Chem 162-2010 homework 8th week H&amp;P Chapter 16

# Chem 162-2010 homework 8th week H&amp;P Chapter 16 -...

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Chem 162-2010 Homework 8 th Week Hill & Petrucci Chapter 16 Homework Problems The Solubility Product Constant, K sp 19. Write a chemical equation representing solubility equilibrium for (a) Hg 2 (CN) 2 , K sp = 5 x 10 -40 ; and (b) Ag 3 AsO 4 , K sp = 1.0 x 10 -22 . (a) Hg 2 (CN) 2 (s) Hg 2 2+ (aq) + 2CN - (aq) (b) Ag 3 AsO 4 (s) 3Ag + (aq) + AsO 4 3- (aq) The Relationship Between Solubility and K sp 23. Can the numerical values of the molar solubility and the solubility product constant of a slightly soluble ionic compound ever be the same? Which of the two is usually the larger value? Explain. They can never be the same. The Ksp is a very small number (less than one), and is a result of the product of two other small numbers (each less than one) representing the solubility of the individual ions. The solubility numbers must be larger than the K sp because the only way to have the product of two numbers, each less than one, become a number less than one, would be if the two numbers are larger than the product. For example: BaSO 4 (s) Ba 2+ + SO 4 2- K sp = 1.1 x 10 -10 Solubility of BaSO 4 = 1.05 x 10 -5 , which is a larger number than the K sp number. (1.05x10 -5 ) 2 =1.1x10 -10 . 27. For each set, determine which slightly soluble solute has the greater molar solubility: (a) AgCl or Ag 2 CrO 4 , and (b) Mg(OH) 2 or MgCO 3 . AgCl Ag + + Cl - K sp = 1.8 x 10 -10 AgCl Ag + + Cl - Initial Y 0 0 Change -X +X +X Equilibrium Y-X +X +X [Ag + ][Cl - ] = 1.8 x 10 -10 [X][X] = 1.8 x 10 -10 X= 1.3 x 10 -5 M = AgCl solubility Ag 2 CrO 4 2Ag + + CrO 4 2- K sp = 1.1 x 10 -12 Ag 2 CrO 4 2Ag + + CrO 4 2- Initial Y 0 0 Change -X +2X +X Equilibrium Y-X +2X +X ([Ag + ] 2 )[CrO 4 2- ] = 1.8 x 10 -10 ([2X] 2 )[X] = 1.1 x 10 -12 (((([2X] 2 ) x [X]) = (1.1 x 10 -12 )),X) X = 6.5 x 10 -5 M = Ag 2 CrO 4 solubility 1

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6.5 x 10 -5 M > 1.3 x 10 -5 M. Therefore, Ag 2 CrO 4 is more soluble. Mg(OH) 2 Mg 2+ + 2OH - K sp = 1.8 x 10 -11 Mg(OH) 2 Mg 2+ + 2OH - Initial Y 0 0 Change -X +X +2X Equilibrium Y-X +X +2X [Mg ++ ]([OH - ] 2 ) = 1.8 x 10 -11 [X]([2X] 2 ) = 1.8 x 10 -11 X = 1.65 x 10 -4 = Mg(OH) 2 solubility MgCO 3 Mg 2+ + CO 3 2- K sp = 3.5 x 10 -8 MgCO 3 Mg 2+ + CO 3 2- Initial Y 0 0 Change -X +X +X Equilibrium Y-X +X +X [Mg ++ ][CO 3 2- ] = 3.5 x 10 -8 [X][X] = 3.5 x 10 -8 X = 1.87 x 10 -4 = MgCO 3 solubility Since 1.87 x 10 -4 > 1.65 x 10 -4 , then MgCO 3 is more soluble than Mg(OH) 2 . 35. Given the K sp values for PbCl 2 of 1.6 x 10 -5 at 25 o C and 3.3 x 10 -3 at 80 o C, if 1.00 mL of saturated PbCl 2 (aq) at 80 o C is cooled to 25 o C, will a sufficient amount of PbCl 2 (s) precipitate to be visible? Assume that you can detect as little as 1 mg of the solid. PbCl 2 Pb 2+ + 2Cl - K sp = 1.6 x 10 -5 at 25 o C Pb(Cl) 2 Pb 2+ + 2Cl - Initial Y 0 0 Change -X +X +2X Equilibrium Y-X +X +2X [Pb ++ ]([Cl - ] 2 ) = 1.6 x 10 -5 [X]([2X] 2 ) = 1.6 x 10 -5 ((([X] x ([2X] 2 )) = (1.6 x 10 -5 )),X) X = 1.59 x 10 -2 = PbCl 2 solubility at 25 o C. PbCl 2 Pb 2+ + 2Cl - K sp = 3.3 x 10 -3 at 80 o C Pb(Cl) 2 Pb 2+ + 2Cl - Initial Y 0 0 Change -X +X +2X 2
Equilibrium Y-X +X +2X [Pb ++ ]([Cl - ] 2 ) = 3.3 x 10 -3 [X]([2X] 2 ) = 3.3 x 10 -3 ((([X] x ([2X] 2 )) = (3.3 x 10 -3 )),X) X = 9.38 x 10 -2 = PbCl 2 solubility at 80 o C.

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