Mg(OH)2 is dissolved in aq solution has a pH of 8.90 at 25 C. Fill Ice table.

Question

Based off of Provide a equation to find the value of Ksp at 1.0 x 10^-31

then find molar solubility of SMg(OH)2= M

Answer 1

Mg(OH)₂(s) <==> Mg²⁺(aq) + 2OH^-(aq)

From pH = 8.90, we can find [OH^-]:

[OH^-] = 1x10^-8.90 = 1.26x10^-9 M

Mg(OH)₂(s) <==> Mg²⁺(aq) + 2OH^-(aq)

1........................0...................0...........Initial

-x.....................+x.................+2x..........Change

1-x....................x...................2x...........Equilibrium

At equilibrium, 2x = 1.26x10^-9 M

Therefore x = [Mg²⁺] = 1/2 (1.26x10^-9 M) = 6.3x10^-10 M

Ksp = [Mg²⁺][OH^-]² = (6.3x10^-10)(1.26x10^-9)²

Ksp = 1.0x10^-27 This is the Ksp @ pH 8.90.

The molar solubility at this pH = 6.3x10^-10 M