1:2. Therefore, the molarity of the Ba(OH) 2 solution is: 0.0180 mol HCOOH 1molBa(OH) 2 1 2molHCOOH 3 19.3 10 L = 0.466 M 16.25 (a) Since the acid is monoprotic, the moles of acid equals the moles of base added. HA (aq) + NaOH (aq) NaA (aq) + H 2 O (l) F 1 L I 0 0633 Moles acid = 18.4 ml HG 1000 ml K J F H G. moli 1 K J L = 0.00116 mol We know the mass of the unknown acid in grams and the number of moles of the unknown acid. 01276. g Molar mass = = 1.10 10 2 g/mol 0. 00116 mol The number of moles of NaOH in 10.0 ml of solution is 10.0 ml F HG 1 L 1000 ml I KJ F H G 00633. mol 1 L I K J = 6.33 10 4 mol

CHAPTER 16: ACID-BASE EQUILIBRIA AND SOLUBILITY EQUILIBRIA 363 The neutralization reaction is: HA (aq) + NaOH (aq) NaA (aq) + H 2 O (l) Initial (mol) 0.00116 6.33 10 4 0 After reaction (mol) 5.3 10 4 0 6.33 10 4 Now, the weak acid equilibrium will be reestablished. The total volume of solution is 35.0 ml. [HA] = [A ] = 4 53. 10 mol 0.035 L 4 6.33 10 mol 0.035 L = 0.015 M = 0.0181 M We can calculate the [H + ] from the ph. [H + ] = 10 ph = 10 5.87 = 1.35 10 6 M HA (aq) H + (aq) + A (aq) Initial (M) 0.015 0 0.0181 Change (M) 1.35 10 6 +1.35 10 6 +1.35 10 6 Equilibrium (M) 0.015 1.35 10 6 0.0181 Substitute the equilibrium concentrations into the equilibrium constant expression to solve for K a. K a = [ + ][ ] H A )(. ) = (1.35 10 6 00181 [ HA] 0015. = 1.6 10 6 16.26 The resulting solution is not a buffer system. There is excess NaOH and the neutralization is well past the equivalence point. Moles NaOH = 0.500 L 0.167 mol = 0.0835 mol 1L Moles CH 3 COOH = 0.500 L 0.100 mol = 0.0500 mol 1L After neutralization, the amount of NaOH remaining in 0.0835 0.0500 = 0.0335 mol. The volume of the resulting solution is 1.00 L [OH ] = 0.0335 mol 1.00 L [Na + ] = [H + ] = 0.0835 mol 1.00 L 14 1.0 10 0.0335 = 0.0335 M = 0.0835 M 13 = 3.0 10 M [CH 3 COO 0.0500 mol ] = = 0.0500 M 1.00 L + 13 [H ][CH3COOH ] (3.0 10 )(0.0500) 10 [CH 3 COOH] = = = 8.3 10 M K 5 a 1.8 10

364 CHAPTER 16: ACID-BASE EQUILIBRIA AND SOLUBILITY EQUILIBRIA 16.27 Since the solution volume is doubled at the equivalence point, the concentration of the conjugate acid from + the salt, CH 3 NH 3, is: 0.20 M 2 = 0.10 M The conjugate acid undergoes hydrolysis. CH 3 NH 3 + + H2 O CH 3 NH 2 + H 3 O + 0.10 M x x x 2 x 0.10 x = 2.3 1011 Assuming that, 0.10 x 0.10 x = [H 3 O + ] = 1.5 10 6 M ph = 5.82 16.28 Concentration of HCOONa at the equivalence point is 0.050 M, since the solution volume doubles (the volume of NaOH will equal the volume of HCOOH since the molarities are equal). HCOO + H 2 O HCOOH + OH 0.050 x x x 2 x 0.050 x Assume 0.050 x 0.050 11 = 5.9 10 x = 1.7 10 6 M = [OH ] poh = 5.77 ph = 8.23 16.31 (a) HCOOH is a weak acid and NaOH is a strong base. Suitable indicators are cresol red and phenolphthalein. HCl is a strong acid and KOH is a strong base. Suitable indicators are all those listed with the exceptions of thymol blue, bromophenol blue, and methyl orange. (c) HNO 3 is a strong acid and CH 3 NH 2 is a weak base. Suitable indicators are bromophenol blue, methyl orange, methyl red, and chlorophenol blue. 16.32 CO 2 in the air dissolves in the solution: The carbonic acid neutralizes the NaOH. CO 2 + H 2 O H 2 CO 3

CHAPTER 16: ACID-BASE EQUILIBRIA AND SOLUBILITY EQUILIBRIA 365 16.33 The weak acid equilibrium is HIn (aq) H + (aq) + In (aq) We can write a K a expression for this equilibrium. Rearranging, K a = [ + ][ H In ] [ HIn] + [ HIn] [ H ] = [ In ] Ka From the ph, we can calculate the H + concentration. [H + ] = 10 ph = 10 4 = 1.0 10 4 M [HIn] [In ] + 4 [ H ] 1.0 10 = = K 6 a 1.0 10 = 100 Since the concentration of HIn is 100 times greater than the concentration of In, the color of the solution will be that of HIn, the nonionized formed. The color of the solution will be red. 16.34 According to Section 16.5 of the text, when [HIn] [In ] the indicator color is a mixture of the colors of HIn and In. In other words, the indicator color changes at this point. When [HIn] [In ] we can write: [In ] Ka = = [HIn] + [H ] 1 [H + ] = K a = 2.0 10 6 ph = 5.70 16.41 (a) The solubility equilibrium is given by the equation AgI(s) Ag + (aq) + I (aq) The expression for K sp is given by K sp = [Ag + ][I ] The value of K sp can be found in Table 16.2 of the text. If the equilibrium concentration of silver ion is the value given, the concentration of iodide ion must be [I K 17 sp 83. 10 ] = = = 9.1 10 9 M + 9 [ Ag ] 91. 10

366 CHAPTER 16: ACID-BASE EQUILIBRIA AND SOLUBILITY EQUILIBRIA The value of K sp for aluminum hydroxide can be found in Table 16.2 of the text. The equilibrium expressions are: Al(OH) 3 (s) Al 3+ (aq) + 3OH (aq) K sp = [Al 3+ ][OH ] 3 Using the given value of the hydroxide ion concentration, the equilibrium concentration of aluminum ion is: 3+ K 33 sp 18. 10 8 [Al ] = = = 7.4 10 [ OH ] 3 M 9 3 (. 29 10 ) What is the ph of this solution? Will the aluminum concentration change if the ph is altered? 16.42 In each case, we first calculate the number of moles of compound dissolved in one liter of solution (the molar solubility). (a) 2 7.3 10 g SrF 2 1 mol SrF2 4 = 5.8 10 mol / L 1 L soln 125.6 g SrF2 Step 1: Write the equilibrium reaction. Then, from the equilibrium equation, write the solubility product expression. SrF 2 (s) Sr 2+ (aq) + 2 F (aq) K sp = [Sr 2+ ][F ] 2 Step 2: The molar solubility is the amount of SrF 2 that dissolves. From the stoichiometry of the equilibrium equation, you should find that and [Sr 2+ ] = [SrF 2 ] = 5.8 10 4 M [F ] = 2[SrF 2 ] = 1.16 10 3 M Step 3: Substitute the equilibrium concentrations of Sr 2+ and F into the solubility product expression to calculate K sp. K sp = [Sr 2+ ][F ] 2 = (5.8 10 4 )(1.16 10 3 ) 2 = 7.8 10 10 3 6.7 10 g Ag3PO 4 1 mol Ag3PO4 5 = 1.6 10 mol / L 1 L soln 418.7 g Ag3PO4 is solved in a similar manner to (a) The equilibrium equation is: Ag 3 PO 4 (s) 3 Ag + 3 (aq) + PO 4 (aq) K sp = [Ag + ] 3 3 5 3 5 18 [PO 4 ] = [3 (1.6 10 )] (1.6 10 ) = 1.8 10

CHAPTER 16: ACID-BASE EQUILIBRIA AND SOLUBILITY EQUILIBRIA 367 16.43 For MnCO 3 dissolving, we write MnCO 3 (s) Mn 2+ (aq) + CO 3 2 (aq) For every mole of MnCO 3 that dissolves, one mole of Mn 2+ 2 will be produced and one mole of CO 3 will be produced. If the molar solubility of MnCO 3 is s mol/l, then the concentrations of Mn 2+ 2 and CO 3 are: [Mn 2+ 2 6 ] = [CO 3 ] = s = 4.2 10 M K sp = [Mn 2+ 2 2 6 2 11 ][CO 3 ] = s = (4.2 10 ) = 1.8 10 16.44 First, we can convert the solubility of MX in g/l to mol/l. 3 4.63 10 g MX 1 mol MX = 1.34 10 5 1 L soln 346 g MX mol/l = s (molar solubility) The equilibrium reaction is: MX (s) M n+ (aq) + X n (aq) Since the mole ratio of MX to each of the ions is 1:1, the equilibrium concentrations of each of the ions can also be represented by s. Solving for K sp, K sp = [M n+ ][X n ] = s 2 = (1.34 10 5 ) 2 = 1.80 10 10 16.45 The charges of the M and X ions are +3 and 2, respectively (are other values possible?). We first calculate the number of moles of M 2 X 3 that dissolve in 1.0 L of water 17 1mol 19 Moles M2X 3 = (3.6 10 g) = 1.3 10 mol 288 g The molar solubility, s, of the compound is therefore 1.3 10 19 M. At equilibrium the concentration of M 3+ must be 2s and that of X 2 must be 3s. (See Table 16.3 of the text.) K sp = [M 3+ ] 2 [X 2 ] 3 = [2s] 2 [3s] 3 = 108s 5 Since these are equilibrium concentrations, the value of K sp can be found by simple substitution K sp = 108s 5 = 108(1.3 10 19 ) 5 = 4.0 10 93 16.46 Step 1: Write the equilibrium reaction. Then, from the equilibrium equation, write the solubility product expression. CaF 2 (s) Ca 2+ (aq) + 2 F (aq) K sp = [Ca 2+ ][F ] 2 Step 2: A certain amount of calcium fluoride will dissociate in solution. Let s represent this amount as s. Since one unit of CaF 2 yields one Ca 2+ ion and two F ions, at equilibrium [Ca 2+ ] is s and [F ] is 2s. We summarize the changes in concentration as follows:

368 CHAPTER 16: ACID-BASE EQUILIBRIA AND SOLUBILITY EQUILIBRIA CaF 2 (s) Ca 2+ (aq) + 2 F (aq) Initial (M): 0 0 Change (M): s +s +2s Equilibrium (M): s 2s Recall, that the concentration of a pure solid does not enter into an equilibrium constant expression. Therefore, the concentration of CaF 2 is not important. Step 3: Substitute the value of K sp and the concentrations of Ca 2+ and F in terms of s into the solubility product expression to solve for s, the molar solubility. K sp = [Ca 2+ ][F ] 2 4.0 10 11 = (s)(2s) 2 4.0 10 11 = 4s 3 s = molar solubility = 2.2 10 4 mol/l The molar solubility indicates that 2.2 10 4 mol of CaF 2 will dissolve in 1 L of an aqueous solution. 16.47 Let s be the molar solubility of Zn(OH) 2. The equilibrium concentrations of the ions are then [Zn 2+ ] = s and [OH ] = 2s K sp = [Zn 2+ ][OH ] 2 = (s)(2s) 2 = 4s 3 = 1.8 10 14 s = F HG I 18 10 14 = 17 10 4 KJ 13 /.. 5 [OH ] = 2s = 3.4 10 5 M and poh = 4.47 ph = 14.00 4.47 = 9.53 If the K sp of Zn(OH) 2 were smaller by many more powers of ten, would 2s still be the hydroxide ion concentration in the solution? 16.48 First we can calculate the OH concentration from the ph. poh = 14.00 ph poh = 14.00 9.68 = 4.32 [OH ] = 10 poh = 10 4.32 = 4.8 10 5 M The equilibrium equation is: MOH (s) M + (aq) + OH (aq) From the balanced equation we know that [M + ] = [OH ] K sp = [M + ][OH ] = (4.8 10 5 ) 2 = 2.3 10 9

CHAPTER 16: ACID-BASE EQUILIBRIA AND SOLUBILITY EQUILIBRIA 369 16.49 According to the solubility rules, the only precipitate that might form is BaCO 3. Ba 2+ 2 (aq) + CO 3 (aq) BaCO3 (s) The number of moles of Ba 2+ present in the original 20.0 ml of Ba(NO 3 ) 2 solution is 2+ 0.10 mol Ba 1 L 3 2+ 20.0 ml = 2.0 10 mol Ba 1 L soln 1000 ml The total volume after combining the two solutions is 70.0 ml. The concentration of Ba 2+ in 70 ml is 3 2+ 2+ 2.0 10 mol Ba 1000 ml 2 [Ba ] = = 2.9 10 M 70.0 ml 1 L soln 2 The number of moles of CO 3 present in the original 50.0 ml Na2 CO 3 solution is 2 0.10 mol CO3 1L 3 2 50.0 ml = 5.0 10 mol CO3 1 L soln 1000 ml 2 The concentration of CO 3 in the 70.0 ml of combined solution is 3 2 2 5.0 10 mol CO3 1000 ml 2 [CO 3 ] = = 7.1 10 M 70.0 ml 1 L soln Now we must compare Q and K sp. From Table 16.2 of the text, the K sp for BaCO 3 is 8.1 10 9. As for Q, Q = [Ba 2+ ][CO 3 2 ] = (2.9 10 2 )(7.1 10 2 ) = 2.1 10 3 Since (2.1 10 3 ) > (8.1 10 9 ), then Q > K sp. Therefore, BaCO 3 will precipitate. 16.50 The net ionic equation is: Sr 2+ (aq) + 2 F (aq) SrF 2 (s) Let s find the limiting reagent in the precipitation reaction. Moles F 1 L 0.060 mol = 75 ml 1000 ml 1 L Moles Sr 2+ 1 L 0.15 mol = 25 ml 1000 ml 1 L = 0.0045 mol = 0.0038 mol From the stoichiometry of the balanced equation, twice as many moles of F are required to react with Sr 2+. This would require 0.0076 mol of F, but we only have 0.0045 mol. Thus, F is the limiting reagent.

370 CHAPTER 16: ACID-BASE EQUILIBRIA AND SOLUBILITY EQUILIBRIA Let s assume that the above reaction goes to completion. Then, we will consider the equilibrium that is established when SrF 2 partially dissociates into ions. Sr 2+ (aq) + 2 F (aq) SrF 2 (s) Initial (mol) 0.0038 0.0045 0 Change (mol) 0.00225 0.0045 +0.00225 After reaction (mol) 0.00155 0 0.00225 Now, let s establish the equilibrium reaction. The total volume of the solution is 100 ml = 0.100 L. Divide the above moles by 0.100 L to convert to molar concentration. SrF 2 (s) Sr 2+ (aq) + 2 F (aq) Initial (M) 0.0225 0.0155 0 Change (M) x +x +2x Equilibrium (M) 0.0225 x 0.0155 + x 2x Write the solubility product expression, then solve for x. K sp = [Sr 2+ ][F ] 2 2.0 10 10 = (0.0155 + x)(2x) 2 (0.0155)(2x) 2 x = 5.7 10 5 M [F ] = 2x = 1.1 10 4 M [Sr 2+ ] = 0.0155 + x = 0.016 M Both sodium ions and nitrate ions are spectator ions and therefore do not enter into the precipitation reaction. [NO 3 ] = 2(0.0038) mol 0.10 L = 0.076 M [Na + ] = 0.0045 mol 0.10 L = 0.045 M 16.51 (a) The solubility product expressions for both substances have exactly the same mathematical form and are therefore directly comparable. The substance having the smaller K sp (AgI) will precipitate first. (Why?) When CuI just begins to precipitate the solubility product expression will just equal K sp (saturated solution). The concentration of Cu + at this point is 0.010 M (given in the problem), so the concentration of iodide ion must be: K sp = [Cu + ][I ] = (0.010)[I ] = 5.1 10 12 [I ] = 12 5.1 10 0.010 = 5.1 10 10 M Using this value of [I ], we find the silver ion concentration [Ag + ] = K 17 sp 8.3 10 = 10 [I ] 5.1 10 = 1.6 10 7 M

CHAPTER 16: ACID-BASE EQUILIBRIA AND SOLUBILITY EQUILIBRIA 371 (c) The percent of silver ion remaining in solution is: % Ag + (aq) = 7 1.6 10 M 0.010 M 100% = 0.0016% or 1.6 10 3 % Is this an effective way to separate silver from copper? 16.52 For Fe(OH) 3, K sp = 1.1 10 36. When [Fe 3+ ] = 0.010 M, the [OH ] value is: or K sp = [Fe 3+ ][OH ] 3 [OH ] = [OH ] = 1 Ksp 3 3+ [Fe ] 1 1.1 10 36 3 0.010 = 4.8 10 12 M This [OH ] corresponds to a ph of 2.68. In other words, Fe(OH) 3 will begin to precipitate from this solution at ph of 2.68. For Zn(OH) 2, K sp = 1.8 10 14. When [Zn 2+ ] = 0.010 M, the [OH ] value is: [OH ] = [OH ] = 1 Ksp 2 2+ [Zn ] 1 1.8 10 14 2 0.010 = 1.3 10 6 M This corresponds to a ph of 8.11. In other words Zn(OH) 2 will begin to precipitate from the solution at ph = 8.11. These results show that Fe(OH) 3 will precipitate when the ph just exceeds 2.68 and that Zn(OH) 2 will precipitate when the ph just exceeds 8.11. Therefore, to selectively remove iron as Fe(OH) 3, the ph must be greater than 2.68 but less than 8.11. 16.55 First let s be the molar solubility of CaCO 3 in this solution. CaCO 3 (s) Ca 2+ (aq) + 2 CO 3 (aq) Initial (M): 0.050 0 Change (M): +s +s Equilibrium (M): (0.050 + s) s K sp = [Ca 2+ ][CO 3 2 ] = (0.050 + s)s = 8.7 10 9 We can assume 0.050 + s 0.050, then 9 8.7 10 7 s = = 1.7 10 M 0.050

372 CHAPTER 16: ACID-BASE EQUILIBRIA AND SOLUBILITY EQUILIBRIA The mass of CaCO 3 can then be found. 7 2 1 L 1.7 10 mol 100.1 g CaCO3 6 3.0 10 ml = 5.1 10 g CaCO 1000 ml 1 L 3 1 mol 16.56 (a) Set up a table to find the equilibrium concentrations in pure water. PbBr 2 (s) Pb 2+ (aq) + 2 Br (aq) Initial (M) 0 0 Change (M) s +s +2s Equilibrium (M) s 2s K sp = [Pb 2+ ][Br ] 2 8.9 10 6 = (s)(2s) 2 s = molar solubility = 0.013 M Set up a table to find the equilibrium concentrations in 0.20 M KBr. KBr is a soluble salt that ionizes completely giving a initial concentration of Br = 0.20 M. PbBr 2 (s) Pb 2+ (aq) + 2 Br (aq) Initial (M) 0 0.20 Change (M) s +s +2s Equilibrium (M) s 0.20 + 2s K sp = [Pb 2+ ][Br ] 2 8.9 10 6 = (s)(0.20 + 2s) 2 8.9 10 6 (s)(0.20) 2 s = molar solubility = 2.2 10 4 M Thus, the molar solubility of PbBr 2 is reduced from 0.013 M to 2.2 10 4 M as a result of the common ion (Br ) effect. (c) Set up a table to find the equilibrium concentrations in 0.20 M Pb(NO 3 ) 2. Pb(NO 3 ) 2 is a soluble salt that dissociates completely giving an initial concentration of [Pb 2+ ] = 0.20 M. PbBr 2 (s) Pb 2+ (aq) + 2 Br (aq) Initial (M): 0.20 0 Change (M): s +s +2s Equilibrium (M): 0.20 + s 2s K sp = [Pb 2+ ][Br ] 2 8.9 10 6 = (0.20 + s)(2s) 2 8.9 10 6 (0.20)(2s) 2 s = molar solubility = 3.3 10 3 M

CHAPTER 16: ACID-BASE EQUILIBRIA AND SOLUBILITY EQUILIBRIA 373 Thus, the molar solubility of PbBr 2 is reduced from 0.013 M to 3.3 10 3 M as a result of the common ion (Pb 2+ ) effect. 16.57 We first calculate the concentration of chloride ion in the solution. 10.0 g CaCl2 1 mol CaCl2 2molCl [Cl ] = = 0.180 M 1 L soln 111.0 g CaCl 2 1 mol CaCl 2 AgCl(s) Ag + (aq) + Cl (aq) Initial (M): 0.000 0.180 Change (M): +s +s Equilibrium (M): s (0.180 + s) If we assume that (0.180 + s) 0.180, then K sp = [Ag + ][Cl ] = 1.6 10 10 10 + Ksp 1.6 10 10 [Ag ] = = = 8.9 10 M = s [Cl ] 0.180 The molar solubility of AgCl is 8.9 10 10 M. 16.58 The equilibrium reaction is: BaSO 4 (s) Ba 2+ (aq) + SO 4 2 (aq) For both parts of the problem: K sp = [Ba 2+ ][SO 4 2 ] = 1.1 10 10 (a) In pure water, let [Ba 2+ ] = [SO 4 2 ] = s K sp = [Ba 2+ 2 ][SO 4 ] 1.1 10 10 = s 2 s = 1.0 10 5 M The molar solubility of BaSO 4 in pure water is 1.0 10 5 mol/l. Assuming the molar solubility of BaSO 4 to be s, then [Ba 2+ 2 ] = s M and [SO 4 ] = (1.0 + s) M 1.0 M K sp = [Ba 2+ 2 ][SO 4 ] 1.1 10 10 = (s)(1.0) s = 1.1 10 10 M Due to the common ion effect, the molar solubility of BaSO 4 decreases to 1.1 10 10 mol/l in 2 5 1.0 M SO 4 (aq) compared to 1.0 10 mol/l in pure water.

374 CHAPTER 16: ACID-BASE EQUILIBRIA AND SOLUBILITY EQUILIBRIA 16.59 When the anion of a salt is a base, the salt will be more soluble in acidic solution because the hydrogen ion decreases the concentration of the anion (Le Chatelier's principle): B (aq) + H + (aq) HB(aq) (a) (c) (d) BaSO 4 will be slightly more soluble because SO 4 2 is a base (although a weak one). The solubility of PbCl 2 in acid is unchanged over the solubility in pure water because HCl is a strong acid, and therefore Cl is a negligibly weak base. Fe(OH) 3 will be more soluble in acid because OH is a base. CaCO 3 will be more soluble in acidic solution because the CO 3 2 ions react with H + ions to form CO2 and H 2 O. The CO 2 escapes from the solution, shifting the equilibrium. Although it is not important in this case, the carbonate ion is also a base. 16.60 SO 4 2 (aq) is a weak base (c) (d) (e) OH (aq) is a strong base C 2 O 4 2 (aq) is a weak base PO 4 3 (aq) is a weak base. The solubilities of the above will increase in acidic solution. Only (a), which contains an extremely weak base (I is the conjugate base of the strong acid HI) is unaffected by the acid solution. 16.61 In water: Mg(OH) 2 Mg 2+ + 2OH s 2s K sp = 4s 3 = 1.2 10 11 s = 1.4 10 4 M In a buffer at ph = 9.0 [H + ] = 1.0 10 9 [OH ] = 1.0 10 5 1.2 10 11 = [s](1.0 10 5 ) 2 s = 0.12 M 16.62 From Table 16.2, the value of K sp for iron(ii) is 1.6 10 14. (a) At ph = 8.00, poh = 14.00 8.00 = 6.00, and [OH ] = 1.0 10 6 M [Fe 2+ ] = K 14 sp 1.6 10 = 2 6 2 [OH ] (1.0 10 ) = 0.016 M The molar solubility of iron(ii) hydroxide at ph = 8.00 is 0.016 M

CHAPTER 16: ACID-BASE EQUILIBRIA AND SOLUBILITY EQUILIBRIA 375 At ph = 10.00, poh = 14.00 10.00 = 4.00, and [OH ] = 1.0 10 4 M [Fe 2+ ] = K 14 sp 1.6 10 6 = = 1.6 10 2 4 2 [OH ] (1.0 10 ) M The molar solubility of iron(ii) hydroxide at ph = 10.00 is 1.6 10 6 M. 16.63 The solubility product expression for magnesium hydroxide is K sp = [Mg 2+ ][OH ] 2 = 1.2 10 11 We find the hydroxide ion concentration when [Mg 2+ ] is 1.0 10 10 M 1/ 2 11 1.2 10 [OH ] = = 0.35 M 10 1.0 10 Therefore the concentration of OH must be slightly greater than 0.35 M. 16.64 We first determine the effect of the added ammonia. Let's calculate the concentration of NH 3. This is a dilution problem. M I V I = M F V F (0.60 M)(2.00 ml) = M 2 (1002 ml) M 2 = 0.0012 M NH 3 Ammonia is a weak base (K b = 1.8 10 5 ). NH 3 + H 2 O + NH 4 + OH initial (M): 0.0012 0 0 change (M): x +x +x equil. (M): 0.0012 x x x K b = + 2 [NH 4 ][OH ] x 5 = = 1.8 10 [NH 3] ( 0.0012 x) Solving the resulting quadratic equation gives x = 0.00014, or [OH ] = 0.00014 M This is a solution of iron(ii) sulfate, which contains Fe 2+ ions. These Fe 2+ ions could combine with OH to precipitate Fe(OH) 2. Therefore, we must use K sp for iron(ii) hydroxide. We compute the value of Q c for this solution. Q = [Fe 2+ ] 0 [OH ] 0 2 = (1.0 10 3 )(0.00014) 2 = 2.0 10 11 Q is larger than K sp [Fe(OH) 2 ] = 1.6 10 14 ; therefore a precipitate of Fe(OH) 2 will form.

376 CHAPTER 16: ACID-BASE EQUILIBRIA AND SOLUBILITY EQUILIBRIA 16.67 First find the molarity of the copper(ii) ion 1mol Moles CuSO4 = 2.50 g = 0.0157 mol 159.6 g 2+ 0.0157 mol [Cu ] = = 0.0174 M 0.90 L As in Example 16.15 of the text, the position of equilibrium will be far to the right. We assume essentially all the copper ion is complexed with NH 3. The NH 3 consumed is 4 0.0174 M = 0.0696 M. The uncombined NH 3 remaining is (0.30 0.0696) M, or 0.23 M. The equilibrium concentrations of Cu(NH 3 ) 4 2+ and NH3 are therefore 0.0174 M and 0.23 M, respectively. We find [Cu 2+ ] from the formation constant expression. K f [ Cu( NH = 3) 4 2+ ] 13 00174. =. = 2+ + [ Cu ][ NH3] 4 50 10 2 4 [ Cu ][ 023. ] [Cu 2+ ] = 1.2 10 13 M 16.68 In solution, Cd 2+ ions will complex with CN ions. The concentration of Cd 2+ will be determined by the following equilibrium Cd 2+ (aq) + 4 CN (aq) Cd(CN) 4 2 K f = 7.1 10 16 Since K f is so large, this equilibrium lies far to the right. We can safely assume that all the Cd 2+ reacts. Step 1: Calculate the initial concentration of Cd 2+ ions. [Cd 2+ ] 0 = 0.50 g 1 mol Cd(NO 3) 2 1 mol Cd 1 = 4.2 10 3 M 236.42 g Cd(NO 3) 2 1 mol Cd(NO 3) 2 0.50 L Step 2: If we assume that the above equilibrium goes to completion, we can write Cd 2+ (aq) + 4 CN 2 (aq) Cd(CN) 4 Initial (M) 4.2 10 3 0.50 0 After reaction (M) 0 0.48 4.2 10 3 Step 3: To find the concentration of free Cd 2+ at equilibrium, use the formation constant expression. K f = 2 [Cd(CN) 4 ] 2+ 4 [Cd ][CN ] Rearranging, [Cd 2+ ] = 2 [Cd(CN) 4 ] 4 Kf [CN ]

CHAPTER 16: ACID-BASE EQUILIBRIA AND SOLUBILITY EQUILIBRIA 377 Substitute the equilibrium concentrations calculated above into the formation constant expression to calculate the equilibrium concentration of Cd 2+. [Cd 2+ ] = 2 [Cd(CN) 4 ] 4 Kf [CN ] = 3 4.2 10 16 4 (7.1 10 )(0.48) = 1.1 10 18 M [CN ] = 0.48 M + (1.1 10 18 M) = 0.48 M [Cd(CN) 4 2 ] = (4.2 10 3 M) (1.1 10 18 ) = 4.2 10 3 M 16.69 The reaction Al(OH) 3 (s) + OH (aq) Al(OH) 4 (aq) is the sum of the two known reactions Al(OH) 3 (s) Al 3+ (aq) + 3OH (aq) K sp = 1.8 10 33 Al 3+ (aq) + 4OH (aq) Al(OH) 4 (aq) Kf = 2.0 10 33 The equilibrium constant is K = K sp K f = (1.8 10 33 )(2.0 10 33 ) = 3.6 = [ Al ( OH ) 4 ] [ OH ] When ph = 14.00, [OH ] = 1.0 M, therefore [Al(OH) 4 ] = K[OH ] = 3.6 1 = 3.6 M This represents the maximum possible concentration of the complex ion at ph 14.00. Since this is much larger than the initial 0.010 M, the complex ion will be the predominant species. 16.70 Silver iodide is only slightly soluble. It dissociates to form a small amount of Ag + and I ions. The Ag + ions then complex with NH 3 in solution to form the complex ion Ag(NH 3 ) 2 +. The balanced equations are: AgI (s) Ag + (aq) + I (aq) Ag + (aq) + 2 NH 3 (aq) Ag(NH 3 ) 2 + (aq) Kf = K sp = [Ag + ][I ] = 8.3 10 17 + [Ag(NH 3) 2 ] = 1.5 7 10 + 2 [Ag ][NH 3] Overall: AgI (s) + 2 NH 3 (aq) Ag(NH 3 ) 2 + (aq) + I (aq) K = Ksp K f = 1.2 10 9 If s is the molar solubility of AgI then, AgI (s) + 2 NH 3 (aq) + Ag(NH 3 ) 2 (aq) + I (aq) initial (M): 1.0 0.0 0.0 change (M): s 2s +s +s equilibrium (M) (1.0 2s) s s Because K f is large, we can assume all of the silver ions exist as Ag(NH 3 ) 2 +. Thus, [Ag(NH 3 ) 2 + ] = [I ] = s

378 CHAPTER 16: ACID-BASE EQUILIBRIA AND SOLUBILITY EQUILIBRIA We can write the equilibrium constant expression for the above reaction, then solve for s. K = 1.2 10 9 = ()() s s ()() s s (1.0 2 ) 2 (1.0) 2 s s = 3.5 10 5 M At equilibrium, 3.5 10 5 moles of AgI dissolves in 1 L of 1.0 M NH 3 solution. 16.71 The balanced equations are: Ag + (aq) + 2NH 3 (aq) Ag(NH 3 ) 2 + (aq) Zn 2+ (aq) + 4NH 3 (aq) Zn(NH 3 ) 4 2+ (aq) Zinc hydroxide forms a complex ion with excess OH and silver hydroxide does not; therefore, zinc hydroxide is soluble in 6 M NaOH. 16.72 (a) The equations are as follows: CuI 2 (s) Cu 2+ (aq) + 2 I (aq) Cu 2+ (aq) + 4 NH 3 (aq) [Cu(NH 3 ) 4 ] 2+ (aq) The ammonia combines with the Cu 2+ ions formed in the first step to form the complex ion [Cu(NH 3 ) 4 ] 2+, effectively removing the Cu 2+ ions, causing the first equilibrium to shift to the right (resulting in more CuI 2 dissolving). Similar to part (a): AgBr (s) Ag + (aq) + Br (aq) Ag + (aq) + 2 CN (aq) [Ag(CN)2 ] (aq) (c) Similar to parts (a) and. HgCl 2 (s) Hg 2+ (aq) + 2Cl (aq) Hg 2+ (aq) + 4Cl (aq) [HgCl4 ] 2 (aq) 16.75 Silver chloride will dissolve in aqueous ammonia because of the formation of a complex ion. Lead chloride will not dissolve; it doesn t form an ammonia complex. 16.76 Since some PbCl 2 precipitates, the solution is saturated. From Table 16.2, the value of K sp for lead(ii) chloride is 2.4 10 4. The equilibrium is: PbCl 2 (aq) Pb 2+ (aq) + 2 Cl (aq)

CHAPTER 16: ACID-BASE EQUILIBRIA AND SOLUBILITY EQUILIBRIA 379 We can write the solubility product expression for the equilibrium. K sp = [Pb 2+ ][Cl ] 2 K sp and [Cl ] are known. Solving for the Pb 2+ concentration, 4 2+ Ksp 2.4 10 [Pb ] = = = 0.011 M 2 2 [Cl ] (0.15) 16.77 Ammonium chloride is the salt of a weak base (ammonia). It will react with strong aqueous hydroxide to form ammonia (Le Chatelier s principle). NH 4 Cl(s) + OH (aq) NH 3 (g) + H 2 O(l) + Cl (aq) The human nose is an excellent ammonia detector. Nothing happens between KCl and strong aqueous NaOH. 16.78 Chloride ion will precipitate Ag + but not Cu 2+. So, dissolve some solid in H 2 O and add HCl. If a precipitate forms, the salt was AgNO 3. A flame test will also work. Cu 2+ gives a green flame test. 16.79 According to the HendersonHasselbalch equation: ph = pk a + log [conjugate base] [acid] If: [conjugate base] [acid] = 10, then: ph = pk a + 1 If: [conjugate base] [acid] = 01., then: ph = pk a 1 Therefore, the range of the ratio is: [conjugate base] 0.1 < < 10 [acid] 16.80 We can use the Henderson-Hasselbalch equation to solve for the ph when the indicator is 90% acid / 10% conjugate base and when the indicator is 10% acid / 90% conjugate base. [conjugate base] ph = p K a + log [acid] Solving for the ph with 90% of the indicator in the HIn form: [10] ph = 3.46 + log = 3.46 0.95 = 2.51 [90]

380 CHAPTER 16: ACID-BASE EQUILIBRIA AND SOLUBILITY EQUILIBRIA Next, solving for the ph with 90% of the indicator in the In form: [90] ph = 3.46 + log = 3.46 + 0.95 = 4.41 [10] Thus the ph range varies from 2.51 to 4.41 as the [HIn] varies from 90% to 10%. 16.81 Referring to Figure 16.4, at the halfequivalence point, [weak acid] = [conjugate base]. Using the Henderson-Hasselbalch equation: [conjugate base] ph = pk a + log [acid] so, ph = pk a 16.82 First, calculate the ph of the 2.00 M weak acid (HNO 2 ) solution before any NaOH is added. K a = + [H ][NO 2 ] [HNO 2] 2 4 x 4.5 10 = 2.00 x x = [H + ] = 0.030 M ph = log(0.030) = 1.52 Since the ph after the addition is 1.5 ph units greater, the new ph = 1.52 + 1.50 = 3.02. From this new ph, we can calculate the [H + ] in solution. [H + ] = 10 ph = 10 3.02 = 9.55 10 4 M When the NaOH is added, we dilute our original 2.00 M HNO 2 solution to: M I V I = M F V F (2.00 M)(400 ml) = M F (600 ml) M F = 1.33 M Since we have not reached the equivalence point, we have a buffer solution. The reaction between HNO 2 and NaOH is: HNO 2 (aq) + NaOH (aq) NaNO 2 (aq) + H 2 O (l) Since the mole ratio between HNO 2 and NaOH is 1:1, the decrease in [HNO 2 ] is the same as the decrease in [NaOH].

CHAPTER 16: ACID-BASE EQUILIBRIA AND SOLUBILITY EQUILIBRIA 381 We can calculate the decrease in [HNO 2 ] by setting up the weak acid equilibrium. From the ph of the solution, we know that the [H + ] at equilibrium is 9.55 10 4 M. HNO 2 (aq) H + (aq) + NO 2 (aq) Initial (M) 1.33 0 0 Change (M) x +x Equilibrium (M) 1.33 x 9.55 10 4 x We can calculate x from the equilibrium constant expression. K a = + [H ][NO 2 ] [HNO 2 ] 4.5 10 4 = x = 0.426 M 4 (9.55 10 )( x) 1.33 x Thus, x is the decrease in [HNO 2 ] which equals the concentration of added OH. However, this is the concentration of NaOH after it has been diluted to 600 ml. We need to correct for the dilution from 200 ml to 600 ml to calculate the concentration of the original NaOH solution. M I V I = M F V F M I (200 ml) = (0.426 M)(600 ml) [NaOH] = M I = 1.28 M 16.83 The K a of butyric acid is obtained by taking the antilog of 4.7 which is 2 10 5. The value of K b is: 14 K 10 K b = w 10. 10 = = 5 10 K 5 a 2 10 16.84 The resulting solution is not a buffer system. There is excess NaOH and the neutralization is well past the equivalence point. We have a solution of sodium acetate and sodium hydroxide. Moles NaOH = 0500. L F HG Moles CH 3 COOH = 0500. L 0167. mol 1 L F HG I KJ = 0100. mol 1 L The reaction between sodium hydroxide and acetic acid is: 00835. mol I KJ = 00500. mol NaOH (aq) + CH 3 COOH (aq) NaCH 3 COO (aq) + H 2 O (l) Initial (mol) 0.0835 0.0500 0 After Reaction (mol) 0.0335 0 0.0500 Since the total volume of the solution is 1.00 L, we can convert the number of moles directly to molarity. [OH ] = 0.0335 M [Na + ] = 0.0335 M + 0.0500 M = 0.0835 M [CH 3 COO ] = 0.0500 M

382 CHAPTER 16: ACID-BASE EQUILIBRIA AND SOLUBILITY EQUILIBRIA We can calculate the [H + ] from the [OH ]. [H + ] = 14 Kw 1.00 10 = [OH ] 0.0335 = 2.99 10 13 M Finally, we can calculate the acetic acid concentration from the K a expression. or, + [H ][CH3COO ] K a = [CH3COOH] + [H ][CH3COO ] [CH3COOH] = Ka 13 (2.99 10 )(0.0500) [CH 3 COOH] = 5 1.8 10 = 8.3 10 10 M 16.85 Most likely the increase in solubility is due to complex ion formation: Cd(OH) 2 (s) + 2OH Cd(OH) 4 2 (aq) 16.86 The number of moles of Ba 2+ present in the original 50.0 ml of solution is: 2+ 1.00 mol Ba 1 L 2+ 50.0 ml = 0.0500 mol Ba 1 L soln 1000 ml 2 The number of moles of SO 4 present in the original 86.4 ml of solution, assuming complete dissociation, is: 2 0.494 mol SO4 1L 2 86.4 ml = 0.0427 mol SO4 1 L soln 1000 ml The complete ionic equation is: Ba 2+ (aq) + 2 OH (aq) + 2 H + 2 (aq) + SO 4 (aq) BaSO4 (s) + H 2 O (l) Initial (mol): 0.0500 0.100 0.0854 0.0427 0 Change (mol) 0.0427 2(0.0427) 2(0.0427) 0.0427 +0.0427 After rxn (mol) 0.0073 0.015 0 0 0.0427 Thus the mass of BaSO 4 formed is: (0.0427 mol BaSO 4) 233.4 g BaSO4 = 9.97 g BaSO 4 1molBaSO4 The ph can be calculated from the excess OH in solution. First, calculate the molar concentration of OH. The total volume of solution is 136.4 ml = 0.1364 L. 0.015 mol [OH ] = 0.1364 L = 0.11 M poh = log(0.11) = 0.96 ph = 14.00 poh = 14.00 0.96 = 13.04

CHAPTER 16: ACID-BASE EQUILIBRIA AND SOLUBILITY EQUILIBRIA 383 16.87 A solubility equilibrium is an equilibrium between a solid (reactant) and its components (products: ions, neutral molecules, etc.) in solution. Only (d) represents a solubility equilibrium. Consider part. Can you write the equilibrium constant for this reaction in terms of K sp for calcium phosphate? 16.88 First, we calculate the molar solubility of CaCO 3. CaCO 3 (s) Ca 2+ 2 (aq) + CO 3 (aq) initial (M):? 0 0 change (M): s +s +s equil. (M):? s s s The moles of CaCO 3 in the kettle is: K sp = [Ca 2+ ][CO 3 2 ] = s 2 = 8.7 10 9 s = 9.3 10 5 M = 9.3 10 5 mol/l 116 g 100.1 g / mol = 1.16 moles CaCO3 The volume of distilled water needed to dissolve 1.16 moles of CaCO 3 is: 1L 4 1.16 mol CaCO3 = 1.2 10 L 5 9.3 10 mol CaCO3 The number of times the kettle would have to be filled is: 4 1.2 10 L 2.0 L per filling = 3 6.0 10 fillings Note that the very important assumption is made that each time the kettle is filled, the calcium carbonate is allowed to reach equilibrium before the kettle is emptied. 16.89 Since equal volumes of the two solutions were used, the initial molar concentrations will be halved. [Ag + ] = [Cl ] = 012. M 2 2014 (. M ) 2 = 0.060 M = 0.14 M Let s assume that the Ag + ions and Cl ions react completely to form AgCl (s). Then, we will reestablish the equilibrium between AgCl, Ag +, and Cl. Ag + (aq) + Cl (aq) AgCl (s) Initial (M) 0.060 0.14 0 Change (M) 0.060 0.060 +0.060 After reaction (M) 0 0.080 0.060

384 CHAPTER 16: ACID-BASE EQUILIBRIA AND SOLUBILITY EQUILIBRIA Now, setting up the equilibrium, AgCl(s) Ag + (aq) + Cl (aq) Initial (M): 0.060 0 0.080 Change (M): s +s +s Equilibrium (M): 0.060 s s 0.080 + s Set up the K sp expression to solve for s. K sp = [Ag + ][Cl ] 1.6 10 10 = (s)(0.080 + s) s = 2.0 10 9 M [Ag + ] = s = 2.0 10 9 M [Cl ] = 0.080 M s = 0.080 M [Zn 2+ 014. M ] = = 0.070 M 2 012. M [NO 3 ] = = 0.060 M 2 16.90 First we find the molar solubility and then convert moles to grams. The solubility equilibrium for silver carbonate is: Ag 2 CO 3 (s) 2Ag + 2 (aq) + CO 3 (aq) K sp = [Ag + ] 2 [CO 3 2 ] = (2s) 2 (s) = 4s 3 = 8.1 10 12 s = 1/3 12 8.1 10 4 4 = 1.3 10 M Converting from mol/l to g/l: 4 1.3 10 mol 275.8 g 1Lsoln 1mol = 0.036 g/l 16.91 For Fe(OH) 3, K sp = 1.1 10 36. When [Fe 3+ ] = 0.010 M, the [OH ] value is or K sp = [Fe 3+ ][OH ] 3 [OH ] = [OH ] = F HG F HG K sp 3+ [ Fe ] I KJ 1 3 36 11. 10 0010. I KJ 1 3 = 4.8 10 12 M This [OH ] corresponds to a ph of 2.68. In other words, Fe(OH) 3 will begin to precipitate from this solution at ph of 2.68.

CHAPTER 16: ACID-BASE EQUILIBRIA AND SOLUBILITY EQUILIBRIA 385 For Zn(OH) 2, K sp = 1.8 10 14. When [Zn 2+ ] = 0.010 M, the [OH ] value is [OH ] = [OH ] = F HG F HG K sp 2+ [ Zn ] I KJ 1 2 14 18. 10 0010. I KJ 1 2 6 = 1.3 10 M This corresponds to a ph of 8.11. In other words Zn(OH) 2 will begin to precipitate from the solution at ph = 8.11. These results show that Fe(OH) 3 will precipitate when the ph just exceeds 2.68 and that Zn(OH) 2 will precipitate when the ph just exceeds 8.11. Therefore, to selectively remove iron as Fe(OH) 3, the ph must be greater than 2.68 but less than 8.11. 16.92 (a) To 2.50 10 3 mol HCl (that is, 0.0250 L of 0.100 M solution) is added 1.00 10 3 mol CH 3 NH 2 (that is, 0.0100 L of 0.100 M solution). After the acid-base reaction, we have 1.50 10 3 mol of HCl remaining. Since HCl is a strong acid, the [H + ] will come from the HCl. The total solution volume is 35.0 ml = 0.0350 L. [H + ] = ph = 1.37 3 1.50 10 mol 0.0350 L = 0.0429 M When a total of 25.0 ml of CH 3 NH 2 is added, we reach the equivalence point. That is, 2.50 10 3 mol HCl reacts with 2.50 10 3 mol CH 3 NH 2 to form 2.50 10 3 mol CH 3 NH 3 Cl. Since there is a total of + 50.0 ml of solution, the concentration of CH 3 NH 3 is: 3 + 2.50 10 mol [CH 3 NH 3 ] = 0.0500 L 2 = 5.00 10 M This is a problem involving the hydrolysis of the weak acid CH 3 NH 3 + K a = 2.3 10 11 = 1.15 10 12 = x 2 + 2 2 [CH3NH 2][H ] x x = + 2 [CH3NH 3 ] 5.00 10 5.00 10 2 ( x) x = 1.07 10 6 M = [H + ] ph = 5.97 (c) When a total of 35.0 ml of 0.100 M CH 3 NH 2 (3.50 10 3 mol) is added to the 25 ml of 0.100 M HCl (2.50 10 3 mol), the acid-base reaction produces 2.50 10 3 mol CH 3 NH 3 Cl with 1.00 10 3 mol of CH 3 NH 2 in excess. Using the Henderson-Hasselbalch equation: [conjugate base] ph = pk a + log [acid] ph = log(2.3 10 11 3 (1.00 10 ) ) + log 3 (2.50 10 ) = 10.24

386 CHAPTER 16: ACID-BASE EQUILIBRIA AND SOLUBILITY EQUILIBRIA 16.93 The equilibrium reaction is: Pb(IO 3 ) 2 (aq) Pb 2+ (aq) + 2 IO 3 (aq) Initial (M) 0 0.10 Change (M) 2.4 10 11 +2.4 10 11 +2(2.4 10 11 ) Equilibrium (M) 2.4 10 11 0.10 Substitute the equilibrium concentrations into the solubility product expression to calculate K sp. K sp = [Pb 2+ ][IO 3 ] 2 K sp = (2.4 10 11 )(0.10) 2 = 2.4 10 13 16.94 The precipitate is HgI 2. Hg 2+ (aq) + 2 I (aq) HgI 2 (s) With further addition of I, a soluble complex ion is formed and the precipitate redissolves. HgI 2 (s) + 2 I (aq) HgI 4 2 (aq) 16.95 K sp = [Ba 2+ ][SO 4 2 ] = 1.1 10 10 [Ba 2+ ] = 1.0 10 5 M In 5.0 L, the number of moles of Ba 2+ is (5.0 L)(1.0 10 5 mol/l) = 5.0 10 5 mol Ba 2+ = 5.0 10 5 mol BaSO 4 The number of grams of BaSO 4 dissolved is 5 233.4 g BaSO4 5.0 10 mol BaSO4 = 0.012 g BaSO4 1molBaSO4 In practice, even less BaSO 4 will dissolve because the BaSO 4 is not in contact with the entire volume of blood. Ba(NO 3 ) 2 is too soluble to be used for this purpose. 16.96 We can use the Henderson-Hasselbalch equation to solve for the ph when the indicator is 95% acid / 5% conjugate base and when the indicator is 5% acid / 95% conjugate base. [conjugate base] ph = p K a + log [acid] Solving for the ph with 95% of the indicator in the HIn form: [5] ph = 9.10 + log = 9.10 1.28 = 7.82 [95] Next, solving for the ph with 95% of the indicator in the In form: [95] ph = 9.10 + log = 9.10 + 1.28 = 10.38 [5] Thus the ph range varies from 7.82 to 10.38 as the [HIn] varies from 95% to 5%.

CHAPTER 16: ACID-BASE EQUILIBRIA AND SOLUBILITY EQUILIBRIA 387 16.97 (a) The solubility product expressions for both substances have exactly the same mathematical form and are therefore directly comparable. The substance having the smaller K sp (AgI) will precipitate first. (Why?) When CuI just begins to precipitate the solubility product expression will just equal K sp (saturated solution). The concentration of Cu + at this point is 0.010 M (given in the problem), so the concentration of iodide ion must be: K sp = [Cu + ][I ] = (0.010)[I ] = 5.1 10 12 [I ] = 12 51. 10 0010. = 5.1 10 10 M Using this value of [I ], we find the silver ion concentration [Ag + ] = K sp [ I ] = 17 83. 10 10 51. 10 = 1.6 10 7 M (c) The percent of silver ion remaining in solution is: % Ag + (aq) = 7 16. 10 M 0010. M 100% = 0.0016% or 1.6 10 3 % Is this an effective way to separate silver from copper? 16.98 (a) We abbreviate the name of cacodylic acid to CacH. We set up the usual table. CacH (aq) Cac (aq) + H + (aq) Initial(M): 0.10 0.00 0.00 Change(M) x +x +x Equilibrium(M): (0.10 x) x x K a = + 2 [H ][Cac ] x 7 = = 6.4 10 CacH 0.10 x [ ] We assume that (0.10 x) 0.10. Then, x = 2.5 10 4 M = [H + ] ph = log(2.5 10 4 ) = 3.60 We set up a table for the hydrolysis of the anion: Cac (aq) + H 2 O (l) CacH (aq) + OH (aq) Initial (M): 0.15 0.00 0.00 Change(M): x +x +x Equilibrium(M): (0.l5 x) x x The ionization constant, K b, for Cac is: K b = Kw Ka 14 1.0 10 8 = = 1.6 10 7 6.4 10

388 CHAPTER 16: ACID-BASE EQUILIBRIA AND SOLUBILITY EQUILIBRIA 2 x 0.15 x 8 = 1.6 10 x = 4.9 10 5 M poh = log(4.9 10 5 ) = 4.31 ph = 14.00 4.31 = 9.69 (c) Number of moles of CacH from (a) is: 0.10 mol CacH 3 50.0 ml CacH = 5.0 10 mol CacH 1000 ml Number of moles of Cac from is: 0.15 mol CacNa 3 25.0 ml CacNa = 3.8 10 mol CacNa 1000 ml At this point we have a buffer solution. 3 [Cac ] 7 3.8 10 ph = pk a + log = log(6.4 10 ) + log = [CacH] 3 5.0 10 6.07 16.99 The initial number of moles of Ag + is Ag + = + 0010. mol Ag 1 L 1 L 50 ml = 5.0 10 4 mol Ag + 1000 ml We can use the counts of radioactivity as being proportional to concentration. Thus, we can use the ratio to determine the quantity of Ag + still in solution. However, since our original 50 ml of solution has been diluted to 500 ml, the counts per ml will be reduced by ten. Our diluted solution would then produce 7402.5 counts per minute if no removal of Ag + had occurred. The number of moles of Ag + that correspond to 44.4 counts are: 44.4 counts 4 + 50. 10 mol Ag 7402. 5 counts = 3.0 10 6 mol Ag + 0030. mol IO The original moles of IO 3 = 3 1 L 1 L 100 ml = 3.0 10 3 mol 1000 ml The quantity of IO 3 remaining after reaction with Ag + : (original moles moles reacted with Ag + ) = (3.0 10 3 mol) (5.0 10 4 mol) = 2.5 10 3 mol IO 3

CHAPTER 16: ACID-BASE EQUILIBRIA AND SOLUBILITY EQUILIBRIA 389 The total final volume is 500 ml or 0.50 L. [Ag + ] = 6 + 30. 10 mol Ag 050. L = 6.0 10 6 M 3 25. 10 mol IO [IO 3 ] = 3 050. L = 5.0 10 3 M K sp = [Ag + ][IO 3 ] = (6.0 10 6 )(5.0 10 3 ) = 3.0 10 8 16.100 (a) MCO 3 + 2HCl MCl 2 + H 2 O + CO 2 HCl + NaOH NaCl + H 2 O Moles of HCl reacted with MCO 3 = Total moles of HCl Moles of excess HCl Total moles of HCl = 20.00 ml 1 L 0.0800 mol = 1.60 10 3 mol HCl 1000 ml 1 L Moles of excess HCl = 1 L 0.1000 mol 4 5.64 ml = 5.64 10 mol HCl 1000 ml 1 L Moles of HCl reacted with MCO 3 = 1.60 10 3 mol 5.64 10 4 = 1.04 10 3 mol HCl Moles of MCO 3 reacted = 3 1molMCO3 1.04 10 mol HCl = 5.20 10 4 mol MCO 3 2molHCl Molar mass of MCO 3 = 0.1022 g 4 5.20 10 mol = 197 g/mol Molar mass of CO 3 = 60.01 Molar mass of M = 197 g/mol 60.01 g/mol = 137 g/mol The metal, M, is Ba! 16.101 (a) H + + OH H 2 O K = 1.0 10 14 H + + NH 3 NH 4 + K = 1 1 = K 56 10 10 a. = 1.8 10 9 (c) CH 3 COOH + OH CH 3 COO + H 2 O Broken into 2 equations: CH 3 COOH CH 3 COO + H + H + + OH H 2 O K a 1/K w K = Ka Kw = 5 18. 10 14 10. 10 = 1.8 10 9