General Chemistry by Whitten, Atwood, Morrison Chapter 20 solutions

General Chemistry by Whitten, Atwood, Morrison Chapter 20 solutions

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222 20 Ionic Equilibrium III: The Solubility Product Principle 20-1 Most salts are more soluble at higher temperatures. Therefore, we should expect K sp values for most salts to increase if the temperature is increased. 20-3 The “concentration of a solid” is constant, depending only on the density of the solid, so it does not appear in the equilibrium constant expression, K sp . In thermodynamic terms, the activity of a pure solid is taken by definition to be one. NOTE: In the chemical equations in this chapter, state designations (solids on left, aqueous on right for solubility equilibria) are often omitted. 20-5 (a) MgF 2 (s) Mg 2+ + 2F K sp = [Mg ][F ] 2 (b) AlPO 4 (s) Al 3+ + PO 4 3– K sp = [Al ][PO 4 3– ] (c) CuCO 3 (s) Cu + CO 3 2 K sp = [Cu ][ CO 3 2 ] (d) Ag 3 PO 4 (s) 3Ag + + PO 4 3– K sp = [Ag + ] 3 [PO 4 3– ] 20-7 The coefficients from the balanced equations become the exponents in equilibrium equations. In the dissolving of Mg(OH) 2 (s) (or [Mg(OH) 2 (s) Mg + 2OH ]) the coefficient of OH is 2 and it becomes the exponent of [OH ] in the K sp expression. The OH is the only species in the dissolving of the 2 salts in this question that has a coefficient other than 1. 20-9 Molar solubility = mol solute dissolved L solution = mmol solute dissolved mL solution (a) Molar solubility of CuBr = 1.0 x 10 -3 g CuBr L x 1 mol CuBr 143.45 g CuBr = 7.0 x 10 –6 M CuBr(s) Cu + + Br 7.0 x 10 –6 M 7.0 x 10 –6 M 7.0 x 10 –6 M K sp = [Cu + ][Br ] = (7.0 x 10 –6 )(7.0 x 10 –6 ) = 4.9 x 10 -11
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223 (b) Molar solubility of AgI = 2.8 x 10 -8 g AgCl 10 mL x 1000 mL L x 1 mol AgI 234.8 g AgI = 1.2 x 10 –8 M AgI(s) Ag + + I 1.2 x 10 –8 M 1.2 x 10 –8 M 1.2 x 10 –8 M K sp = [Ag + ][I ] = (1.2 x 10 –8 )(1.2 x 10 –8 ) = 1.4 x 10 -16 (c) Molar solubility of Pb 3 (PO 4 ) 2 = 6.2 x 10 -7 g Pb 3 4 ) 2 L x 1 mol Pb 3 4 ) 2 811.6 g Pb 3 4 ) 2 = 7.6 x 10 –10 M Pb 3 4 ) 2 (s) 3Pb 2+ + 2PO 4 3– 7.6 x 10 –10 M 3(7.6 x 10 –10 ) M 2(7.6 x 10 –10 ) M = 2.3 x 10 –9 M = 1.5 x 10 –9 M K sp = [Pb ] 3 [PO 4 3– ] 2 = (2.3 x 10 –9 ) 3 (1.5 x 10 –9 ) 2 = 2.7 x 10 -44 (d) Molar solubility of Ag 2 SO 4 = 5.0 mg Ag 2 4 mL x 1 mmol Ag 2 4 311.8 mg Ag 2 4 = 1.6 x 10 –2 M 2 4 (s) 2Ag + + 4 2– 1.6 x 10 –2 M 2(1.6 x 10 –2 ) M 1.6 x 10 –2 M = 3.2 x 10 –2 M K sp = [Ag + ] 2 [SO 4 2– ] = (3.2 x 10 –2 ) 2 (1.6 x 10 –2 ) = 1.6 x 10 -5 20-11 Compound Molar solubility K sp CuBr 7.0 x 10 –6 M K sp = [Cu + ][Br ] = 4.9 x 10 –11 AgI 1.2 x 10 –8 M K sp = [Ag + ][I ] = 1.5 x 10 –16 3 4 ) 2 7.6 x 10 –10 M K sp = [Pb ] 3 4 3– ] 2 = 2.7 x 10 –44 2 4 1.6 x 10 –2 M K sp = [Ag + ] 2 4 2– ] = 1.6 x 10 –5 (a) 2 4 has the highest molar solubility. (b) Pb 3 4 ) 2 has the lowest molar solubility. (c) 2 4 has the largest K sp . (d) Pb 3 4 ) 2 has the smallest K sp . 20-13 Solubility Zn 3 4 ) 2 = 1.18 x 10 -4 g 2.5 L = 1 mol Zn 3 4 ) 2 386.11 g Zn 3 (PO 4 ) 2 = 1.2 x 10 –7 M Zn 3 4 ) 2 (s) 3Zn + 2PO 4 3- [Zn ] = 3.6 x 10 –7 M [PO 4 ] = 2.4 x 10 –7 M K sp for Zn 3 4 ) 2 = [Zn ] 3 4 ] 2 = (3.6 x 10 –7 ) 3 (2.4 x 10 –7 ) 2 = 2.7 x 10 -33
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224 20-15 (a) From Appendix H, K sp for Cd(CN) 2 = 1.0 x 10 –8 Let x = molar solubility of Cd(CN) 2 Cd (CN) 2 (s) Cd 2+ + 2CN -
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General Chemistry by Whitten, Atwood, Morrison Chapter 20 solutions

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