Question

Mg(OH)2 is a sparingly soluble salt 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. What is the ratio of solubility of Mg(OH)2 dissolved in pure H2O to Mg(OH)2dissolved in a 0.160 M NaOH solution?

Express your answer numerically as the ratio of molar solubility in H2O to the molar solubility in NaOH.

Answer 1

In pure water:

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

initiak:                                                              0              0

at equilibrium:                                               S              2S

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

5.61*10^-11 = S* (2S)^2

5.61*10^-11 = 4S^3

S = 2.4*10^-4

In NaOH:

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

initiak:                                                              0              0.16

at equilibrium:                                               S              0.16+2S

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

5.61*10^-11 = S* (0.16+2S)^2

Since Ksp is very small, S will be very small and it can ignored as compared to 0.16

5.61*10^-11 = S* (0.16)^2

S = 2.2*10^-9

molar solubility in H2O to the molar solubility in NaOH = 2.4*10^-4 / (2.2*10^-9)

                             =1.1*10^5

Answer:    1.1*10^5