Calculate the solubility at 25 °C of Ni(OH)2 in pure water and in a 0.0170M NaOH solution.

Question

Calculate the solubility at 25 °C of Ni(OH)₂ in pure water and in a 0.0170M NaOH solution.

Ni(OH)₂ K_sp = 5.48 x 10^-16

Round both of your answers to 2 significant digits.

solubility in pure water: ______g/L

solubility in 0.0170 M NaOH solution: ______g/L

Answer 1

Again, this is another common ion effect, and we expect the solubility to be LESS in 0.0170 M NaOH than in pure water.

Ni(OH)₂(s) <==> Ni²⁺(aq) + 2OH^-(aq)

Ksp = [Ni²⁺][OH^-]²

5.48x10^-16 = (x)(2x)² = 4x³

x³ = 1.37x10^-16

x = 5.2x10^-6 M

converting to g/L we have...

5.2x10^-6 mol/L x 92.7 g/mol = 4.8x10^-4 g/L = solubility in pure water

The common ion in this case is the OH- ion. We can assume the final [OH-] come totally from the NaOH and is 0.0170 M. We can neglect the [OH-] contributed from the Ni(OH)₂ which is only 1.0x10^-5 M

Ksp = [Ni²⁺][OH-]²

5.48x10^-16 = (x)(0.0170)²

5.48x10^-16 = 2.89x10^-4x

x = 1.9x10^-12 mol/L = [OH^-]

[Ni(OH)2] = 1/2 x 1.9x10^-12 mol/L = 9.5x10^-13 mol/L

converting to g/L we have...

9.5x10^-13 mol/L x 92.7 g/mol = 8.8x10^-11 g/L = solubility in 0.0170 M NaOH