Question

100 mL of a saturated solution of magnesium palmitate, Mg(C16H31O2)2, is prepared at 25 °C. The Ksp is 3.3 × 10^-12 . To this solution is added 0.015 g of magnesium palmitate and the temperature is raised to 50 °C. The Ksp is 4.8 × 10^-12. How much of the solid magnesium palmitate remains?

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I understand that the lower the Ksp, the less soluble a compound (lower molar solubility). I have a final on tuesday morning and would like to learn how to approach this question. Thanks so much!

Answer 1

ksp of Mg(C16H31O2)2 = [Mg2+][C16H31O2]^2

                                               = S*(2S)^2 = 4S^3

AT 25 C   KSP = 4S^3 = 3.3*10^-12

       S = solubility = 9.38*10^-5 M

at 50c , ksp = 4s^3 = 4.8*10^-12

                    S = solubility = 1.062*10^-4 M

change in solubility = 1.062*10^-4 - 9.38*10^-5 = 1.24*10^-5 M

solubility in g/L = 1.24*10^-5*58.32 = 7.23*10^-4 g/L

in 100 ml solubility , = 7.23*10^-4*100/1000 = 7.23*10^-5 grams

solid magnesium palmitate remains , = 0.015 - (7.23*10^-5) = 0.01493 g