Chemistry 122 [Tyvoll] Spring 2008

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1 Chemistry 122 [Tyvoll] Spring 2008 Chapter 17 Homework Solutions for Moore Problems # 1, 2, 5, 13, 16, 19, 21, 24, 34, 36, 39, 40, 47, 51, 57, 62, 65, 67, 68, 79, 82, 88, The buffer capacity of a buffer solution is related to the concentrations of the buffer s acid and base. The buffer solution is able to resist changes in ph only until the total amount of added acid or added base completely neutralizes one component of the buffer solution. The higher the buffer concentrations are, the higher the buffer capacity is. 2. A weak acid-base conjugate pair must be in the solution. The weak base can neutralize added acid making more of the weak base, and the weak acid can neutralize added base making more of the weak acid; hence, the pair together makes a buffer solution capable of resisting ph. (a) A strong acid, HCl, and a weak acid, CH 3 COOH, together in a system are able to neutralize added base, but are not able to neutralize added acid, so this pair does not form a buffer. (b) The dissolved NaH 2 PO 4 salt contains the weak acid H 2 PO 4, and the dissolved Na 2 HPO 4 salt contains its conjugate weak base, HPO 4 2, so this pair would form a buffer. (c) The solution contains the weak acid H 2 CO 3 and a dissolved salt, NaHCO 3, containing its conjugate weak base, HCO 3, so this pair would form a buffer. 5. The end point is the term used during an acid base titration describing the volume of added titrant at which an appropriate indicator being used changes color indicators colors changes occur only based on the solution s ph. The equivalence point is the point during a titration describing the volume of added titrant at which the moles of reactant acid are equal to the moles of reactant base. With the proper choice of indicator, these two points should coincide, though they usually will not be exactly the same. The ph changes rapidly near the equivalence point, so the indicator will changes color close to the ph of the equivalence point with almost the same amount of titrant. 13. The ion product and solubility product expression have the same functional form. Both of them are the mathematical product of the concentrations of the ions in a solution. However, in the ion product, the concentrations need not be equilibrium concentrations. In the solubility product expression, the concentrations must be equilibrium concentrations. 16. To determine the ph of a buffer, look up the pk a (Table 17.1) or look up the K a (Table 16.2) and calculate the pk a (pk a = log K a ). The pk a closest to the desired ph is the best buffer, since close to equal quantities of the acid and base would be used, giving the solution approximately equal ability to neutralize added acid or added base. (a) The CH 3 COOH/NaCH 3 COO buffer system has pk a = (b) The acid in HCl/NaCl is HCl. It has Ka = very large. This is not a buffer. (c) The NH 3 /NH 4 Cl buffer system has pk a = The combination that would make the best ph 9 buffer system is (c) NH 3 /NH 4 Cl.

2 19. Compare the [H 3 O + ] to the values of K a in Table 17.1, since that is equal to the [H 3 O + ] in an 1:1 buffer solution. The K a closest to the desired [H 3 O + ] is the most suitable. (a) [H 3 O+] = 4.5 x 10 3 M, needs a lactic acid/lactate buffer (K a = 1.4 x 10 4 ). (b) [H 3 O+] = 5.2 x 10 8 M, needs a dihydrogen phosphate/hydrogen phosphate buffer (K a = 6.2 x 10 8 ). (c) [H 3 O+] = 8.3 x 10 6 M, needs an acetic acid/acetate buffer (K a = 1.8 x 10 5 ). (d) [H 3 O+] = 9.7 x M, needs a hydrogen carbonate/carbonate buffer (Ka = 4.8 x ). 21. Use ph = pka + log {[base]/[acid]} 4.57 = log {[base]/(0.150)} 0.17 = log {[base]/(0.150)} = = {[base]/(0.150)} [base] = (0.676)( M) = 0.10 M g NaCH 3 COO = (0.500 L)(0.10 mol/l)( g/mol) = 4.2 g 24. Calculate the initial concentration of NH 4 +, the conjugate acid, where C(NH 4 + ) = (5.15 g/0.10 L)(1 mol/80.04 g)(1 mol NH 4 + /1 mol NH 4 NO 3 = 0.64 M NH 4 + Table 17.1 gives pk a of ammonium/ammonia buffer as 9.25, and the initial concentration of NH 3, the conjugate base, is 0.15 M. so, ph = log {0.15/0.64) = ( 0.63) = At the equivalence point, the solution contains the conjugate base of the weak acid. Weaker acids have stronger conjugate bases. Stronger bases have higher ph, so the titration of a weaker acid will have a more basic equivalence point. 36. The color change needs to occur as close as possible to the ph of the equivalence point. (a) The strong base, NaOH, titrated with a strong acid, HClO 4, has a neutral equivalence point. It would be best to choose bromthymol blue, which is shown changing color in Figure 17.5 at a ph near 7. In practice, any of them would be suitable, because of the extreme change in the ph of the solution very close to the equivalence point. (b) The weak acid, CH 3 COOH, titrated with a strong base, KOH, has a basic equivalence point, due to the presence of the weak base CH 3 COO in the solution. It would be best to choose phenolphthalein, which is shown changing color in Figure 17.5 at a ph near 9. (c) The weak base, NH 3, titrated with a strong acid, HBr, has an acidic equivalence point, due to the presence of the weak acid NH 4 + in the solution. It would be best to choose methyl red, which is shown changing color in Figure 17.8 at a ph near 5.

3 36. (d) The strong base, KOH, titrated with a strong acid, HNO 3, has a neutral equivalence point. It would be best to choose bromthymol blue, which is shown changing color in Figure 17.5 at a ph near 7. In practice, any of them would be suitable, because of the extreme change in the ph of the solution very close to the color in Figure 17.8 at a ph near M NaOH = {(12.4 ml H 2 SO 4 /20.0 ml NaOH)(1 L H 2 SO 4 /1000 ml H 2 SO 4 )( x (1000 ml NaOH/1 L NaOH)(0.205 mol H 2 SO 4 /1 L H 2 SO 4 )(2 mol H + /1 mol H 2 SO 4 ) x (1 mol NaOH/1 mol H + )} = mol NaOH/L NaOH = M 40. g C 6 H 8 O 6 = {(24.4 ml NaOH)(1 L/1000 ml)(0.110 mol NaOH/1 L NaOH) x (1 mol C 6 H 8 O 6 /1 mol NaOH)( g C 6 H 8 O 6 /1 mol C 6 H 8 O 6 ) = g C 6 H 8 O The ph in rain always contains some dissolved CO2 from the air. As described in Section 17.3, atmospheric carbon dioxide dissolves and reacts with rainwater to form a solution of weak carbonic acid. Rainwater in equilibrium with dissolved CO 2 has a ph of 5.6. Any water less acidic than this, even if it is acidic by the chemical definition, is still not called acid rain. 51. Adapt the method developed in Problem-Solving Example (a) BaCrO 4 (s) Ba 2+ (aq) + CrO 2 4 (aq) Ksp = [Ba 2+ ][CrO 2 4 ] (b) Mn(OH) 2 (s) Mn 2+ (aq) + 2 OH (aq) Ksp = [Mn 2+ ][OH ] 2 (c) Ag 2 SO 4 (s) 2 Ag + (aq) + SO 2 4 (aq) Ksp = [Ag + ] 2 [SO 2 4 ] 57. Write the chemical equation and the equilibrium expression for the dissociation of the solute: SrF 2 (s) Sr 2+ (aq) + 2 F (aq) K sp = [Sr 2+ ][F ] 2 At equilibrium, the moles of solid that dissolve per liter is M = {(0.011 g SrF 2 /100. ml)(1 mol SrF 2 / g SrF 2 )(1000 ml/1 L) = 8.8 x 10 4 M The stoichiometry of the equation shows that the concentration of strontium ion is the same as the moles of solid that dissolve per liter, 8.8 x 10 4 M, and the fluoride ion concentration is two times that value, or, 2 (8.8 x 10 4 M) = 1.8 x 10 4 M. K sp = (8.8 x 10 4 )(1.8 x 10 4 ) 2 = 2.7 x Use the equilibrium expression for K sp and the known values to determine the unknown value. Get the value of Ksp from Appendix H. Na 2 CO 3 produces carbonate ions in solution, so we use the base hydrolysis reaction where CO 3 2 (aq) + H 2 O (l) HCO 3 (aq) + OH (aq) ICE Table [CO 2 3 ] (aq)] [HCO 3 ] (aq)] [OH ] (aq)] I C x +x +x E 0.25 x x x

4 62. K b = (x)(x)/(0.25 x) = 2.1 x 10 4 x = [OH ] = M and [CO 3 2 ] = 0.25 x = = 0.24 M Finally, Ksp = [Zn 2+ ][CO 3 2 ] = [Zn 2+ ](0.24) = 1.5 x 10 11, [Zn 2+ ] = mol/l 65. (a) Write the chemical equation and the equilibrium expression for the dissociation of the solute: Fe (OH) 2 (s) Fe 2+ (aq) + 2 OH (aq) K sp = [Fe 2+ ][OH ] 2 At equilibrium, the moles of solid per liter are: M = (6.0 x 10 1 mg Fe(OH) 2 /L)(1 g/1000 mg)(1 mol Fe(OH) 2 / g Fe(OH) 2 ) M = 6.7 x 10 6 M The stoichiometry of the equation shows that the iron(ii) ion concentration is the identical to the number of moles of solid that dissolve per liter, 6.7 x 10 6 M, and the hydroxide ion concentration is two times that value, 2(6.7 x 10 6 M) = M, so Ksp = (6.7 x 10 6 )(1.3 x 10 5 )2 = 1.1 x (b) First we determine the concentration of the iron (II) ion in the Fe 2+ ion solution: [Fe 2+ ] = (1.0 μg Fe 2+ /L)(10 6 g/1 μg)(1 mol Fe 2+ / g Fe 2+ ) = 1.8 x 10 8 M Next, use the calculated K sp from (a) to calculate the hydroxide ion concentration in this solution: Ksp = 1.1 x = (1.8 x 10 8 )[OH ] 2 [OH ] = 2.6 x 10 4 M An equilibrium [OH ] of 2.6 x 10 4 M or higher will keep the [Fe 2+ ] at or below 1.0 μg/l 67. Write the chemical equation and the equilibrium expression for the dissociation of the solute: AgCl (s) Ag + (aq) + Cl (aq) Ksp = [Ag + ][Cl ] Get the value of K sp = 1.8 x from Table 17.2 or Appendix H. The soluble AgNO 3 salt produces silver ions in the solution. [Ag + ] = 0.05 mol/l. Ksp = 1.8 x = (0.05)[Cl ] [Cl ] = 4 x 10 9 M

5 68. Given the ph, we can determine the concentration of the aqueous ion, OH. The precipitation of sparingly soluble Zn(OH) 2 will occur above a certain concentration of Zn 2+. [OH ] = 10 poh = 10 ph = = 1.0 x 10 4 M K sp = [Zn 2+ ][OH ] 2 = 4.5 x (from Appendix H) K sp = 4.5 x = [Zn 2+ ](1.0 x 10 4 ) 2 [Zn 2+ ] = 4.5 x 10 9 M An equilibrium [Zn 2+ ] of 4.5 x 10 9 M or lower can exist in a solution with ph Above that concentration, Zn(OH) 2 will precipitate. 79. Use ph = pka + log {[base]/[acid]} 9.00 = log {[0.93]/ [NH 4 + ]} Table 17.1 gives pk a of ammonium/ammonia buffer as The ph is The initial concentration of NH 3, the conjugate base, is 0.93 M. The conjugate acid, NH 4 +, is being added, but its initial concentration is unknown, so use [NH 4 + ]to identify that quantity = 0.25 = log {(0.93 M)/ [NH 4 + ]} = {(0.93 M)/ [NH 4 + ]} [NH 4 + ] = (0.93 M)( = 1.7 M 82. [NO 2 ] = (7.50 g KNO 2 /L)(1 mol KNO 2 /85.11 g KNO 2 )(1 mol NO 2 /1 mol KNO 2 ) [NO 2 ] = M In Table 16.2, the K a of dihygydrogen phosphate is given: 4.5 x As described in Section 17.1, calculate the pk a : pk a = logk a = log (4.5 x 10 4 ) = 3.35 The ph is The initial concentration of NO 2, the conjugate base, is M. The conjugate acid, HNO 2, is being added, but its initial concentration is unknown, so use [HNO 2 ] to identify it. Use ph = pka + log {[base]/[acid]} 4.00 = = 0.65 = log {[ M]/[HNO 2 ]} = {[ M]/[HNO 2 ]} = 4.47 [HNO 2 ] = {[ M]/ 4.47} = M 88. There is no effect on the equilibrium state if more solid is added, since the quantity of that solid present does not affect the equilibrium state. 95. The ph 10 buffer is made with a hydrogen carbonate/carbonate buffer system (according to Table 17.1). The lab technician wrote down the wrong acid/base conjugate pair: carbonic acid/hydrogen carbonate, H 2 CO 3 /HCO 3. ph = pka + log {[base]/[acid]} = log {[CO 3 2 ]/[HCO 3 ] 0.32 = log {[CO 3 2 ]/[HCO 3 ]

6 95. If the technician put equimolar quantities (a 1:1 ratio) of HCO 3 and CO 3 2 into the same solution, the concentration ratio of [CO 3 2 ]/[HCO 3 ] would be ph = log {[CO 3 2 ][HCO 3 ]} = log (1.00) = 10.32