Answer:
5.3 × 10⁻¹⁷ mol·L⁻¹
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
[tex]s^{3}= \frac{6.1 \times 10^{-49}}{4} = 1.52 \times 10^{-49}[/tex]
[tex]s = \sqrt[3]{1.52 \times 10^{-49}} \text{ mol/L} = 5.3 \times 10^{-17} \text{ mol/L}[/tex]
