86 There is no effect on the equilibrium state if more solid is added since the

86 there is no effect on the equilibrium state if

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86 . 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. 87. Follow the method developed in the answers to Questions 51 - 58 CaSO 4 (s) Ca 2+ (aq) + SO 4 2– (aq) conc. initial (M) 0 0 change conc. (M) + S + S equilibrium conc. (M) S S At equilibrium 2.03 g CaSO 4 1 L × 1 mol CaSO 4 136.14 g CaSO 4 = 0.0149 M = S K sp = [Ca 2+ ][SO 4 2– ] = (S)(S) = S 2 = (0.0149) 2 = 2.22 × 10 –4 88 . Follow the method developed in the answer to Question 51 - 58. Ag 2 CrO 4 (s) 2 Ag + (aq) + CrO 4 2– (aq) conc. initial (M) 0 0 change conc. (M) + 2S + S equilibrium conc. (M) 2S S At equilibrium, 2.7 × 10 3 g Ag 2 CrO 4 100. mL × 1 mol Ag 2 CrO 4 331.8 g Ag 2 CrO 4 × 1000 mL 1 L = 8.1 × 10 –5 M = S K sp = [Ag + ] 2 [CrO 4 2– ] = (2S) 2 (S) = 4S 3 = 4 × (8.1 × 10 –5 ) 3 = 2.2 × 10 –12 Applying Concepts 89. Calculate the molar Ca 2+ concentration. 5.33 g Ca 2 + L × 1 mol Ca 2 + 40.078 g Ca 2 + = 0.133 M Ca 2+ (a) Use K sp for CaC 2 O 4 to calculate the necessary C 2 O 4 2– concentration. K sp = [Ca 2+ ][C 2 O 4 2– ] 2.3 × 10 –9 = (0.133)[C 2 O 4 2– ] [C 2 O 4 2– ] = 1.7 × 10 –8 M (b) Use K sp for Ca 3 (PO 4 ) 2 to calculate the necessary PO 4 3– concentration. K sp = [Ca 2+ ] 3 [PO 4 3– ] 2 1 × 10 –25 = (0.133) 3 [PO 4 3– ] 2 [PO 4 3– ] = 7 × 10 –12 M
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Chapter 17: Additional Aqueous Equilibria 751 90 . The tiny amount of base (CH 3 COO ) present is insufficient to prevent the pH from changing dramatically if a strong acid is introduced into the solution. 91. Use the method described in the answers to Questions 38 - 41. 33.5 mL NaOH 5.00 mLvinegar × 0.100 mol NaOH 1 L NaOH × 1 mol CH 3 COOH 1 mol NaOH = 0.670 M CH 3 COOH in the sample. Yes, this vinegar meets the legal limit of 4 % (0.67 M). 92 . When exactly half of the acid in the original solution has been neutralized, that means that equal quantities of the acid and base are present in the solution. As described in the answer to Question 18, when equimolar quantities are present, pH = pK a . pH = 3.64 = pK a K a = 10 –pKa = 10 –3.64 = 2.3 × 10 –4 93. 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. pH = pK a + log [base] [acid] 10 = 10.25 + log [CO 3 2 ] [HCO 3 ] If he used the H 2 CO 3 /HCO 3 buffer components instead, he would have to use pK a = 6.38 (from Table 17.1) pH = 6.38 + log [CO 3 2 ] [HCO 3 ] = 6.38 + log(1) = 6.38 The resulting buffer pH would be 6.38. 94 . Blood pH decreases because of an increase in H 2 CO 3 , which leads to acidosis, acidification of the blood. 95. The dissolution reaction is given by the following chemical equation. As the solid dissolves, one mole of fluoride ions are created for each half mole of calcium ion. CaF 2 (s) Ca 2+ F (aq) conc. init. (M) solid 0.070 0 change conc. (M) + 1 2 S + S eq conc. (M) solid 0.070 + 1 2 S + S Use the solubility product and the calcium ion concentration to determine the concentration of fluoride ion at equilibrium.
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  • Spring '08
  • Kerber
  • Chemistry, pH, Additional Aqueous Equilibria

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