Question

Mg(OH)2 is a sparingly soluble compound, in this case a base, with a solubility product, Ksp, of 5.61×10−11. It is used to control the pH and provide nutrients in the biological (microbial) treatment of municipal wastewater streams.

Part A:

Based on the given value of the Ksp, calculate the ratio of solubility of Mg(OH)2 dissolved in pure H2O to Mg(OH)2 dissolved in a 0.160 M NaOH solution.

Part B:

What is the pH change of a 0.300 M solution of citric acid (pKa=4.77) if citrate is added to a concentration of 0.140 M with no change in volume?​

Answer 1

Part A)

Ksp of Mg(OH)2 = 5.61 x 10^-11

Mg(OH)2 (s)   ------------> Mg2+ (aq) +   2 OH- (aq)

                                              S                  2S

Ksp = [Mg2+][OH-]^2

5.61 x 10^-11 = 4 S^3

S = 2.41 x 10^-4

solubility in H2O = 2.41 x 10^-4 M

in NaOH :

Mg(OH)2 (s)   ------------> Mg2+ (aq) +   2 OH- (aq)

                                              S                  0.160

Ksp = [Mg2+][OH-]^2

5.61 x 10^-11 = S x (0.160)^2

S = 2.19 x 10^-9

solubility = 2.19 x 10^-9 M

ratio = 2.41 x 10^-4 / 2.19 x 10^-9

ratio = 1.10 x 10^5

Part B)

pKa = 4.77

pH = 1/2 (pKa - log C)

     = 1/2 (4.77 - log 0.300)

pH = 2.646

pH = pKa + log [citrate / citric acid]

    = 4.77 + log [0.140 / 0.300]

pH = 4.439

pH change = 4.439 - 2.646 = 1.79

pH change = 1.79