What is the solubility of Cr(OH)₃ at a pH of 10.70? (Ksp Cr(OH)₃ is 6.70 × 10⁻³¹)

Question

What is the solubility of Cr(OH)₃ at a pH of 10.70? (Ksp Cr(OH)₃ is 6.70 × 10⁻³¹)

Answer 1

Cr(OH)₃(s) <==> Cr³⁺(aq) + 3OH^-(aq) ... Ksp = 6.70x10^-31

Ksp = [Cr³⁺][OH^-]³

From the pH, we can find the [OH-]:

pH = 10.70

pOH = 14 - 10.70 = 3.30

[OH-] = 1x10^-3.30 = 5.0x10^-4 M

Substituting in the Ksp expression we have...

6.70x10^-31 = [Cr³⁺][5x10^-4]³

[Cr³⁺] = 5.36x10^-21 M = solubility of Cr(OH)₃ at pH 10.7