The 𝐾sp of manganese(II) hydroxide, Mn(OH)2, is 2.00×10−13 .

Question

Calculate the solubility of this compound in grams per liter.

Answer 1

To calculate solubility, solve the Ksp equation.

Mn(OH)₂(s) <==> Mn²⁺⁽aq) + 2OH^-(aq)

Ksp = [Mn²⁺][OH^-]²

let x = [Mn²⁺] and then 2x = [OH^-] since you get 2 OH^- for each 1 Mn²⁺

2.00x10^-13 = (x)(2x)² = 4x³

x³ = 5x10^-14

x = 3.7x10^-5 M = [Mn²⁺] = molar solubility of Mn(OH)₂