What is the solubility of Ca(OH)₂ in 0.810 M Ba(OH)₂? Ksp for Ca(OH)₂ is 6.50 × 10⁻⁶. Ba(OH)₂ is a strong base.

Answer 1

This is a common ion problem, the common ion being OH^-.

Ca(OH)₂(s) <==> Ca²⁺(aq) + 2OH^-(aq)

Ksp = [Ca²⁺][OH^-]²

Since the solution already contains 0.810 M Ba(OH)₂ (a soluble hydroxide), we can use the 0.810 M as the concentration of OH^- and then solve for the solubility of Ca(OH)₂.

Ksp = [Ca²⁺][OH^-]²

6.50x10^-6 = [Ca²⁺][0.810]² = [Ca²⁺][0.656]

[Ca²⁺] = 9.91x10^-6 M = solubility of Ca(OH)₂