Answer:
0.0349 g/L
Explanation:
Let s = the molar solubility.
Ag₂CO₃(s) ⇌ 2Ag⁺(aq) + CO₃²⁻(aq); [tex]K_{sp}[/tex] = 8.10 × 10⁻¹²
E/mol·L⁻¹: 2s s
[tex]K_{sp}[/tex] =[Ag⁺]²[CO₃²⁻] = (2s)²×s = 4s^3 = 8.10 × 10⁻¹²
[tex]s^{3}= \frac{8.10 \times 10^{-12}}{4} = 2.025 \times 10^{-12}[/tex]
[tex]s = \sqrt[3]{2.025\times 10^{-12}} \text{ mol/L}= 1.27 \times 10^{-4} \text{ mol/L}[/tex]
[tex]s = 1.27 \times 10^{-4} \text{ mol/L} \times 275.75 \text{ g/mol} = \text{0.0349 g/L}[/tex]
