Question

Chemistry!! Please help

What is the solubility in mol/L of copper(I) sulfide, Cu2S? Its Ksp value is 6.1 × 10-49 .

Answer 1

5.3 × 10⁻¹⁷ mol·L⁻¹

Step-by-step explanation:

Let s = the molar solubility.

Cu₂S(s) ⇌ 2Cu⁺(aq) + S²⁻(aq); K_{sp} = 6.1 × 10⁻⁴⁹

E/mol·L⁻¹: 2s s

K_{sp} =[Cu⁺]²[S²⁻] = (2s)²×s = 4s^3 = 6.1 × 10⁻⁴⁹

`s^(3)= (6.1 * 10^(-49))/(4) = 1.52 * 10^(-49)`

`s = \sqrt[3]{1.52 * 10^(-49)} \text{ mol/L} = 5.3 * 10^(-17) \text{ mol/L}`