Question

The solubility product of BaCl₂ is 4.0 × 10^-8. What will be its molar solubility in mol dm^-3?

Answer 1

Given: Solubility product (K_sp) = 4.0 × 10^-8

To find: Molar solubility, S

Formula: K_sp = x^x y^y S^x+y

Calculation: 

K_sp ​= [Ba²⁺][Cl⁻]²

K_sp​ = (s)(2s)² = s ⋅ 4s² = 4s³

4s³ = 4.0 × 10⁻⁸

`s^3 = (4.0 xx 10^-8)/4 = 1.0 xx 10^-8`

`s = root3(1.0 xx 10^-8)`

= 2.15 × 10^-3 mol dm⁻³