Question

The aqueous solubility product of nickel hydroxide, Ni(OH)2 is Ksp = 2.0 x 10-15.What is the solubility in a solution with pH = 12.2 in grams per Liter of solution?

Answer 1

Ni(OH)_2(s) --------------> Ni²⁺ + 2OH^-

With a pH = 12.2 let's calculate the concentration of [OH]:

pOH = 14 - 12.2 = 1.8

[OH^-] = 10^-1.8 = 0.0158 M

With this we can calculate the solubility:

Ni(OH)_2(s) --------------> Ni²⁺ + 2OH^-

i. s 0 0.0158

e. 0 s 0.0158

We can assume that this solution is buffered with that pH therefore, there's no change in the concentration of OH, now let's solve for s:

Ksp = s [OH^-]²

s = Ksp / [OH^-]²

s = 2x10^-15 / (0.0158)²

s = 1.26x10^-13 mol/L

The molar mass of Ni(OH)₂ is: (58.69) + (2*16) + (2*1) = 92.69 g/mol

s = 1.26x10^-13 mol/L * 92.69 g/mol = 1.17x10^-11 g/L

Hope this helps