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SQSM-08

# SQSM-08 - Chapter 8 Equilibria Containing Metal Ions 1...

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Chapter 8 Equilibria Containing Metal Ions 8-1 1. Explain why Fe(H 2 O) 6 3+ is a stronger acid than Fe(H 2 O) 6 2+ . The higher oxidation state on the Fe in Fe(H 2 O) 6 3+ results in more electron density being withdrawn from the O-H bonds in H 2 O, which makes the water a stronger acid. 3. Write the chemical equation that explains the acidity of an aqueous CuSO 4 solution and calculate the pH of a 0.20 M CuSO 4 solution. CuSO 4 completely dissociates in water to form the copper hydrate, a weak acid. The K a can be found in Table 8.1, and the reaction table for the reaction is Cu(H 2 O) 6 2+ + H 2 O U Cu(H 2 O) 5 OH 1+ + H 3 O + Initial: 0.20 - 0 0 - x + x + x Final: 0.20 –x x x 1+ 1+ 2 -8 2 5 3 a 2+ 2 6 [Cu(H O) OH ][H O ] (x)(x) x K 3 10 = [Cu(H O) OH ] 0.20-x 0.20 = × = = The above simplification is possible because K a for the reaction is so small that x will be negligible relative to 0.20 M. Therefore, -8 -5 x= (3 x 10 )(0.20) 8 x 10 M = pH = -log[H 3 O 1+ ] = 4.1 5. Write the chemical equation for the dissolution process and the K sp expression in terms of the molar solubility for each of the following substances: a) CoS CoS(s) U Co 2+ + S 2- K sp = [Co 2+ ][S 2- ] b) HgI 2 HgI 2 (s) U Hg 2+ + 2I 1- K sp = [Hg 2+ ][I 1- ] 2 c) Al(OH) 3 Al(OH) 3 (s) U Al 3+ + 3OH 1- K sp = [Al 3+ ][OH 1- ] 3 7. Express the K sp expression of each of the compounds in Exercise 5 in terms of its molar solubility (x). a) CoS K sp = [Co 2+ ][S 2- ] = (x)(x) = x 2 b) HgI 2 K sp = [Hg 2+ ][I 1- ] 2 = (x)(2x) 2 = 4x 3 c) Al(OH) 3 K sp = [Al 3+ ][OH 1- ] 3 = (x)(3x) 3 = 27x 4 9. Write the chemical equations for the dissolution of each of the following substances and determine their molar solubilities. a) AgI (a 1:1 salt) AgI(s) U Ag 1+ + I 1- K sp = 8.3 x 10 -17 = x 2 -17 -9 sp x = K = 8.3 10 = 9.1 10 M × × b) CaF 2 (a 1:2 salt) CaF 2 (s) U Ca 2+ + 2F 1- K sp = 3.9 x 10 -11 = 4x 3 3 -11 sp -4 3 K 3.9 10 x = = = 2.1 10 M 4 4 × × 11. The solubility of mercury(I) chloride is 0.0020 g/L. What is its K sp ? Note: mercury(I) exists as Hg 2 2+ ions. Hg 2 Cl 2 (s) U Hg 2 2+ + 2Cl 1- M m (Hg 2 Cl 2 ) = 472 g/mol 2+ -6 2 2+ 1- 2 3 -6 3 -16 sp 2 0.0020 g/L x = [Hg ] = = 4.2 10 M 472 g/mol K = [Hg ][Cl ] = (x)(2x) = 4x = 4(4.2 10 ) = 3.0 10 × × × 13. The gold ion concentration in a saturated solution of gold(III) chloride is 33 µ M. What is the solubility-product constant of AuCl 3 ? The dissolution reaction is AuCl 3 (s) U Au 3+ + 3Cl 1- . K sp = [Au 3+ ][Cl 1- ] 3 = (x)(3x) 3 = 27x 4 . x = [Au 3+ ] = 33 x 10 -6 M, so K sp = 27(33 x 10 -6 ) 4 = 3.2 x 10 -17

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Equilibria Containing Metal Ions 8-2 15. What is the pH of saturated barium hydroxide? Dissolution reaction: Ba(OH) 2 U Ba 2+ + 2OH 1- K sp = 5.0 x 10 -3 from Appendix D -3 sp 1- 3 3 K 5.0 10 x = = = 0.11 M; [OH ] = 2x = 0.22 M pOH = 0.67 4 4 × and pH = 14.00 – pOH = 13.33 17. A 386-mg sample of PbCl 2 is washed with 10.0 mL of 0.10 M HCl. What is the maximum fraction of PbCl 2 that can dissolve in the wash? PbCl 2 (s) U Pb 2+ + 2Cl 1- , K sp = [Pb 2+ ][Cl 1- ] 2 We are given that the chloride ion concentration is 0.10 M, and asked to determine the solubility of the PbCl 2 . This is a common ion solubility, so we solve for the lead(II) concentration: -5 sp 2+ -3 1- 2 2 K 1.7 10 [Pb ] = = = 1.7 10 M [Cl ] (0.10) × × where we have assumed that the amount of chloride obtained by the dissolution is negligible compared to 0.10 M. The mass of PbCl 2 that must dissolve to produce that concentration is -3 2+ 2 2 2 2+ 2 1.7 10 mmol Pb 1 mmol PbCl 278.1 mg PbCl
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SQSM-08 - Chapter 8 Equilibria Containing Metal Ions 1...

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