A saturated solution of Be(OH)2 was found to contain 1.8 ✕ 10−8 mol OH − in 100.0 mL. Calculate the Ksp of Be(OH)2.

Question

A saturated solution of Be(OH)₂ was found to contain 1.8 ✕ 10⁻⁸ mol OH⁻ in 100.0 mL. Calculate the K_sp of Be(OH)₂. 

Answer 1

Be(OH)₂(s) <==> Be²⁺(aq) + 2OH^-(aq)

The Ksp expression is Ksp = [Be²⁺][OH-]²

Given 100.0 ml contains 1.8x10^-8 mol OH-, we can find the [OH-]: 1.8x10^-8/0.1 L = 1.8x10^-7 M = [OH-]

The [Be^2+] will be 1/2 the [OH-] (see dissociation reaction above): [Be²⁺] = 9x10^-8 M]

Ksp = (9x10^-8)(1.8x10^-7)²

Ksp = 2.9x10^-21