18. SOUBIITY AND COMPEX-ION EQUIIBRIA Solutions to Exercises Note on significant figures: If the final answer to a solution needs to be rounded off, it is given first with one nonsignificant figure, and the last significant figure is underlined. The final answer is then rounded to the correct number of significant figures. In multiple-step problems, intermediate answers are given with at least one nonsignificant figure; however, only the final answer has been rounded off. 18.1 a. BaSO 4 (s) Ba + (aq) + SO 4 - (aq) K sp [Ba + ][SO 4 - ] b. Fe(OH) 3 (s) Fe 3+ (aq) + 3OH - (aq) K sp [Fe 3+ ][OH - ] 3 c. Ca 3 (PO 4 ) (s) 3Ca + (aq) + PO 4 3- (aq); K sp [Ca + ] 3 [PO 4 3- ] 18. Calculate the molar solubility. Then, assemble the usual concentration table, and substitute the equilibrium concentrations from it into the equilibrium-constant expression. (Because no concentrations can be given for solid AgCl, dashes are written; in later problems, similar spaces will be left blank.) -3 1.9 x 10 g 1 x 1 mol 143 g 1.33 x 10-5 M (continued)
SOUBIITY AND COMPEX-ION EQUIIBRIA 679 Set up the table as usual. Conc. (M) AgCl(s) Ag + + Cl - Starting 0 0 Change +1.33 x 10-5 +1.33 x 10-5 Equilibrium 1.33 x 10-5 1.33 x 10-5 K sp [Ag + ][Cl - ] (1.33 x 10-5 )(1.33 x 10-5 ) 1.768 x 10-10 1.8 x 10-10 18.3 Calculate the molar solubility. Then, assemble the usual concentration table and substitute from it into the equilibrium-constant expression. (Because no concentrations can be given for solid Pb 3 (AsO 4 ), spaces are left blank.) -5 3.0 x 10 g 1 x 1 mol 899 g 3.34 x 10-8 M Conc. (M) Pb 3 (AsO 4 ) (s) 3Pb + + 3- AsO 4 Starting 0 0 Change +3(3.34 x 10-8 ) +(3.34 x 10-8 ) Equilibrium 3(3.34 x 10-8 ) (3.34 x 10-8 ) K sp [Pb + ] 3 [AsO 3-4 ] [3 x (3.34 x 10-8 )] 3 ( x 3.34 x 10-8 ) 4.489 x 10-36 4.5 x 10-36 18.4 Assemble the usual concentration table. et x equal the molar solubility of CaSO 4. When x mol CaSO 4 dissolves in one of solution, x mol Ca + and x mol SO 4 - form. Conc. (M) CaSO 4 (s) Ca + + SO 4 - Starting 0 0 Change +x +x Equilibrium x x Substitute the equilibrium concentrations into the equilibrium-constant expression and solve for x. Then, convert to g CaSO 4 per. (continued)
680 CHAPTER 18 [Ca + ][SO 4 - ] K sp (x)(x) x.4 x 10-5 -5 x (.4 x 10 ) 4.89 x 10-3 M -3 4.89 x 10 mol x 136 g 1mol 0.6664 0.67 g/ 18.5 a. et x equal the molar solubility of BaF. Assemble the usual concentration table, and substitute from the table into the equilibrium-constant expression. Conc. (M) BaF (s) Ba + + F - Starting 0 0 Change +x +x Equilibrium x x [Ba + ][F - ] K sp (x)(x) 4x 3 1.0 x 10-6 x 3-6 1.0 x 10 4 6.99 x 10-3 6.3 x 10-3 M b. At the start, before any BaF dissolves, the solution contains 0.15 M F -. At equilibrium, x mol of solid BaF dissolves to yield x mol Ba + and x mol F -. Assemble the usual concentration table, and substitute the equilibrium concentrations into the equilibriumconstant expression. As an approximation, assume x is negligible compared to 0.15 M F -. Conc. (M) BaF (s) Ba + + F - Starting 0 0.15 Change +x +x Equilibrium x 0.15 + x [Ba + ][F - ] K sp (continued)
(x)(0.15 + x) (x)(0.15) 1.0 x 10-6 SOUBIITY AND COMPEX-ION EQUIIBRIA 681 x -6 1.0 x 10 (0.15) 4.444 x 10-5 4.4 x 10-5 M Note that adding x to 0.15 M will not change it (to two significant figures), so x is negligible compared to 0.15 M. The solubility of 4.4 x 10-5 M in 0.15 M NaF is lower than the solubility of 6.3 x 10-3 M in pure water. 18.6 Calculate the ion product, Q c, after evaporation, assuming no precipitation has occurred. Compare it with the K sp. Q c [Ca + ][SO 4 - ] Q c ( x 0.005)( x 0.0041) 8.58 x 10-5 Since Q c > K sp (.4 x 10-5 ), precipitation occurs. 18.7 Calculate the concentrations of Pb + and SO 4 -, assuming no precipitation. Use a total volume of 0.456 + 0.55, or 0.711. [Pb + ] [SO 4 - ] 0.00016 mol x 0.55 0.711 0.0003 mol x 0.456 0.711 5.74 x 10-5 M 1.48 x 10-4 M Calculate the ion product, and compare it to K sp. Q c [Pb + ][SO 4 - ] (5.74 x 10-5 )(1.48 x 10-4 ) 8.49 x 10-9 Because Q c is less than the K sp of 1.7 x 10-8, no precipitation occurs, and the solution is unsaturated.
68 CHAPTER 18 18.8 The solubility of AgCN would increase as the ph decreases because the increasing concentration of H 3 O + would react with the CN - to form the weakly ionized acid HCN. As CN - is removed, more AgCN dissolves to replace the cyanide: AgCN(s) Ag + (aq) + CN - (aq) [+ H 3 O + HCN + H O] In the case of AgCl, the chloride ion is the conjugate base of a strong acid and would, therefore, not be affected by any amount of hydrogen ion. 18.9 Because K f 4.8 x 10 1 and because the starting concentration of NH 3 is much larger than that of the Cu + ion, you can make a rough assumption that most of the copper(ii) is converted to Cu(NH 3 ) 4 + ion. This ion then dissociates slightly to give a small concentration of Cu + and additional NH 3. The amount of NH 3 remaining at the start after reacting with 0.015 M Cu + is [0.100 M - (4 x 0.015 M)] 0.040 M starting NH 3 Assemble the usual concentration table using this starting concentration for NH 3 and assuming the starting concentration of Cu + is zero. Conc. (M) Cu(NH 3 ) 4 + Cu + + 4NH 3 Starting 0.015 0 0.040 Change -x +x +4x Equilibrium 0.015 - x x 0.040 + 4x Even though this reaction is the opposite of the equation for the formation constant, the formation-constant expression can be used. Simply substitute all exact equilibrium concentrations into the formation-constant expression; then, simplify the exact equation by assuming x is negligible compared to 0.015 and 4x is negligible compared to 0.040. K f + [Cu(NH 3) 4 ] + 4 [Cu ][NH 3 ] (0.015 - x) (x)(0.040 + 4x) 4 (0.015) (x)(0.040) 4 4.8 x 10-1 Rearrange and solve for x: x [Cu + ] (0.015) [(4.8 x 10 1 )(0.040) 4 ] 1. x 10-9 1. x 10-9 M
SOUBIITY AND COMPEX-ION EQUIIBRIA 683 18.10 Start by calculating the [Ag + ] in equilibrium with the Ag(CN) - formed from Ag + and CN -. Then, use the [Ag + ] to decide whether or not AgI will precipitate by calculating the ion product and comparing it with the K sp of 8.3 x 10-17 for AgI. Assume all the 0.0045 M Ag + reacts with CN - to form 0.0045 M Ag(CN) -, and calculate the remaining CN -. Use these as starting concentrations for the usual concentration table. [0.0 M KCN - ( x 0.0045 M)] 0.191 M starting CN - Conc. (M) Ag(CN) - Ag + + CN - Starting 0.0045 0 0.191 Change -x +x +x Equilibrium 0.0045 - x x 0.191 + x Even though this reaction is the opposite of the equation for the formation constant, the formation-constant expression can be used. Simply substitute all exact equilibrium concentrations into the formation-constant expression; then, simplify the exact equation by assuming x is negligible compared to 0.0045 and x is negligible compared to 0.191. K f - + - [ Ag(CN) ] [Ag ][CN ] (0.0045 - x) (x)(0.191 + x) (0.0045) (x)(0.191) 5.6 x 10 18 Rearrange and solve for x: x [Ag + ] (0.0045) [(5.6 x 10 18 )(0.191) ].0 x 10-0 M Now, calculate the ion product for AgI: Q c [Ag + ][I - ] (.0 x 10-0 )(0.15) 3.30 x 10-1 3.3 x 10-1 Because Q c is less than the K sp of 8.3 x 10-17, no precipitate will form, and the solution is unsaturated.
684 CHAPTER 18 18.11 Obtain the overall equilibrium constant for this reaction from the product of the individual equilibrium constants of the two individual equations whose sum gives this equation: AgBr(s) Ag + (aq) + Br - (aq) K sp 5.0 x 10-13 Ag + (aq) + S O 3 - (aq) Ag(S O 3 ) 3- (aq) K f.9 x 10 13 AgBr(s) + S O 3 - (aq) Ag(S O 3 ) 3- (aq) + Br - (aq) K c K sp x K f 14.5 Assemble the usual table using 1.0 M as the starting concentration of S O 3 - and x as the unknown concentration of Ag(S O 3 ) 3- formed. Conc. (M) AgBr(s) + S O 3 - Ag(S O 3 ) 3- + Br - Starting 1.0 0 0 Change -x +x +x Equilibrium 1.0 - x x x The equilibrium-constant expression can now be used. Simply substitute all exact equilibrium concentrations into the equilibrium-constant expression. The solution can be obtained without using the quadratic equation. K c 3-3 - - 3 [Ag(S O ) ][Br ] [S O ] (x) (1.0 - x) 14.5 Take the square root of both sides of the two right-hand terms, and solve for x: x (1.0 - x) 3.808 x 3.808 (1.0 - x) 7.6x + x 3.808 x 0.4417 0.44 M (Molar solubility of AgBr in 1.0 M Na S O 3 )
Answers to Concept Checks SOUBIITY AND COMPEX-ION EQUIIBRIA 685 18.1 Solubility and K sp are related, although not directly. You can compare K sp 's for a series of salts, however, if they have the same number of cations and anions in each of their formulas. (In that case, K sp and solubility are related in the same way for each salt.) In this problem, each of the lead(ii) compounds has one Pb + cation and one anion, so you can compare the K sp 's directly. ead(ii) sulfate has the largest K sp and, therefore, is the most soluble of these lead(ii) compounds. 18. et's look at each compound in turn. NaNO 3 has no ion in common with PbSO 4, so it should have little effect on its solubility. Na SO 4 is a soluble compound and provides the common ion SO 4 -, which would repress the solubility of PbSO 4. PbS has an ion in common with PbSO 4 (Pb + ), but the compound is so insoluble that very little of the Pb + ion is available. Because of this, the solubility of PbSO 4 is little affected by the PbS. Therefore, the NaNO 3 solution will dissolve the most PbSO 4. 18.3 If NaCl were added to a saturated AgCl solution, the equilibrium would shift to consume the added chloride ion, and some AgCl would precipitate. After the addition of two Cl - ions, there would be five Cl - ions, three Ag + ions, and two AgCl molecules. The solution would look like the following (Na + not shown for clarity). AgCl Cl - Ag + 18.4 If you compare K sp s for magnesium oxalate (8.5 x 10-5 ) and calcium oxalate (.3 x 10-9 ), you can see magnesium oxalate is much more soluble in water solution than the calcium salt (the K sp is larger). This means it provides a greater concentration of oxalate ion. In water solution, some of the magnesium oxalate dissolves, giving an oxalate ion concentration that tends to repress the dissolution of calcium oxalate (common-ion effect). The addition of acid tends to remove oxalate ion, but this is replenished by the dissolution of more magnesium oxalate. Therefore, you would expect the magnesium oxalate to be more likely to dissolve.
686 CHAPTER 18 Answers to Review Questions 18.1 The solubility equation is Ni(OH) (s) Ni + (aq) + OH - (aq). If the molar solubility of Ni(OH) x molar, the concentrations of the ions in the solution must be x M Ni + and x M OH -. Substituting into the equilibrium-constant expression gives K sp [Ni + ][OH - ] (x)(x) 4x 3 18. Calcium sulfate is less soluble in a solution containing sodium sulfate because the increase in sulfate from the sodium sulfate causes the equilibrium composition in the equation below to shift to the left: CaSO 4 (s) Ca + (aq) + SO 4 - (aq) The result is a decrease in both the calcium ion and the calcium sulfate concentrations. 18.3 Substitute the 0.10 M concentration of chloride into the solubility product expression and solve for [A + ]: [Ag + ] Ksp - [Cl ] -10 1.8 x 10 0.10 1.80 x 10-9 1.8 x 10-9 M 18.4 In order to predict whether or not PbI will precipitate when lead nitrate and potassium iodide are mixed, the concentrations of Pb + and I - after mixing would have to be calculated first (if the concentrations are not known or are not given). Then, the value of Q c, the ion product, would have to be calculated for PbI. Finally, Q c would have to be compared with the value of K sp. If Q c > K sp, then a precipitate will form at equilibrium. If Q c is than K sp, no precipitate will form. 18.5 Barium fluoride, normally insoluble in water, dissolves in dilute hydrochloric acid because the fluoride ion, once it forms, reacts with the hydronium ion to form weakly ionized HF: BaF (s) Ba + (aq) + F - (aq) [+ H 3 O + HF + H O] 18.6 Metal ions such as Pb + and Zn + are separated by controlling the [S - ] in a solution of saturated H S by means of adjusting the ph correctly. Because the K sp of.5 x 10-7 for PbS is smaller than the K sp of 1.1 x 10-1 for ZnS, the ph can be adjusted to make the [S - ] just high enough to precipitate PbS without precipitating ZnS.
SOUBIITY AND COMPEX-ION EQUIIBRIA 687 18.7 When NaCl is first added to a solution of Pb(NO 3 ), a precipitate of PbCl forms. As more NaCl is added, the excess chloride reacts further with the insoluble PbCl, forming soluble complex ions of PbCl 3 - and PbCl 4 - : Pb + (aq) + Cl - (aq) PbCl (s) PbCl (s) + Cl - (aq) PbCl 3 - (aq) PbCl 3 - (aq) + Cl - (aq) PbCl 4 - (aq) 18.8 When a small amount of NaOH is added to a solution of Al (SO 4 ) 3, a precipitate of Al(OH) 3 forms at first. As more NaOH is added, the excess hydroxide ion reacts further with the insoluble Al(OH) 3, forming a soluble complex ion of Al(OH) 4 -. 18.9 The Ag +, Cu +, and Ni + ions can be separated in two steps: (1) Add HCl to precipitate just the Ag + as AgCl, leaving the others in solution. () After pouring the solution away from the precipitate, add 0.3 M HCl and H S to precipitate only the CuS away from the Ni + ion, whose sulfide is soluble under these conditions. 18.10 By controlling the ph through the appropriate buffer, one can control the [CO 3 - ] using the equilibrium reaction: H 3 O + (aq) + CO 3 - (aq) HCO 3 - (aq) + H O(l) Calcium carbonate is much more insoluble than magnesium carbonate and thus will precipitate in weakly basic solution, whereas magnesium carbonate will not. Magnesium carbonate will precipitate only in highly basic solution. (There is the possibility that Mg(OH) might precipitate, but the [OH - ] is too low for this to occur.) Answers to Conceptual Problems 18.11 a. Since both compounds contain the same number of ions, the salt with the larger K sp value will be more soluble. The K sp for AgCl is 1.8 x 10-10 and for AgI is 8.3 x 10-17. Therefore, silver chloride (AgCl) is more soluble. b. Since both compounds contain the same number of ions, the salt with the larger K sp value will be more soluble. The K sp for Mg(OH) is 1.8 x 10-11 and for Cu(OH) is.6 x 10-19. Therefore, magnesium hydroxide (Mg(OH) ) is more soluble. 18.1 Dissolve a sample of each mineral in water in a separate container. NaCl is soluble in water, and CaF is insoluble in water (K sp 3.4 x 10-11 ), so the sample that dissolves is NaCl, and the sample that doesn t dissolve is CaF. 18.13 0.1 M NaCl will be most effective in causing a precipitate from a saturated solution of PbCl, since it is completely soluble and contains the common ion Cl -. Therefore, the answer is a.
688 CHAPTER 18 18.14 The beaker on the left depicts all the AgCl as individual formula units in solution. This implies AgCl is a soluble nonelectrolye, which is not the case since AgCl as a slightly soluble ionic compound. The center beaker depicts AgCl as a soluble ionic compound that completely dissolves in solution, leaving no AgCl(s). Once again, this cannot be correct since AgCl is slightly soluble. The beaker on the right indicates there are ions of Ag + and Cl - present in the solution along with solid AgCl. This is consistent with AgCl being a slightly soluble ionic compound. 18.15 The beaker on the left depicts all the NaCl as individual formula units in solution. This implies NaCl is a soluble nonelectrolye, which is not the case since NaCl is a very soluble ionic compound that produces ions in solution. The center beaker depicts NaCl as a soluble ionic compound that completely dissolves in solution, producing as only Na + (aq) and Cl - (aq). This must be correct since NaCl is a very soluble ionic compound. The beaker on the right indicates there are ions of Na + and Cl - present in the solution along with solid NaCl. The presence of solid is not consistent with NaCl being a very soluble ionic compound. 18.16 When first added to a solution of copper(ii) nitrate, the ammonia causes a pale blue precipitate of Cu(OH) to form (K sp.6 x 10-19 ). After enough ammonia has been added, however, the complex ion Cu(NH 3 ) 4 + forms (K f 4.8 x 10 1 ), and the precipitate dissolves. 18.17 Add just enough Na SO 4 to precipitate all the Ba + ; filter off the BaSO 4 ; add more Na SO 4 to precipitate all the Ca + ; filter off the CaSO 4 ; Mg + remains in the solution. 18.18 When a precipitate fails to form when HCl is added to the solution, this indicates no silver ion is present. When a precipitate fails to form when the solution is acidified and H S is added, this indicates no copper(ii) ion is present in the solution. Solutions to Practice Problems Note on significant figures: If the final answer to a solution needs to be rounded off, it is given first with one nonsignificant figure, and the last significant figure is underlined. The final answer is then rounded to the correct number of significant figures. In multiple-step problems, intermediate answers are given with at least one nonsignificant figure; however, only the final answer has been rounded off. 18.19 a. NaBr is soluble (Group IA salts are soluble). b. PbI is insoluble (PbI is an insoluble iodide). c. BaCO 3 is insoluble (Carbonates are generally insoluble). d. (NH 4 ) SO 4 is soluble (All ammonium salts are soluble).
SOUBIITY AND COMPEX-ION EQUIIBRIA 689 18.0 a. Ca(NO 3 ) is soluble (Nitrate salts are soluble). b. AgBr is insoluble (AgBr is an insoluble bromide salt). c. MgI is soluble (Iodide salts are generally soluble). d. PbSO 4 is insoluble (PbSO 4 is an insoluble sulfate salt). 18.1 a. K sp [Mg + ][OH - ] b. K sp [Sr + ][CO 3 - ] c. K sp [Ca + ] 3 [AsO 4 3- ] d. K sp [Fe 3+ ][OH - ] 3 18. a. K sp [Ba + ] 3 [PO 4 3- ] b. K sp [Fe 3+ ][PO 4 3- ] c. K sp [Pb + ][I - ] d. K sp [Ag + ] [S - ] 18.3 Calculate molar solubility, assemble the usual table, and substitute the equilibrium concentrations from it into the equilibrium-constant expression. (Dashes are given for AgBrO 3 (s); in later problems, the dashes are omitted.) -3 7. x 10 g x 1 mol 36 g 3.05 x 10-5 M Conc. (M) AgBrO 3 (s) Ag + + - BrO 3 Starting 0 0 Change +3.05 x 10-5 +3.05 x 10-5 Equilibrium 3.05 x 10-5 3.05 x 10-5 K sp [Ag + ][BrO - 3 ] (3.05 x 10-5 )(3.05 x 10-5 ) 9.30 x 10-10 9.3 x 10-10 18.4 Assemble the usual concentration table, and substitute from it the equilibrium concentrations into the equilibrium-constant expression. Conc. (M) MgC O 4 (s) Mg + + C O 4 - Starting 0 0 Change +9.3 x 10-3 +9.3 x 10-3 Equilibrium 9.3 x 10-3 9.3 x 10-3 K sp [Mg + ][C O 4 - ] (9.3 x 10-3 )(9.3 x 10-3 ) 8.649 x 10-5 8.6 x 10-5
690 CHAPTER 18 18.5 Calculate the molar solubility. Then, assemble the usual concentration table, and substitute the equilibrium concentrations from it into the equilibrium-constant expression. 0.13 g 0.100 x 1 mol 413 g 3.147 x 10-3 M Conc. (M) Cu(IO 3 ) (s) Cu + + IO 3 - Starting 0 0 Change +3.147 x 10-3 + x 3.147 x 10-3 Equilibrium 3.147 x 10-3 x 3.147 x 10-3 K sp [Cu + ][IO 3 - ] (3.147 x 10-3 )[(3.147 x 10-3 )] 1.47 x 10-7 1. x 10-7 18.6 Calculate the molar solubility. Then, assemble the usual concentration table, and substitute the equilibrium concentrations from it into the equilibrium-constant expression. 0.0 g x 1 mol 33 g 6.6 x 10-5 M Conc. (M) Ag CrO 4 (s) Ag + + - CrO 4 Starting 0 0 Change +(6.6 x 10-5 ) 6.6 x 10-5 Equilibrium (6.6 x 10-5 ) 6.6 x 10-5 K sp [Ag + ] [CrO - 4 ] [(6.6 x 10-5 )] (6.6 x 10-5 ) 1.16 x 10-1 1. x 10-1 18.7 Calculate the poh from the ph, and then convert poh to [OH - ]. Then, assemble the usual concentration table, and substitute the equilibrium concentrations from it into the equilibrium-constant expression. poh 14.00-10.5 3.48 [OH - ] antilog (-poh) antilog (-3.48) 3.311 x 10-4 M [Mg + ] [OH - ] (3.311 x 10-4 ) 1.655 x 10-4 M Conc. (M) Mg(OH) (s) Mg + + OH - Starting 0 0 Change +1.655 x 10-4 +3.311 x 10-4 Equilibrium 1.655 x 10-4 3.311 x 10-4 K sp [Mg + ][OH - ] (1.655 x 10-4 )(3.311 x 10-4 ) 1.81 x 10-11 1.8 x 10-11
SOUBIITY AND COMPEX-ION EQUIIBRIA 691 18.8 Calculate the poh from the ph, and then convert poh to [OH - ]. poh 14.00-1.35 1.65 [OH - ] antilog (-poh) antilog (-1.65).38 x 10 - M [Ca + ] [OH - ] (.38 x 10 - ) 1.119 x 10 - M Assemble the usual concentration table, and substitute the equilibrium concentrations from it into the equilibrium-constant expression. Conc. (M) Ca(OH) (s) Ca + + OH - Starting 0 0 Change +1.119 x 10 - +.38 x 10 - Equilibrium 1.119 x 10 -.38 x 10 - K sp [Ca + ][OH - ] (1.119 x 10 - )(.38 x 10 - ) 5.60 x 10-6 5.6 x 10-6 18.9 Assemble the usual concentration table. et x equal the molar solubility of SrCO 3. When x mol SrCO 3 dissolves in one of solution, x mol Sr + and x mol CO 3 - form. Conc. (M) SrCO 3 (s) Sr + + CO 3 - Starting 0 0 Change +x +x Equilibrium x x Substitute the equilibrium concentrations into the equilibrium-constant expression and solve for x. Then, convert to grams SrCO 3 per liter. [Sr + ][CO 3 - ] K sp (x)(x) x 9.3 x 10-10 -10 x (9.3 x 10 ) 3.04 x 10-5 M -5 3.04 x 10 mol x 147.63 g 1mol SrCO 3 4.50 x 10-3 0.0045 g/
69 CHAPTER 18 18.30 Assemble the usual concentration table. et x equal the molar solubility of MgCO 3. When x mol MgCO 3 dissolves in one of solution, x mol Mg + and x mol CO 3 - form. Conc. (M) MgCO 3 (s) Mg + + CO 3 - Starting 0 0 Change +x +x Equilibrium x x Substitute the equilibrium concentrations into the equilibrium-constant expression and solve for x. Then, convert to grams MgCO 3 per liter. [Mg + ][CO 3 - ] K sp (x)(x) x 1.0 x 10-5 -5 x (1.0 x 10 ) 3.16 x 10-3 M -3 3.16 x 10 mol x 84.3 g 1mol MgCO 3 0.666 0.7 g/ 18.31 et x equal the molar solubility of PbF. Assemble the usual concentration table, and substitute from the table into the equilibrium-constant expression. Conc. (M) PbF (s) Pb + + F - Starting 0 0 Change +x +x Equilibrium x x [Pb + ][F - ] K sp (x)(x) 4x 3.7 x 10-8 x 3-8.7 x 10 4 1.88 x 10-3 1.9 x 10-3 M
SOUBIITY AND COMPEX-ION EQUIIBRIA 693 18.3 et x equal the molar solubility of MgF. Assemble the usual concentration table. Conc. (M) MgF (s) Mg + + F - Starting 0 0 Change +x +x Equilibrium x x Substitute from the table into the equilibrium-constant expression [Mg + ][F - ] K sp (x)(x) 4x 3 7.4 x 10-11 x 3-11 7.4 x 10 4.644 x 10-4.6 x 10-4 M 18.33 At the start, before any SrSO 4 dissolves, the solution contains 0.3 M SO 4 -. At equilibrium, x mol of solid SrSO 4 dissolves to yield x mol Sr + and x mol SO 4 -. Assemble the usual concentration table, and substitute the equilibrium concentrations into the equilibrium-constant expression. As an approximation, assume x is negligible compared to 0.3 M SO 4 -. Conc. (M) SrSO 4 (s) Sr + + SO 4 - Starting 0 0.3 Change +x +x Equilibrium x 0.3 + x [Sr + ][SO 4 - ] K sp (x)(0.3 + x) (x)(0.3).5 x 10-7 x -7.5 x 10 0.3 1.086 x 10-6 M -6 1.086 x 10 mol x 183.69 g 1 mol SrSO 1.99 x 10-4.0 x 10-4 g/ 4 Note that adding x to 0.3 M will not change it (to two significant figures), so x is negligible compared to 0.3 M.
694 CHAPTER 18 18.34 et x the molar solubility of PbCrO 4. At the start, before any PbCrO 4 dissolves, the solution contains 0.13 M CrO 4 -. At equilibrium, x mol of solid PbCrO 4 dissolves to yield x mol Pb + and x mol CrO 4 -. Assemble the usual concentration table, and substitute the equilibrium concentrations into the equilibrium-constant expression. As an approximation, assume x is negligible compared to 0.13 M CrO 4 -. Conc. (M) PbCrO 4 (s) Pb + + CrO 4 - Starting 0 0.13 Change +x +x Equilibrium x 0.13 + x [Pb + ][CrO 4 - ] K sp (x)(0.13 + x) (x)(0.13) 1.8 x 10-14 x -14 1.8 x 10 0.13 1.38 x 10-13 M -13 1.38 x 10 mol x 33. g 1molPbCrO 4.47 x 10-11 4.5 x 10-11 g/ 4 Note that adding x to 0.13 M will not change it (to two significant figures), so x is negligible compared to 0.13 M. 18.35 Calculate the value of K sp from the solubility using the concentration table. Then, using the common-ion calculation, assemble another concentration table. Use 0.00 M NaF as the starting concentration of F - ion. Substitute the equilibrium concentrations from the table into the equilibrium-constant expression. As an approximation, assume x is negligible compared to 0.00 M F - ion. 0.016 g 1molMgF x 6.31 g.567 x 10-4 M Conc. (M) MgF (s) Mg + + F - Starting 0 0 Change +.567 x 10-4 + x.567 x 10-4 Equilibrium.567 x 10-4 x.567 x 10-4 (continued)
SOUBIITY AND COMPEX-ION EQUIIBRIA 695 K sp [Mg + ][F - ] (.567 x 10-4 )[(.567 x 10-4 )] 6.77 x 10-11 Now, use K sp to calculate the molar solubility of MgF. Conc. (M) MgF (s) Mg + + F - Starting 0 0.00 Change +x +x Equilibrium x 0.00 + x [Mg + ][F - ] K sp (x)(0.00 + x) (x)(0.00) 6.77 x 10-11 x -11 6.77 x 10 (0.00) 1.693 x 10-7 M -7 1.693 x 10 mol x 6.31g 1mol MgF 1.054 x 10-5 1.1 x 10-5 g/ 18.36 Calculate the value of K sp from the solubility using the concentration table. Then, using the common-ion calculation, assemble another concentration table. Use 0.45 M Na SO 4 as the starting concentration of SO 4 - ion. Substitute the equilibrium concentrations from the table into the equilibrium-constant expression. As an approximation, assume x is negligible compared to 0.45 M SO 4 - ion. 8.0 g 1 mol Ag x SO4 311.81 g 0.0565 M Conc. (M) Ag SO 4 (s) Ag + + - SO 4 Starting 0 0 Change + x 0.0565 +0.0565 Equilibrium x 0.0565 0.0565 K sp [Ag + ] [SO - 4 ] ( x 0.0565) (0.0565) 6.755 x 10-5 Now, use K sp to calculate the molar solubility of Ag SO 4. (continued)
696 CHAPTER 18 Conc. (M) Ag SO 4 (s) Ag + + - SO 4 Starting 0 0.45 Change +x +x Equilibrium x 0.45 + x [Ag + ] [SO 4 - ] K sp (x) (0.45 + x) (x) (0.45) 6.755 x 10-5 x -5 ( 6.755 x 10 ) 4 x 0.45 6.15 x 10-3 M -3 6.15 x 10 mol x 311.81 g 1 mol AgSO 4 1.910 1.9 g/ 18.37 The concentration table follows. Conc. (M) MgC O 4 (s) Mg + + - C O 4 Starting 0 0.00 Change +x +x Equilibrium x 0.00 + x The equilibrium-constant equation is K sp [Mg + ][C O - 4 ] 8.5 x 10-5 x(0.00 + x) x + 0.00x - (8.5 x 10-5 ) 0 Solving the quadratic equation gives x - 0.00 ± (0.00) + -5 4(8.5 x 10 ) 0.0036 M The solubility in grams per liter is 0.0036 mol x 11 g 1mol 0.403 0.4 g/
SOUBIITY AND COMPEX-ION EQUIIBRIA 697 18.38 The concentration table follows. Conc. (M) SrSO 4 (s) Sr + + - SO 4 Starting 0 0.0015 Change +x +x Equilibrium x 0.0015 + x K sp.5 x 10-7 x(0.0015 + x) x + 0.0015x - (.5 x 10-7 ) 0 Solving the quadratic equation gives x -0.0015 ± -7 (0.0015) + 4(.5 x 10 ) 1.51 x 10-4 M The solubility in grams per liter is -4 1.51x 10 mol x 183.69 g 1 mol SrSO 4 0.077 0.03 g/ 18.39 a. Calculate Q c, the ion product of the solution, using the concentrations in the problem as the concentrations present after mixing and assuming no precipitation. Then, compare Q c with K sp to determine whether precipitation has occurred. Start by defining the ion product with brackets as used for the definition of K sp ; then, use parentheses for the concentrations. Q c [Ba + ][F - ] Q c (0.00)(0.015) 4.50 x 10-6 Since Q c > K sp (1.0 x 10-6 ), precipitation will occur. b. Calculate Q c, the ion product of the solution, using the concentrations in the problem as the concentrations present after mixing and assuming no precipitation. Then, compare Q c with K sp to determine whether precipitation has occurred. Start by defining the ion product with brackets as used for the definition of K sp ; then, use parentheses for the concentrations. Q c [Pb + ][Cl - ] Q c (0.0035)(0.15) 7.87 x 10-5 Since Q c > K sp (1.6 x 10-5 ), precipitation will occur.
698 CHAPTER 18 18.40 a. Calculate Q c, the ion product of the solution, using the concentrations in the problem as the concentrations present after mixing and assuming no precipitation. Then, compare Q c with K sp to determine whether precipitation has occurred. Start by defining the ion product with brackets as used for the definition of K sp ; then, use parentheses for the concentrations. Q c [Sr + ][CrO 4 - ] Q c (0.01)(0.0015) 1.80 x 10-5 Since Q c < K sp (3.5 x 10-5 ), no precipitation will occur. b. Calculate Q c, the ion product of the solution, using the concentrations in the problem as the concentrations present after mixing and assuming no precipitation. Then, compare Q c with K sp to determine whether precipitation has occurred. Start by defining the ion product with brackets as used for the definition of K sp ; then, use parentheses for the concentrations. Q c [Mg + ][CO 3 - ] Q c (0.001)(0.041) 8.61 x 10-5 Since Q c > K sp (1.0 x 10-5 ), precipitation will occur. 18.41 Calculate the ion product, Q c, after preparation of the solution, assuming no precipitation has occurred. Compare it with the K sp. Q c [Pb + ][CrO 4 - ] Q c (5.0 x 10-4 )(5.0 x 10-5 ).50 x 10-8 M (> K sp of 1.8 x 10-14 ) The solution is supersaturated before equilibrium is reached. At equilibrium, precipitation occurs, and the solution is saturated. 18.4 Calculate the ion product, Q c, after preparation of the solution and assuming no precipitation has occurred. Compare it with the K sp. Q c [Pb + ][SO 4 - ] Q c (5.0 x 10-4 )(1.0 x 10-5 ) 5.00 x 10-9 M (< K sp of 1.7 x 10-8 ) At equilibrium, no precipitation occurs, and the solution is unsaturated.
SOUBIITY AND COMPEX-ION EQUIIBRIA 699 18.43 Calculate the concentrations of Mg + and OH -, assuming no precipitation. Use a total volume of 1.0 + 1.0, or.0. (Note the concentrations are halved when the volume is doubled.) [Mg + ] 0.000 mol x 1.0.0 1.00 x 10-3 M [OH - ] 0.00010 mol.0 x 1.0 5.00 x 10-5 M Calculate the ion product, and compare it to K sp. Q c (Mg + )(OH - ) (1.00 x 10-3 )(5.00 x 10-5 ).50 x 10-1 Because Q c is less than the K sp of 1.8 x 10-11, no precipitation occurs, and the solution is unsaturated. 18.44 Calculate the concentrations of Ca + and SO 4 -, assuming no precipitation. Use a total volume of 0.045 + 0.055, or 0.100. [Ca + ] [SO 4 - ] 0.015 mol x 0.045 0.100 0.010 mol x 0.055 0.100 0.00675 M 0.00550 M Calculate the ion product, and compare it to K sp. Q c [Ca + ][SO 4 - ] (0.00675)(0.00550) 3.71 x 10-5 Because Q c is greater than the K sp of.4 x 10-5, the solution is supersaturated before equilibrium is reached. At equilibrium, precipitation occurs, and the solution is saturated.
700 CHAPTER 18 18.45 Calculate the concentrations of Ba + and F -, assuming no precipitation. Use a total volume of 0.045 + 0.075, or 0.10. [Ba + ] 0.0015 mol x 0.045 0.10 5.65 x 10-4 M [F - ] 0.005 mol x 0.075 0.10 1.56 x 10-3 M Calculate the ion product and compare it to K sp. Q c [Ba + ][F - ] (5.65 x 10-4 )(1.56 x 10-3 ) 1.36 x 10-9 Because Q c is less than the K sp of 1.0 x 10-6, no precipitation occurs, and the solution is unsaturated. 18.46 Calculate the concentrations of Pb + and Cl -, assuming no precipitation. Use a total volume of 0.040 + 0.065, or 0.105. [Pb + ] [Cl - ] 0.010 mol x 0.065 0.105 0.035 mol x 0.040 0.105 6.19 x 10-3 M 1.33 x 10 - M Calculate the ion product and compare it to K sp. Q c [Pb + ][Cl - ] (6.19 x 10-3 )(1.33 x 10 - ) 1.09 x 10-6 Because Q c is less than the K sp of 1.6 x 10-5, no precipitation occurs, and the solution is unsaturated.
SOUBIITY AND COMPEX-ION EQUIIBRIA 701 18.47 A mixture of CaCl and K SO 4 can only precipitate CaSO 4 because KCl is soluble. Use the K sp expression to calculate the [Ca + ] needed to just begin precipitating the 0.00 M SO 4 - (in essentially a saturated solution). Then, convert to moles. [Ca + ][SO 4 - ] K sp.4 x 10-5 [Ca + ] Ksp - [SO 4 ] -5.4 x 10 -.0 x 10 1.0 x 10-3 M The number of moles in 1.5 of this calcium-containing solution is (mol CaCl mol Ca + ): 1.5 x (1.0 x 10-3 ) mol/ 1.80 x 10-3 0.0018 mol CaCl 18.48 A mixture of MgSO 4 and NaOH can only precipitate Mg(OH) because Na SO 4 is soluble. Use the K sp expression to calculate the [Mg + ] needed to just begin precipitating the 0.040 M OH - (in essentially a saturated solution). Then, convert to moles and finally to grams. [Mg + ][OH - ] K sp 1.8 x 10-11 [Mg + ] Ksp - [OH ] -11 1.8 x 10 (0.040) 1.15 x 10-8 M The number of moles, and grams, in 0.456 (456 m) of this magnesium-containing solution is (mol MgSO 4 mol Mg + ). 0.456 x (1.15 x 10-8 mol/) 5.13 x 10-9 mol MgSO 4 (5.13 x 10-9 mol MgSO 4 ) x (10 g MgSO 4 /1 mol MgSO 4 ) 6.156 x 10-7 6. x 10-7 g MgSO 4 18.49 Because the AgNO 3 solution is relatively concentrated, ignore the dilution of the solution of Cl - and I - from the addition of AgNO 3. The [Ag + ] just as the AgCl begins to precipitate can be calculated from the K sp expression for AgCl using [Cl - ] 0.015 M. Therefore, for AgCl [Ag + ][Cl - ] K sp (continued)
70 CHAPTER 18 [Ag + ][0.015] 1.8 x 10-10 [Ag + ] -10 1.8 x 10 0.015 1.0 x 10-8 M The [I - ] at this point can be obtained by substituting the [Ag + ] into the K sp expression for AgI. Therefore, for AgI [Ag + ][I - ] K sp [1.0 x 10-8 ][I - ] 8.3 x 10-17 [I - ] -17 8.3 x 10-8 1. x 10 6.91 x 10-9 6.9 x 10-9 M 18.50 Because the problem asks you to find [Cl - ] at the point when Ag CrO 4 just begins to precipitate, we start with the K sp expression for Ag CrO 4, and ignore temporarily what may have happened to the Cl - ion. Also, because the AgNO 3 solution is relatively concentrated, ignore the dilution of the solution of Cl - and CrO 4 - from the addition of AgNO 3. The [Ag + ] just as the Ag CrO 4 begins to precipitate can be calculated from the K sp expression for Ag CrO 4 using [CrO 4 - ] 0.015 M. Thus, for Ag CrO 4 [Ag + ] [CrO 4 - ] K sp [Ag + ] [0.015] 1.1 x 10-1 [Ag + ] -1 1.1 x 10 0.015 8.56 x 10-6 M The [Cl - ] at this point can be obtained by substituting the [Ag + ] into the K sp expression for AgCl. Therefore, for AgCl [Ag + ][Cl - ] K sp [8.56 x 10-6 ][Cl - ] 1.8 x 10-10 [Cl - ] -10 1.8 x 10-6 8.56 x 10.10 x 10-5.1 x 10-5 M Because this Cl - concentration is extremely small compared to the initial 0.015 M concentration, you can also deduce that essentially all the chloride ion precipitates before the Ag CrO 4 begins to precipitate.
SOUBIITY AND COMPEX-ION EQUIIBRIA 703 18.51 The net ionic equation is BaF (s) + H 3 O + (aq) Ba + (aq) + HF(aq) + H O(l) 18.5 The net ionic equation is PbCO 3 (s) + H 3 O + (aq) Pb + (aq) + CO (g) + H O(l) 18.53 Calculate the value of K for the reaction of H 3 O + with both the SO 4 - and F - anions as they form by the slight dissolving of the insoluble salts. These constants are the reciprocals of the K a values of the conjugate acids of these anions. F - (aq) + H 3 O + (aq) HF(aq) + H O(l) K 1 K a 1-4 6.8 x 10 1.47 x 10 3 SO 4 - (aq) + H 3 O + (aq) HSO 4 - (aq) + H O(l) K 1 K a 1-1.1x 10 90.9 Because K for the fluoride ion is relatively larger, more BaF will dissolve in acid than BaSO 4. 18.54 Calculate the value of K for the reaction of H 3 O + with both the SO 4 - and PO 4 3- anions as they form by the slight dissolving of the insoluble salts. These constants are the reciprocals of the K a values of the conjugate acids of these anions. PO 4 3- (aq) + H 3 O + (aq) HPO 4 - (aq) + H O(l) K 1 K a3 1-13 4.8 x 10.08 x 10 1 SO 4 - (aq) + H 3 O + (aq) HSO 4 - (aq) + H O(l) K 1 K a 1-1.1x 10 90.9 Because K for the phosphate ion is relatively larger, more Ca 3 (PO 4 ) will dissolve in acid than CaSO 4.
704 CHAPTER 18 18.55 The equation is Cu + (aq) + CN - (aq) Cu(CN) - (aq) The K f expression is K f - [Cu(CN) ] + - [Cu ] [CN ] 1.0 x 10 16 18.56 The equation is Ni + (aq) + 6 NH 3 (aq) Ni(NH 3 ) 6 + (aq) The K f expression is K f + [Ni(NH3 ) 6 ] + 6 [Ni ] [NH3 ] 5.6 x 10 8 18.57 Assume the only [Ag + ] is that in equilibrium with the Ag(CN) - formed from Ag + and CN -. (In other words, assume all the 0.015 M Ag + reacts with CN - to form 0.015 M Ag(CN) -.) Subtract the CN - that forms the 0.015 M Ag(CN) - from the initial 0.100 M CN -. Use this as the starting concentration of CN - for the usual concentration table. [0.100 M NaCN - ( x 0.015 M)] 0.070 M starting CN - Conc. (M) Ag(CN) - Ag + + CN - Starting 0.015 0 0.070 Change -x +x +x Equilibrium 0.015 - x x 0.070 + x Even though this reaction is the opposite of the equation for the formation constant, the formation-constant expression can be used. Simply substitute all the exact equilibrium concentrations into the formation-constant expression; then, simplify the exact equation by assuming x is negligible compared to 0.015 and x is negligible compared to 0.070. K f - [ Ag(CN) ] + - [Ag ][CN ] (0.015 - x) (x)(0.070 + x) (0.015) (x)(0.070) 5.6 x 10 18 Rearrange and solve for x [Ag + ]: x (0.015) [(5.6 x 10 18 )(0.070) ] 5.46 x 10-19 5.5 x 10-19 M
SOUBIITY AND COMPEX-ION EQUIIBRIA 705 18.58 You can make a rough assumption that most of the Zn(OH) 4 - that dissociates forms Zn + and OH - ions. Assemble the usual concentration table, using 0.0 M as the starting concentration for Zn(OH) 4 - and assuming the starting concentrations of Zn + and OH - ions are zero. Conc. (M) Zn(OH) 4 - Zn + + 4OH - Starting 0.0 0 0 Change -x +x +4x Equilibrium 0.0 - x x 4x Even though this reaction is the opposite of the equation for the formation constant, the formation-constant expression can be used. Simply substitute all the exact equilibrium concentrations into the formation-constant expression; then, simplify the exact equation by assuming x is negligible compared to 0.0. K f - [Zn(OH) 4 ] + - 4 [Zn ][OH ] (0.0 - x) 4 (x)(4x) (0.0) 4 (x)(4x).8 x 10 15 Rearrange and solve for x [Zn + ]: (0.0) x 5 1. 94 x 10-4 1.9 x 10-4 M 56 x (.8 x 10 15 ) 18.59 Start by calculating the [Cd + ] in equilibrium with the Cd(NH 3 ) 4 + formed from Cd + and NH 3. Then, use the [Cd + ] to decide whether or not CdC O 4 will precipitate by calculating the ion product and comparing it with the K sp of 1.5 x 10-8 for CdC O 4. Assume all the 0.000 M Cd + reacts with NH 3 to form 0.000 M Cd(NH 3 ) 4 +, and calculate the remaining NH 3. Use these as the starting concentrations for the usual concentration table. [0.10 M NH 3 - (4 x 0.000 M)] 0.09 M starting NH 3 Conc. (M) Cd(NH 3 ) 4 + Cd + + 4NH 3 Starting 0.000 0 0.09 Change -x +x +4x Equilibrium 0.000 - x x 0.09 + 4x (continued)
706 CHAPTER 18 Even though this reaction is the opposite of the equation for the formation constant, the formation-constant expression can be used. Simply substitute all the exact equilibrium concentrations into the formation-constant expression; then, simplify the exact equation by assuming x is negligible compared to 0.000 and 4x is negligible compared to 0.09. K f + [Cd(NH 3) 4 ] + 4 [Cd ][NH 3 ] (0.000 - x) 4 (x)(0.09 + 4x) (0.000) 4 (x)(0.09) 1.0 x 10 7 Rearrange and solve for x: x (0.000) [(1.0 x 10 7 )(0.09) 4 ].79 x 10-6 M [Cd + ] Now, calculate the ion product for CdC O 4 : Q c [Cd + ][C O 4 - ] (.79 x 10-6 )(0.010).79 x 10-8 Because Q c is greater than the K sp of 1.5 x 10-8, the solution is supersaturated before equilibrium. At equilibrium, a precipitate will form, and the solution will be saturated. 18.60 Start by calculating the [Ni + ] in equilibrium with the Ni(NH 3 ) 6 + formed from Ni + and NH 3. Then, use the [Ni + ] to decide whether or not Ni(OH) will precipitate by calculating the ion product and comparing it with the K sp of.0 x 10-15 for Ni(OH). Assume all the 0.000 M Ni + reacts with NH 3 to form 0.000 M Ni(NH 3 ) 6 +, and calculate the remaining NH 3. Use these as starting concentrations for the usual concentration table. [0.10 M NH 3 - (6 x 0.000 M)] 0.088 M starting NH 3 Conc. (M) Ni(NH 3 ) 6 + Ni + + 6NH 3 Starting 0.000 0 0.088 Change -x +x +6x Equilibrium 0.000 - x x 0.088 + 6x Even though this reaction is the opposite of the equation for the formation constant, the formation-constant expression can be used. Simply substitute all the exact equilibrium concentrations into the formation-constant expression; then, simplify the exact equation by assuming that x is negligible compared to 0.000 and 6x is negligible compared to 0.088. (continued)
SOUBIITY AND COMPEX-ION EQUIIBRIA 707 K f + [Ni(NH 3) 6 ] + 6 [Ni ][NH 3 ] (0.000 - x) 6 (x)(0.088 + 6x) (0.000) 6 (x)(0.088) 5.60 x 10 8 Rearrange and solve for x: x (0.000) [5.6 x 10 8 x (0.088) 6 ] 7.69 x 10-6 M [Ni + ] Now, calculate the ion product for Ni(OH) : Q c [Ni + ][OH - ] (7.69 x 10-6 )(0.010) 7.69 x 10-10 Because Q c is greater than the K sp of.0 x 10-15, the solution is supersaturated before equilibrium. At equilibrium, a precipitate will form, and the solution will be saturated. 18.61 Using the rule from Chapter 15, obtain the overall equilibrium constant for this reaction from the product of the individual equilibrium constants of the two individual equations whose sum gives this equation. CdC O 4 (s) Cd + (aq) + C O 4 - (aq) K sp 1.5 x 10-8 Cd + (aq) + 4NH 3 (aq) Cd(NH 3 ) 4 + (aq) K f 1.0 x 10 7 CdC O 4 (s) + 4NH 3 (aq) Cd(NH 3 ) 4 + (aq) + C O 4 - (aq) K c K sp x K f 0.15 Assemble the usual table using 0.10 M as the starting concentration of NH 3 and x as the unknown concentration of Cd(NH 3 ) 4 + formed. Conc. (M) CdC O 4 (s) + 4NH 3 Cd(NH 3 ) 4 + + C O 4 - Starting 0.10 0 0 Change -4x +x +x Equilibrium 0.10-4x x x The equilibrium-constant expression can now be used. Simply substitute all the exact equilibrium concentrations into the equilibrium-constant expression. Kc + - [Cd(NH3 ) 4 ][C O4 ] 4 [NH3] (x) 4 (0.10-4x) 0.15 (continued)
708 CHAPTER 18 Take the square root of both sides of the two right-hand terms, rearrange into a quadratic equation, and solve for x: x (0.10-4x ) 0.387 16x - 3.38x + 0.010 0 x 3.38 (-3.38) - 4(16)(0.010) ± (16) 0.08 (too large) and 3.001 x 10-3 Using the smaller root, x 3.001 x 10-3 3.0 x 10-3 M (molar solubility of CdC O 4 ) 18.6 Using the rule from Chapter 15, obtain the overall equilibrium constant for this reaction from the product of the individual equilibrium constants of the two individual equations whose sum gives this equation. NiS(s) Ni + (aq) + S - (aq) K sp 3 x 10-19 Ni + (aq) + 6NH 3 (aq) Ni(NH 3 ) 6 + (aq) K f 5.6 x 10 8 NiS(s) + 6NH 3 (aq) Ni(NH 3 ) 6 + (aq) + S - (aq) K c K sp x K f 1.7 x 10-10 Assemble the usual table using 0.10 M as the starting concentration of NH 3 and x as the unknown concentration of Ni(NH 3 ) 6 + formed. Conc (M) NiS(s) + 6NH 3 Ni(NH 3 ) 6 + + S - Starting 0.10 0 0 Change -6x +x +x Equilibrium 0.10-6x x x The equilibrium-constant expression can now be used. Simply substitute all the exact equilibrium concentrations into the equilibrium-constant expression. It will be necessary to simplify the exact equation to obtain a solution without using a higherorder equation. + - [Ni(NH3 ) 6 ][S ] (x) -10 K c 1.7 x 10 6 6 [NH3 ] (0.10-6x) (continued)
SOUBIITY AND COMPEX-ION EQUIIBRIA 709 Take the square root of both sides of the two right-hand terms. Note that the resulting equation contains an x 3 term and an x term. x 3 (0.10-6x) 1.3 x 10-5 Assume 6x is negligible compared to 0.10 M, and solve the above equation for x: x 3 (0.10) 1.3 x 10-5 x 1.3 x 10-8 1 x 10-8 M (molar solubility of NiS in 0.10 M NH 3 ) 18.63 The Pb +, Cd +, and Sr + ions can be separated in two steps: (1) Add HCl to precipitate only the Pb + as PbCl, leaving the others in solution. () After pouring the solution away from the precipitate, add 0.3 M HCl and H S to precipitate only the CdS away from the Sr + ion, whose sulfide is soluble under these conditions. 18.64 The Hg +, Ca +, and Na + ions can be separated in two steps: (1) Add 0.3 M HCl and H S to precipitate only the HgS, leaving the others in solution. () After pouring the solution away from the precipitate, add NH 3 and (NH 4 ) HPO 4 to precipitate only the Ca 3 (PO 4 ) away from the Na + ion, whose phosphate is soluble under these conditions. 18.65 a. Ag + is not possible because no precipitate formed with HCl. b. Ca + is not possible if the compound contains only one cation because Mn + is indicated by the evidence. However, if the compound consists of two or more cations, Ca + is possible because no reactions were described involving Ca +. c. Mn + is possible because a precipitate was obtained with basic sulfide ion. d. Cd + is not possible because no precipitate was obtained with acidic sulfide solution. 18.66 PbCrO 4 is not the compound formed from H S because it doesn't contain sulfide ion. CdS is possible because it contains sulfide ion and precipitates in acidic sulfide (Analytical Group II). Neither MnS nor Ag S is possible because neither falls into Analytical Group II.
710 CHAPTER 18 Solutions to General Problems 18.67 Assemble the usual concentration table. et x equal the molar solubility of PbSO 4. When x mol PbSO 4 dissolves in one of solution, x mol Pb + and x mol SO 4 - form. Conc. (M) PbSO 4 (s) Pb + + SO 4 - Starting 0 0 Change +x +x Equilibrium x x Substitute the equilibrium concentrations into the equilibrium-constant expression and solve for x. [Pb + ][SO 4 - ] K sp (x)(x) x 1.7 x 10-8 -8 x (1.7 x 10 ) 1.303 x 10-4 1.3 x 10-4 M 18.68 Assemble the usual concentration table. et x equal the molar solubility of HgS. When x mol HgS dissolves in one of solution, x mol Hg + and x mol S - form. Conc. (M) HgS(s) Hg + + S - Starting 0 0 Change +x +x Equilibrium x x Substitute the equilibrium concentrations into the equilibrium-constant expression and solve for x. [Hg + ][S - ] K sp (x)(x) x 1.6 x 10-5 -5 x (1.6 x 10 ) 1.6 x 10-6 1.3 x 10-6 M
SOUBIITY AND COMPEX-ION EQUIIBRIA 711 18.69 et x equal the molar solubility of Hg Cl. Assemble the usual concentration table, and substitute from the table into the equilibrium-constant expression. Conc. (M) Hg Cl (s) Hg + + Cl - Starting 0 0 Change +x +x Equilibrium x x [Hg + ][Cl - ] K sp (x)(x) 4x 3 1.3 x 10-18 x 3 1.3 x 10 4-18 6.87 x 10-7 M a. Molar solubility 6.87 x 10-7 6.9 x 10-7 M b. -7 6.87 x 10 mol x 47.1 g HgCl 1mol 3.4 x 10-4 3. x 10-4 g/ 18.70 et x equal the molar solubility of MgNH 4 PO 4. Assemble the usual concentration table, and substitute from the table into the equilibrium-constant expression. Conc. (M) MgNH 4 PO 4 (s) Mg + + NH 4 + + PO 4 3- Starting 0 0 0 Change +x +x +x Equilibrium x x x [Mg + ][NH 4 + ][PO 4 3- ] K sp (x)(x)(x) x 3.5 x 10-13 x 3-13.5 x 10 6.99 x 10-5 M a. Molar solubility 6.99 x 10-5 6.3 x 10-5 M b. 6.9-5 9 x 10 mol x 137.3 g MgNH4PO4 1mol 8.648 x 10-3 8.6 x 10-3 g/
71 CHAPTER 18 18.71 et x equal the molar solubility of Ce(OH) 3. Assemble the usual concentration table, and substitute from the table into the equilibrium-constant expression. Conc. (M) Ce(OH) 3 (s) Ce 3+ + 3OH - Starting 0 0 Change +x +3x Equilibrium x 3x [Ce 3+ ][OH - ] 3 K sp (x)(3x) 3 7x 4.0 x 10-0 x 4.0 x 10 7-0 5.16 x 10-6 M a. Molar solubility 5.16 x 10-6 5. x 10-6 M b. [OH - ] 3x 1.56 x 10-5 M poh 4.8054 4.81 18.7 et x equal the molar solubility of Cu Fe(CN) 6. Assemble the usual concentration table, and substitute from the table into the equilibrium-constant expression. Conc. (M) Cu Fe(CN) 6 (s) Cu + + Fe(CN) 6 4- Starting 0 0 Change +x +x Equilibrium x x [Cu + ] [Fe(CN) 6 4- ] K sp (x) (x) 4x 3 1.3 x 10-16 x 3 1.3 x 10 4-16 3.19 x 10-6 M a. Molar solubility 3.19 x 10-6 3. x 10-6 M b. 3-6.19 x 10 mol x 339.0gCuFe(CN) 1mol 6 1.08 x 10-3 1.1 x 10-3 g/
SOUBIITY AND COMPEX-ION EQUIIBRIA 713 18.73 Calculate the poh from the ph, and then convert poh to [OH - ]. Then, assemble the usual concentration table, and substitute the equilibrium concentrations from it into the equilibrium-constant expression. poh 14.00-8.80 5.0 [OH - ] antilog (-poh) antilog (-5.0) 6.309 x 10-6 M Conc. (M) Mg(OH) (s) Mg + + OH - Starting 0 0 Change +x Equilibrium x 6.309 x 10-6 K sp [Mg + ][OH - ] (x)(6.309 x 10-6 ) 1.8 x 10-11 x 1.8 x 10 (6.309 x 10-11 -6 ) 0.45 0.45 M 0.45 mol x 58.3 g Mg(OH) 1mol 6.3 6 g/ 18.74 Calculate the poh from the ph, and then convert poh to [OH - ]. Then, assemble the usual concentration table, and substitute the equilibrium concentrations from it into the equilibrium-constant expression. et x equal the molar solubility of the Ag + ion because you are seeking x, the molar solubility of Ag O. poh 14.00-10.50 3.50 [OH - ] antilog (-poh) antilog (-3.50) 3.16 x 10-4 M Conc. (M) Ag O(s) + H O Ag + + OH - Starting 0 0 Change x +x Equilibrium x 3.16 x 10-4 K c [Ag + ] [OH - ] (x) (3.16 x 10-4 ).0 x 10-8 (continued)
714 CHAPTER 18 x -8.0 x 10-4 4 x (3.16 x 10 ) 0.4 0. M 0.4 mol x 31.7 g AgO 1mol 51.9 5 g/ 18.75 et x equal the change in M of Mg + and 0.10 M equal the starting OH - concentration. Then, assemble the usual concentration table, and substitute the equilibrium concentrations from it into the equilibrium-constant expression. Assume x is negligible compared to 0.10 M, and perform an approximate calculation. Conc. (M) Mg(OH) (s) Mg + + OH - Starting 0 0.10 Change +x +x Equilibrium x 0.10 + x K sp [Mg + ][OH - ] (x)(0.10 + x) (x)(0.10) 1.8 x 10-11 x -11 1.8 x 10 (0.10) 1.80 x 10-9 1.8 x 10-9 M 18.76 et x equal the change in M of Al 3+ and 0.0010 M equal the starting OH - concentration. Then, assemble the usual concentration table, and substitute the equilibrium concentrations from it into the equilibrium-constant expression. Assume 3x is negligible compared to 0.0010 M, and perform an approximate calculation. Conc. (M) Al(OH) 3 (s) Al 3+ + 3OH - Starting 0 0.0010 Change +x +3x Equilibrium x 0.0010 + 3x K sp [Al 3+ ][OH - ] 3 (x)(0.0010 + 3x) 3 (x)(0.0010) 3 4.6 x 10-33 x -33 4.6 x 10 3 (0.0010) 4.60 x 10-4 4.6 x 10-4 M
SOUBIITY AND COMPEX-ION EQUIIBRIA 715 18.77 To begin precipitation, you must add just slightly more sulfate ion than that required to give a saturated solution. Use the K sp expression to calculate the [SO 4 - ] needed to just begin precipitating the 0.0030 M Ca +. [Ca + ][SO 4 - ] K sp.4 x 10-5 [SO 4 - ] Ksp + [Ca ] -5.4 x 10-3 3.0 x 10 8.00 x 10-3 M When the sulfate ion concentration slightly exceeds 8.0 x 10-3 M, precipitation begins. 18.78 To begin precipitation, you must add just slightly more chromate ion than that required to give a saturated solution. Use the K sp expression to calculate the [CrO 4 - ] needed to just begin precipitating the 0.005 M Sr +. [Sr + ][CrO 4 - ] K sp 3.5 x 10-5 [CrO 4 - ] Ksp + [Sr ] -5 3.5 x 10-3.5 x 10 1.40 x 10 - M When the chromate ion concentration slightly exceeds 1.4 x 10 - M, precipitation begins. 18.79 Calculate the concentrations of Pb + and Cl -. Use a total volume of 3.0 + 0.80, or 4.00. [Pb + ] -3 1.5 x 10 mol 4.00 x 3.0 1.00 x 10-3 M [Cl - ] -1 5.0 x 10 mol x 0.80 4.00 1.00 x 10-1 M Calculate the ion product, and compare it to K sp. Q c [Pb + ][Cl - ] (1.00 x 10-3 )(1.00 x 10-1 ) 1.00 x 10-5 M 3 Because the Q c is less than the K sp of 1.6 x 10-5, no precipitation occurs, and the solution is not saturated.
716 CHAPTER 18 18.80 Rounded answer: Q c equals 1.8 x 10-11 (equals K sp, so solution is saturated). Calculate the concentrations of Mg + and OH -. Use a total volume of 0.150 + 0.050, or 0.00. [Mg + ] [OH - ] -5.4 x 10 mol x 0.150 0.00-3 4.0 x 10 mol x 0.050 0.00 1.80 x 10-5 M 1.00 x 10-3 M Calculate the ion product and compare it to K sp. Q c [Mg + ][OH - ] (1.80 x 10-5 )(1.00 x 10-3 ) 1.80 x 10-11 Because Q c equals the K sp of 1.8 x 10-11, no precipitation occurs, and the solution is saturated. 18.81 A mixture of AgNO 3 and NaCl can precipitate only AgCl because NaNO 3 is soluble. Use the K sp expression to calculate the [Cl - ] needed to prepare a saturated solution (just before precipitating the 0.0015 M Ag + ). Then, convert to moles and finally to grams. [Ag + ][Cl - ] K sp 1.8 x 10-10 [Cl - ] Ksp + [Ag ] -10 1.8 x 10-3 1.5 x 10 1.0 x 10-7 M The number of moles and grams in 0.785 (785 m) of this chloride-containing solution is 0.785 x (1.0 x 10-7 mol/) 9.4 x 10-8 mol Cl - 9.4 x 10-8 mol NaCl (9.4 x 10-7 mol NaCl) x (58.5 g NaCl/1mol) 5.51 x 10-6 5.5 x 10-6 g NaCl This amount of NaCl is too small to weigh on a balance.
SOUBIITY AND COMPEX-ION EQUIIBRIA 717 18.8 A mixture of BaCl and Na SO 4 can precipitate only BaSO 4 because NaCl is soluble. Use the K sp expression to calculate the [SO 4 - ] needed to prepare a saturated solution (just before precipitating the 0.008 M Ba + ). Then, convert to moles and finally to grams. [Ba + ][SO 4 - ] K sp 1.1 x 10-10 [SO 4 - ] Ksp + [Ba ] -10 1.1 x 10-3.8 x 10 3.93 x 10-8 M The number of moles and grams in 0.435 of this sulfate-containing solution is (0.435 x 3.93 x 10-8 mol/) 1.709 x 10-8 mol SO 4-1.709 x 10-8 mol Na SO 4 (1.709 x 10-8 mol Na SO 4 ) x (14 g/1 mol Na SO 4 ).4 x 10-6.4 x 10-6 g Na SO 4 18.83 From the magnitude of K f, assume Fe 3+ and SCN - react essentially completely to form.00 M Fe(SCN) + at equilibrium. Use.00 M as the starting concentration of Fe(SCN) + for the usual concentration table. Conc. (M) Fe 3+ + SCN - FeSCN + Starting 0 0.00 Change +x +x -x Equilibrium x x.00 - x Substitute all the exact equilibrium concentrations into the formation-constant expression; then, simplify the exact equation by assuming x is negligible compared to.00 M. + [FeSCN ] (.00 - x) (.00) K f 9.0 x 10 3+ - (x)(x) [Fe ][SCN ] x Rearrange and solve for x: x.00 9.0 x 10 4.71 x 10-4.7 x 10 - M Fraction dissociated [(4.71 x 10 - ) (.00)] x 100% 0.035 (< 0.03, so acceptable)
718 CHAPTER 18 18.84 From the magnitude of K f, assume Co + and SCN - react essentially completely to form.00 M Co(SCN) + at equilibrium. Use.00 M as the starting concentration of Co(SCN) + for the usual concentration table. Conc. (M) Co + + SCN - CoSCN + Starting 0 0.00 Change +x +x -x Equilibrium x x.00 - x Substitute all the exact equilibrium concentrations into the formation-constant expression; then, simplify the exact equation by assuming x is negligible compared to.00 M. + [CoSCN ] (.00 - x) (.00) K f 1.0 x 10 + - (x)(x) [Co ][SCN ] x Rearrange and solve for x: x.00 1.0 x 10 0.141 0.14 M The value of x 0.14 affects the term.00 - x slightly in the second significant figure. From the quadratic formula, you obtain x 0.137, which also rounds to 0.14. 18.85 Obtain the overall equilibrium constant for this reaction from the product of the individual equilibrium constants of the two individual equations whose sum gives this equation: AgBr(s) Ag + (aq) + Br - (aq) K sp 5.0 x 10-13 Ag + (aq) + NH 3 (aq) Ag(NH 3 ) + (aq) K f 1.7 x 10 7 AgBr(s) + NH 3 (aq) Ag(NH 3 ) + (aq) + Br - (aq) K c K sp x K f 8.50 x 10-6 Assemble the usual table using 5.0 M as the starting concentration of NH 3 and x as the unknown concentration of Ag(NH 3 ) + formed. (continued)
SOUBIITY AND COMPEX-ION EQUIIBRIA 719 Conc. (M) AgBr(s) + NH 3 Ag(NH 3 ) + + Br - Starting 5.0 0 0 Change -x +x +x Equilibrium 5.0 - x x x The equilibrium-constant expression can now be used. Simply substitute all the exact equilibrium concentrations into the equilibrium-constant expression; it will not be necessary to simplify the equation because taking the square root of both sides removes the x term. + - [Ag(NH3 ) ][Br ] x -6 K c 8.50 x 10 [NH3 ] (5.0 - x) Take the square root of both sides of the two right-hand terms and solve for x. x (5.0 - x) -3.915 x 10 x + (5.8 x 10-3 ) x 1.4575 x 10 - x 1.449 x 10-1.4 x 10 - M (the molar solubility of AgBr in 5.0 M NH 3 ) 18.86 Obtain the overall equilibrium constant for this reaction from the product of the individual equilibrium constants of the two individual equations whose sum gives this equation: AgI(s) Ag + (aq) + I - (aq) K sp 8.3 x 10-17 Ag + (aq) + NH 3 (aq) Ag(NH 3 ) + (aq) K f 1.7 x 10 7 AgI(s) + NH 3 (aq) Ag(NH 3 ) + (aq) + I - (aq) K c K sp x K f 1.41 x 10-9 Assemble the usual table using.0 M as the starting concentration of NH 3 and x as the unknown concentration of Ag(NH 3 ) + formed. (continued)
70 CHAPTER 18 Conc. (M) AgI(s) + NH 3 Ag(NH 3 ) + + I - Starting.0 0 0 Change -x +x +x Equilibrium.0 - x x x The equilibrium-constant expression can now be used. Simply substitute all the exact equilibrium concentrations into the equilibrium-constant expression; it will not be necessary to simplify the equation because taking the square root of both sides removes the x term. + - [Ag(NH3 ) ][I ] x -9 K c 1.41x 10 [NH3 ] (.0 - x) Take the square root of both sides of the two right-hand terms and solve for x. x (.0 - x) -5 3.75 x 10 x + (7.5 x 10-5 )x 7.50 x 10-5 x 7.499 x 10-5 7.5 x 10-5 M (the molar solubility of AgI in.0 M NH 3 ) 18.87 Start by recognizing that, because each zinc oxalate produces one oxalate ion, the solubility of zinc oxalate equals the oxalate concentration: ZnC O 4 (s) + 4NH 3 Zn(NH 3 ) 4 + + C O 4 - Thus, the [C O 4 - ] 3.6 x 10-4 M. Now, calculate the [Zn + ] in equilibrium with the oxalate ion using the K sp expression for zinc oxalate. K sp [Zn + ][C O 4 - ] 1.5 x 10-9 [Zn + ] -9 1.5 x 10-4 3.6 x 10 4.16 x 10-6 4. x 10-6 M (continued)
SOUBIITY AND COMPEX-ION EQUIIBRIA 71 To calculate K f, the [Zn(NH 3 ) 4 + ] term must be calculated. This can be done by recognizing that the molar solubility of ZnC O 4 is the sum of the concentration of Zn + and Zn(NH 3 ) 4 + ions: Molar solubility of ZnC O 4 [Zn + ] + [Zn(NH 3 ) 4 + ] 3.6 x 10-4 (4.16 x 10-6 ) + [Zn(NH 3 ) 4 + ] [Zn(NH 3 ) 4 + ] (3.6 x 10-4 ) - (4.16 x 10-6 ) 3.558 x 10-4 M Now, the [NH 3 ] term must be calculated by subtracting the ammonia in [Zn(NH 3 ) 4 + ] from the starting NH 3 of 0.0150 M: [NH 3 ] 0.0150-4 [Zn(NH 3 ) 4 + ] 0.0150-4(3.558 x 10-4 ) 0.01357 M Solve for K f by substituting the known concentrations into the K f expression: K f + [Zn(NH 3) 4 ] + 4 [Zn ][NH 3 ] -4 3.558 x 10-6 4 (4.16 x 10 )(0.01357).5 x 10 9.5 x 10 9 18.88 Start by recognizing that, because each cadmium oxalate produces one oxalate ion, the solubility of cadmium oxalate equals the oxalate concentration: CdC O 4 (s) + 4NH 3 Cd(NH 3 ) 4 + + C O 4 - Thus, the [C O 4 - ] 6.1 x 10-3 M. Now, calculate the [Cd + ] in equilibrium with the oxalate ion using the K sp expression for cadmium oxalate. K sp [Cd + ][C O 4 - ] 1.5 x 10-8 [Cd + ] -8 1.5 x 10-3 6.1x 10.46 x 10-6.5 x 10-6 M To calculate K f, the [Cd(NH 3 ) 4 + ] term must be calculated. This can be done by recognizing that the molar solubility of CdC O 4 is the sum of the concentration of Cd + and Cd(NH 3 ) 4 + ions: Molar solubility of CdC O 4 [Cd + ] + [Cd(NH 3 ) 4 + ] 6.1 x 10-3 (.46 x 10-6 ) + [Cd(NH 3 ) 4 + ] (continued)
7 CHAPTER 18 [Cd(NH 3 ) 4 + ] (6.1 x 10-3 ) - (.46 x 10-6 ) 6.1 x 10-3 M Now the [NH 3 ] term must be calculated by subtracting the ammonia in [Cd(NH 3 ) 4 + ] from the starting NH 3 of 0.0150 M: [NH 3 ] 0.150-4[Cd(NH 3 ) 4 + ] 0.150-4(6.1 x 10-3 ) 0.156 M Solve for K f by substituting the known concentrations into the K f expression: K f + [Cd(NH 3) 4 ] + 4 [Cd ][NH 3 ] -3 6.1 x 10-6 4 (.46 x 10 )(0.156) 9.96 x 10 6 1.0 x 10 7 18.89 The OH - formed by ionization of NH 3 (to NH 4 + and OH - ) is a common ion that will precipitate Mg + as the slightly soluble Mg(OH) salt. The simplest way to treat the problem is to calculate the [OH - ] of 0.10 M NH 3 before the soluble Mg + salt is added. As the soluble Mg + salt is added, the [Mg + ] will increase until the solution is saturated with respect to Mg(OH) (the next Mg + ions to be added will precipitate). Calculate the [Mg + ] at the point at which precipitation begins. To calculate the [OH - ] of 0.10 M NH 3, let x equal the mol/ of NH 3 that ionize, forming x mol/ of NH 4 + and x mol/ of OH - and leaving (0.10 - x) M NH 3 in solution. We can summarize the situation in tabular form: Conc. (M) NH 3 + H O NH 4 + + OH - Starting 0.10 ~0 0 Change -x +x +x Equilibrium 0.10 - x x x The equilibrium-constant equation is: + - [NH4 ][OH ] x Kc [NH ] (0.10 - x) 3 x (0.10) The value of x can be obtained by rearranging and taking the square root: - [OH ] -5 1.8 x 10 x 0.10 1.34-3 x 10 M (continued)
SOUBIITY AND COMPEX-ION EQUIIBRIA 73 Note that (0.10 - x) is not significantly different from 0.10, so x can be ignored in the (0.10 - x) term. Now, use the K sp of Mg(OH) to calculate the [Mg + ] of a saturated solution of Mg(OH), which essentially will be the Mg + ion concentration when Mg(OH) begins to precipitate. [Mg + ] K sp [OH ] -11 1.8 x 10-3 (1.34 x 10 ) 1.00 x 10-5 1.0 x 10-5 M 18.90 The OH - formed by ionization of N H 4 (to N H 5 + and OH - ) is a common ion that will precipitate Mg + as the slightly soluble Mg(OH) salt. The simplest way to treat the problem is to calculate the [OH - ] of 0.0 M N H 4 before the soluble Mg + salt is added. As the soluble Mg + salt is added, the [Mg + ] will increase until the solution is saturated with respect to Mg(OH) (the next Mg + ions to be added will precipitate). Calculate the [Mg + ] at the point at which precipitation begins. To calculate the [OH - ] of 0.0 M N H 4, let x equal the mol/ of N H 4 that ionizes, forming x mol/ of N H 5 and x mol/ of OH - and leaving (0.0 - x) M N H 4 in solution. We can summarize the situation in tabular form: Conc. (M) N H 4 + H O N H 5 + + OH - Starting 0.0 ~0 0 Change -x +x +x Equilibrium 0.0 - x x x The equilibrium-constant equation is: + - [NH5 ][OH ] x Kc [N H ] (0.0 - x) 4 x (0.0) The value of x can be obtained by rearranging and taking the square root: - [OH ] -6 1.7 x 10 x 0.0 5.83-4 x 10 M (continued)
74 CHAPTER 18 Note that (0.0 - x) is not significantly different from 0.10, so x can be ignored in the (0.0 - x) term. Now, use the K sp of Mg(OH) to calculate the [Mg + ] of a saturated solution of Mg(OH), which essentially will be the Mg + ion concentration when Mg(OH) begins to precipitate. [Mg + ] K sp [OH ] -11 1.8 x 10-4 (5.83 x 10 ) 5.9 x 10-5 5.3 x 10-5 M 18.91 a. Use the solubility information to calculate K sp. The reaction is Cu(IO 3 ) (s) Cu + (aq) + IO - 3 (aq).7 x 10-3 x (.7 x 10-3 ) K sp [Cu + ][IO - 3 ] [.7 x 10-3 ][ x (.7 x 10-3 )] 7.87 x 10-8 Set up an equilibrium. The reaction is Cu(IO 3 ) (s) Cu + (aq) + IO - 3 (aq) y 0.35 + y 0.35 K sp [y][0.35] 7.87 x 10-8 Molar solubility y 6.4 x 10-7 6.4 x 10-7 M b. Set up an equilibrium. The reaction is Cu(IO 3 ) (s) Cu + (aq) + IO 3 - (aq) y + 0.35 0.35 y K sp [0.35][y] 7.87 x 10-8 Molar solubility y.37 x 10-4.4 x 10-4 M c. Yes, Cu(IO 3 ) is a 1: electrolyte. It takes two IO 3 - ions to combine with one Cu + ion. The IO 3 - ion is involved as a square term in the K sp expression.
SOUBIITY AND COMPEX-ION EQUIIBRIA 75 18.9 a. Use the solubility information to calculate K sp. The reaction is Pb(IO 3 ) (s) Pb + (aq) + IO - 3 (aq) 4.0 x 10-5 M x (4.0 x 10-5 ) M K sp [Pb + ][IO - 3 ] [4.0 x 10-5 ][ x (4.0 x 10-5 )].56 x 10-13 Set up an equilibrium. The reaction is Pb(IO 3 ) (s) Pb + (aq) + IO - 3 (aq) y + 0.15 0.15 y K sp [0.15][y].56 x 10-13 Molar solubility y 6.53 x 10-7 6.5 x 10-7 M b. Because of the common ion effect or a consideration of e Chatelier s principle, the molar solubility would be less. 18.93 a. Set up an equilibrium. The reaction and equilibrium-constant expression are PbI (s) Pb + (aq) + I - (aq) K sp [Pb + ][I - ] 6.5 x 10-9 The Pb + ion concentration is 0.0150 M. Plug this in, and solve for the iodine ion concentration. - [I ] -9 6.5 x 10 0.0150-4 6.58 x 10-4 6.6 x 10 M b. Solve the equilibrium-constant expression for the lead ion concentration. + Pb ] -9 6.5 x 10-3 (.0 x 10 ) [ 1.63 x 10-3 M The percent of the lead(ii) ion remaining in solution is Percent Pb + remaining -3 1.63 x 10 0.0150 x 100% 10.9 11 percent
76 CHAPTER 18 18.94 a. Set up an equilibrium. The reaction and equilibrium-constant expression are CaF (s) Ca + (aq) + F - (aq) K sp [Ca + ][F - ] 3.4 x 10-11 Solve the equilibrium-constant expression for the fluoride ion concentration. The calcium ion concentration is 0.00750 M. [F - ] -11 3.4 x 10 0.00750 6.73 x 10-5 6.7 x 10-5 M b. Solve the equilibrium-constant expression for the calcium ion concentration. The fluoride ion concentration is 9.5 x 10-4 M. -11 + 3.4 x 10 [ Ca ] 3.77 x 10-5 M -4 ( 9.5 x 10 ) The percent of the calcium ion precipitated is Percent Ca + precipitated -5 0.00750-3.77 x 10 0.00750 x 100% 99.4 99 percent 18.95 a. The reaction and equilibrium-constant expression are Co(OH) (s) Co + (aq) + OH - (aq) K sp [Co + ][OH - ] From the molar solubility, [Co + ] 5.4 x 10-6 M and [OH - ] x (5.4 x 10-6 ) M. Therefore, K sp [5.4 x 10-6 ][ x (5.4 x 10-6 )] 6.30 x 10-16 6.3 x 10-16 b. From the poh (14 - ph), the [OH - ] 10-3.57.69 x 10-4 M. The molar solubility is equal to the cobalt ion concentration at equilibrium. [Co + ] K sp - [ OH ] -16 6.3 x 10-4 (.69 x 10 ) 8.71 x 10-9 8.7 x 10-9 M c. The common ion effect (OH - ) in part (b) decreases the solubility of Co +.
18.96 a. The reaction and equilibrium-constant expression are SOUBIITY AND COMPEX-ION EQUIIBRIA 77 Be(OH) (s) Be + (aq) + OH - (aq) K sp [Be + ][OH - ] From the molar solubility, [Be + ] 8.6 x 10-7 M and [OH - ] x (8.6 x 10-7 ) M. Therefore, K sp [8.6 x 10-7 ][ x (8.6 x 10-7 )].54 x 10-18.5 x 10-18 b. This is a buffer system. The reaction is NH 3 + H O NH 4 + + OH - K b 1.8 x 10-5 Set up the equilibrium, and calculate the OH - ion concentration. K b + - [NH4 ][ OH ] [NH3 ] - ( 0.5 )[ OH ] (1.50 ) -5 1.8 x 10 [OH - ] 1.08 x 10-4 M The molar solubility is equal to the beryllium ion concentration at equilibrium. [Be + ] K sp - [ OH ] -18.5 x 10-4 (1.08 x 10 ).14 x 10-10.1 x 10-10 M c. The common ion effect (a e Chatelier stress) decreases the solubility of Be +. 18.97 a. AgCl(s) Ag + (aq) + Cl - (aq) K sp 1.8 x 10-10 Ag + (aq) + NH 3 (aq) Ag(NH 3 ) + (aq) K f 1.7 x 10 7 AgCl(s) + NH 3 (aq) Ag(NH 3 ) + (aq) + Cl - (aq) K K sp x K f (1.8 x 10-10 )(1.7 x 10 7 ) 3.06 x 10-3 3.1 x 10-3 (continued)
78 CHAPTER 18 b. The equilibrium-constant expression is + - [ Ag(NH3 ) ][ Cl ] [NH3 ] y ( 0.80 ) -3 3.06 x 10 y [Ag(NH 3 ) + ] 0.0443 0.044 M mol AgCl dissolved mol Ag(NH 3 ) + 0.044 mol/ x 1.00 0.044 mol mol NH 3 reacted 0.0443 mol Ag(NH 3 ) + x molnh3 + 1molAg(NH 3) 0.0886 mol The total moles of NH 3 added 0.80 M x 1.00 + 0.0886 0.889 0.89 mol 18.98 a. AgBr(s) Ag + (aq) + Br - (aq) K sp 5.0 x 10-13 Ag + (aq) + S O 3 - (aq) Ag(S O 3 ) 3- K f.9 x 10 13 AgBr(s) + S O 3 - (aq) Ag(S O 3 ) 3- (aq) + Br - (aq) K K sp x K f (5.0 x 10-13 )(.9 x 10 13 ) 14.5 15 mol AgBr.5 g AgBr x 1 mol AgBr 187.77 g AgBr 0.0133 mol mol Ag(S O 3 ) 3- mol AgBr 0.0133 mol - - [ Ag( SO3 ) ][Br ] - [ SO3 ] ( 0.0133 ) - [ SO3 ] 14.5 Take the square root of both sides to obtain the moles of S O 3 - at equilibrium. 0.0133 - [ SO3 ] 14.5 [S O 3 - ] 0.00349 mol/ (continued)
SOUBIITY AND COMPEX-ION EQUIIBRIA 79 The moles of S O 3 - in Ag(S O 3 ) 3- x 0.0133 0.066 mol Total moles of S O 3-0.066 + 0.00349 0.0301 0.030 mol The moles of Na S O 3 added moles of S O 3-0.030 mol 18.99 a. The moles of NH 4 Cl (molar mass 53.49 g/mol) are given by 1molNHCl mol NH 4 Cl 6.7 g x 4 53.49gNHCl 4 0.499 mol The reaction and equilibrium-constant expression are NH 3 + H O NH 4 + + OH - K b + - [NH4 ][OH ] 1.8 x 10-5 [NH3 ] Solve the equilibrium-constant expression for [OH - ]. The molarity of the NH 4 Cl is 0.499 mol/1.0 0.499 M. - [ OH ] -5 (1.8 x 10 )(4. ) ( 0.499 ) 1.51 x 10-4 1.5 x 10-4 M b. The reaction and equilibrium-constant expression are Mg(OH) Mg + + OH - K sp [Mg + ][OH - ] 1.8 x 10-11 Solving for [Mg + ] gives -11 + (1.8 x 10 ) [ Mg ] 7.89 x 10-4 7.9 x 10-4 M -4 (1.51x 10 ) The percent Mg + that has been removed is given by Percent Mg + removed -4 0.075-7.89 x 10 0.075 x 100% 98.9 99 percent
730 CHAPTER 18 18.100 a. The moles of (NH 4 ) SO 4 (molar mass 13.15 g/mol) are given by 1mol(NH mol (NH 4 ) SO 4 75.8 g x 4)SO 4 13.15 g (NH ) SO 4 4 0.5736 mol The moles of NH + 4 x 0.5736 1.147 mol. The reaction and equilibriumconstant expression are NH 3 + H O NH 4 + + OH - K b + - [NH 4 ][OH ] [NH 3 ] 1.8 x 10-5 Solve the equilibrium-constant expression for [OH - ]. The molarity of the NH 4 Cl is 1.147 mol/1.0 1.147 M. - [ OH ] -5 (1.8 x 10 )(1.6 ) (1.147 ).51 x 10-5.5 x 10-5 M b. The reaction and equilibrium-constant expression are Mn(OH) Mn + + OH - K sp [Mn + ][OH - ] 4.6 x 10-14 Solving for [Mn + ] gives -14 + ( 4.6 x 10 ) [ Mn ] 7.30 x 10-5 7.3 x 10-5 M -5 (.51x 10 ) The percent Mn + that has been removed is given by Percent Mn + removed -5 0.058-7.30 x 10 0.058 x 100% 99.9 percent Nearly 100 percent of the Mn + has been removed. Solutions to Cumulative-Skills Problems 18.101 Using the K sp of 1.1 x 10-1 for ZnS, calculate the [S - ] needed to maintain a saturated solution (without precipitation): [S - ] -1 1.1x 10-4 + 1.5 x 10 M Zn 7.33 x 10-18 M (continued)
SOUBIITY AND COMPEX-ION EQUIIBRIA 731 Next, use the overall H S ionization expression to calculate the [H 3 O + ] needed to achieve this [S - ] level: [H 3 O + ] -0 1.1x 10 (0.10 M) -18 7.33 x 10 M 1.5 x 10 - M Finally, calculate the buffer ratio of [SO 4 - ]/[HSO 4 - ] from the H SO 4 K a expression where K a has the value 1.1 x 10 - : - [SO 4 ] - [HSO4 ] K a + [H3O ] - 1.1x 10-1.5 x 10 0.8979 1.000 If [HSO 4 - ] 0.0 M, then [SO 4 - ] 0.8979 x 0.0 M 0.1795 0.18 M 18.10 Using the K sp of 4 x 10-1 for CoS, calculate the [S - ] needed to maintain a saturated solution (without precipitation): [S - ] -1 4 x 10-4 + 1.8 x 10 M Co. x 10-17 M Next, use the overall H S ionization expression to calculate the [H 3 O + ] needed to achieve this [S - ] level: [H 3 O + ] -0 1.1x 10 (0.10 M) -17. x 10 M 7.04 x 10-3 M Finally, calculate the buffer ratio of [SO 4 - ]/[HSO 4 - ] from the H SO 4 K a expression where K a has the value 1.1 x 10 - : - [SO 4 ] - [HSO4 ] K a + [H3O ] - 1.1x 10-3 7.04 x 10 1.56 1.000 If [HSO 4 - ] 0.0 M, then [SO 4 - ] 1.56 x 0.0 M 0.31 0.3 M
73 CHAPTER 18 18.103 Begin by solving for [H 3 O + ] in the buffer. Ignoring changes in [HCHO ] as a result of ionization in the buffer, you obtain [H 3 O + ] 1.7 x 10-4 x 0.45 M 0.0 M 3.85 x 10-4 M You should verify that this approximation is valid (you obtain the same result from the Henderson-Hasselbalch equation). The equilibrium for the dissolution of CaF in acidic solution is obtained by subtracting twice the acid ionization of HF from the solubility equilibrium of CaF : CaF (s) Ca + (aq) + F - (aq) K sp H 3 O + (aq) + F - (aq) HF(aq) + H O(l) 1/(K a ) H 3 O + (aq) + CaF (s) Ca + (aq) + HF(aq) + H O(l) K c K sp /(K a ) Therefore, K c (3.4 x 10-11 ) (6.8 x 10-4 ) 7.35 x 10-5. In order to solve the equilibrium-constant equation, you require the concentration of HF, which you obtain from the acid-ionization constant for HF. K a + - [H3O ][F ] [HF] 6.8 x 10-4 3.85 x 10-4 x - [F ] [HF] - [F ] [HF] 1.778, or [F - ] 1.778 [HF] et x be the solubility of CaF in the buffer. Then, [Ca + ] x and [F - ] + [HF] x. Substituting from the previous equation, you obtain x 1.778[HF] + [HF].778[HF], or [HF] x/.778 You can now substitute for [H 3 O + ] and [HF] into the equation for K c. + [ Ca ][HF ] x( x/.778 ) -5 K c 7.35 x 10 + -4 [H3O ] ( 3.85 x 10 ) 7.35 x 10-5 (3.543 x 10 6 )x 3 x 3.075 x 10-11 x.748 x 10-4.7 x 10-4 M
SOUBIITY AND COMPEX-ION EQUIIBRIA 733 18.104 Begin by solving for [H 3 O + ] in the buffer. Ignoring changes in [HC H 3 O ] as a result of ionization in the buffer, you obtain [H 3 O + ] 1.7 x 10-5 x 0.45 M 0.0 M 3.85 x 10-5 M You should verify that this approximation is valid (you obtain the same result from the Henderson-Hasselbalch equation). The equilibrium for the dissolution of MgF in acidic solution is obtained by subtracting twice the acid ionization of HF from the solubility equilibrium of MgF : MgF (s) Mg + (aq) + F - (aq) K sp H 3 O + (aq) + F - (aq) HF(aq) + H O(l) 1/(K a ) H 3 O + (aq) + MgF (s) Mg + (aq) + HF(aq) + H O(l) K c K sp /(K a ) Therefore, K c (6.5 x 10-9 ) (6.8 x 10-4 ) 1.406 x 10 -. In order to solve the equilibrium-constant equation, you require the concentration of HF, which you obtain from the acid-ionization constant for HF. K a + - [H3O ][F ] [HF] 6.8 x 10-4 3.85 x 10-5 x - [F ] [HF] - [F ] [HF] 17.78, or [F - ] 17.78 [HF] et x be the solubility of MgF in the buffer. Then, [Mg + ] x and [F - ] + [HF] x. Substituting from the previous equation, you obtain x 17.78[HF] + [HF] 18.78[HF], or [HF] x/18.78 You can now substitute for [H 3 O + ] and [HF] into the equation for K c. K c + [Mg ][HF ] + [H3O ] x( x/18.78 ) -5 ( 3.85 x 10 ) 1.406 x 10-1.406 x 10 - (7.75 x 10 6 )x 3 x 3 1.81 x 10-9 x 1.19 x 10-3 1. x 10-3 M
734 CHAPTER 18 18.105 The net ionic equation is Ba + (aq) + OH - (aq) + Mg + (aq) + SO 4 - (aq) BaSO 4 (s) + Mg(OH) (s) Start by calculating the mol/ of each ion after mixing and before precipitation. Use a total volume of 0.0450 + 0.0670 0.11. M of SO 4 - and Mg + (0.350 mol/ x 0.0670 ) 0.11 0.094 M M of Ba + M of OH - (0.50 mol/ x 0.0450 ) 0.11 0.1004 M ( x 0.50 mol/ x 0.0450 ) 0.11 0.0089 M Assemble a table showing the precipitation of BaSO 4 and Mg(OH). Conc. (M) Ba + + SO 4 - + Mg + + OH - BaSO 4 (s) + Mg(OH) (s) Starting 0.1004 0.094 0.094 0.0089 Change -0.1004-0.1004-0.1004-0.0089 Equilibrium 0.0000 0.1090 0.1090 0.0000 To calculate [Ba + ], use the K sp expression for BaSO 4. [Ba + ] -10 1.1x 10-0.1090 M SO4 1.009 x 10-9 1.0 x 10-9 M [SO 4 - ] 0.1090 0.109 M Calculate the [OH - ] using the K sp expression for Mg(OH) : [OH - ] -11 1.8 x 10 + 0.1090 M Mg 1.8 x 10-5 1.3 x 10-5 M [Mg + ] 0.1090 0.109 M
SOUBIITY AND COMPEX-ION EQUIIBRIA 735 18.106 The net ionic equation is Pb + (aq) + Cl - (aq) + Ag + (aq) + SO 4 - (aq) PbSO 4 (s) + AgCl(s) Start by calculating the mol/ of each ion after mixing and before precipitation. Use a total volume of 0.050 + 0.0500 0.0750. M of SO 4 - M of Pb + (0.0150 mol/ x 0.0500 ) 0.0750 0.01000 M (0.0100 mol/ x 0.050 ) 0.0750 0.003333 M M of Ag + M of Cl - ( x 0.0150 mol/ x 0.0500 ) 0.0750 0.0000 M ( x 0.0100 mol/ x 0.050 ) 0.07500 0.006666 M Assemble a table showing the precipitation of PbSO 4 and AgCl. Conc. (M) Pb + + SO 4 - + Ag + + Cl - PbSO 4 (s) + AgCl(s) Starting 0.003333 0.01000 0.0000 0.006667 Change -0.003333-0.003333-0.006667-0.006667 Equilibrium 0.000000 0.006667 0.013333 0.000000 To calculate [Pb + ], use the K sp expression for PbSO 4. [Pb + ] -8 1.7 x 10-0.006667 M SO4.549 x 10-6.50 x 10-6 M [SO 4 - ] 0.006667 0.0067M Calculate the [Cl - ] using the K sp expression for AgCl: [Cl - ] -10 1.8 x 10 + 0.01333 M Ag 1.3503 x 10-8 1.4 x 10-8 M [Ag + ] 0.01333 0.0133 M
Filename: chapter18.doc Directory: D:\hmco\chemistry\general\ebbing\general_chem\8e\instructors\ solutions_manual Template: C:\Documents and Settings\willsoj\Application Data\Microsoft\Templates\Normal.dot Title: 17 Subject: Author: David Bookin Keywords: Comments: Creation Date: 1/8/000 10:6 AM Change Number: 87 ast Saved On: 1/5/003 3:06 PM ast Saved By: David Bookin Total Editing Time: 718 Minutes ast Printed On: 1/9/004 3:30 PM As of ast Complete Printing Number of Pages: 58 Number of Words: 11,71 (approx.) Number of Characters: 66,810 (approx.)