Chem 162 2016 Chapter 18 Acid base solubility equilibria lecture notes 48 CHEM

Chem 162 2016 chapter 18 acid base solubility

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Chem 162-2016 Chapter 18 Acid-base & solubility equilibria lecture notes 48 CHEM 162-2010 EXAM III Chapter 16 - Equilibria Solubility product ET note: Ksp of varied substances only related to solubility if molar ratio of ions are identical. K sp values vs. Relative Solubility 18. Which of the following compounds would have the greatest molar solubility? A. AgCl (K sp = 1.8 x 10 -10 ) B. AgBr (K sp = 5.0 x 10 -13 ) C. PbCO 3 (K sp = 7.4 x 10 -14 ) D. BaSO 4 (K sp = 1.1 x 10 -10 ) E . Ag 2 CrO 4 (K sp = 2.4 x 10 -12 ) AgCl(s) Ag + (aq) + Cl - (aq) AgBr(s) Ag + (aq) + Br - (aq) PbCO 3 (s) Pb 2+ (aq) + CO 3 2- (aq) BaSO 4 (s) Ba 2+ (aq) + SO 4 2- (aq) Ag 2 CrO 4 (s) 2Ag + (aq) + CrO 4 2- (aq) AgCl(s) Ag + (aq) + Cl - (aq) AgCl(s) Ag + (aq) + Cl - (aq) Initial Y 0 0 Change -X +X +X Equilibrium Y-X +X +X [Ag + ][Cl - ] = K sp = 1.8 x 10 -10 [X][X] = 1.8 x 10 -10 [X] = 1.34 x 10 -5 = solubility of AgCl PbCO 3 (s) Pb 2+ (aq) + CO 3 2- (aq) PbCO 3 (s) Pb 2+ (aq) + CO 3 2- (aq) Initial Y 0 0 Change -X +X +X Equilibrium Y-X +X +X [Pb 2+ ][CO 3 2- ] = K sp = 7.4 x 10 -14 [X][X] = 7.4 x 10 -14 [X] = 2.72 x 10 -7 = solubility of PbCO 3 The AgCl, AgBr, PbCO 3 , BaSO 4 all have the same molar ratio of ions, that is, 1:1. Therefore, they all use the same ICE table format, and therefore the same formula for finding the solubility, which is X 2 = K sp . Hence, the most soluble of these would be the one with the largest K sp , which is AgCl. But Ag 2 CrO 4 has a different ICE table format, and therefore a different formula for determining its solubility. So we need to quantitatively compare AgCl to Ag 2 CrO 4 . Ag 2 CrO 4 (s) 2Ag + (aq) + CrO 4 2- (aq) Ag 2 CrO 4 (s) 2Ag + (aq) + CrO 4 2- (aq) Initial Y 0 0 Change -X +2X +X Equilibrium Y-X +2X +X [Ag + ] 2 [CrO 4 2- ] = K sp = 2.4 x 10 -12 [2X] 2 [X] = 2.4 x 10 -12 [X] = 8.44 x 10 -5 = solubility of Ag 2 CrO 4 = greatest molar solubility, despite not having highest K sp .
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Chem 162-2016 Chapter 18 Acid-base & solubility equilibria lecture notes 49 SKIP (CONCEPT COVERED ON PREVIOUS PAGE) K sp values vs. Relative Solubility ET note: Relationship of Ksp value and solubility Given the K sp’s of CaF 2 , Mg(OH) 2 and FeCO 3 (1.46 x 10 -10 , 2.06 x 10 -13 and 3.07 x 10 -11 , respectively), can the relative solubilities be determined? CaF 2 (s) Ca 2+ (aq) + 2F - (aq) CaF 2 (s) Ca 2+ (aq) + 2F - (aq) Initial Y 0 0 Change -X X 2X Equilibrium Y-X +X +2X [Ca 2+ ][F - ] 2 = K sp [X][2X] 2 = 1.46 x 10 -10 4X 3 = 1.46 x 10 -10 X = 3.32 x 10 -4 M Mg(OH) 2 (s) Mg 2+ (aq) + 2OH - (aq) Mg(OH) 2 (s) Mg 2+ (aq) + 2OH - (aq) Initial Y 0 0 Change -X X 2X Equilibrium Y-X +X +2X [Mg 2+ ][OH - ] 2 = K sp [X][2X] 2 = 2.06 x 10 -13 4X 3 = 2.06 x 10 -13 X = 3.72 x 10 -5 M FeCO 3 (s) Fe 2+ (aq) + CO 3 2- (aq) FeCO 3 (s) Fe 2+ (aq) + CO 3 2- (aq) Initial Y 0 0 Change -X +X +X Equilibrium Y-X +X +X [Fe 2+ ][ CO 3 2- ] = K sp [X][X] = 3.07 x 10 -11 X 2 = 3.07 x 10 -11 X = 5.54 x 10 -6 M Compound K sp Equil. Expr. Solubility (M) CaF 2 1.46 x 10 -10 [X][2X] 2 = K sp 3.32 x 10 -4 Mg(OH) 2 2.06 x 10 -13 [X][2X] 2 = K sp 3.72 x 10 -5 FeCO 3 3.07 x 10 -11 [X][X] = K sp 5.54 x 10 -6 Conclusion: K sp values are directly related to solubility only if the stoichiometric coefficients in the equation, and correspondingly, the forms of equilibrium expressions, are the same. Hence, since CaF 2 has a larger K sp than Mg(OH) 2 , and has the same form of equilibrium expression, then CaF 2 is more soluble than Mg(OH) 2 .
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