Question

The common-ion effect and solubility The solubility of a slightly soluble compound can be greatly affected by the addition of a soluble compound with a common ion, that is, with one of the ions in the added soluble compound being identical to one of the ions of the slightly soluble compound. The general result of the addition of the common ion is to greatly reduce the solubility of the slightly soluble compound. In other words, the addition of the common ion results in a shift in the equilibrium of the slightly soluble compound. Part B - Calculate the molar solubility in NaOH Based on the given value of the Ksp, what is the molar solubility of Mg(OH)2 in 0.110 M NaOH?

Answer 1

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

Initial                                                    0           0.110

Change                                                 x             2x

Equilibrium                                             x          (2x + 0.110)

Ksp (Mg(OH)2) = [Mg2+] [OH-]2 = (x)(2x)² = 5.61 x 10^-11 (in water)

4x³ = 5.61 x 10^-11

x = 2.41 x 10^-4 M

Now in the presence of NaOH,

Ksp = (x)(2x + 0.110)²

2x is neglected in comparison of 0.110 as value of x is very small (2x + 0.110 0.110)

Ksp = (x) (0.110)² = 5.61 x 10^-11

x = molar solubility = 4.63 x 10^-9 M