Question

If $K_{sp}$ of ${Ag}_2{CO}_3$ is $8\times {10}^{-12},$ the molar solubility of ${Ag}_2{CO}_3$ in $0.1MAgN{O}_3$ is:

• (A) 8 × 10^-12 M
• (B) 8 × 10^-10 M
• (C) 8 × 10^-11 M
• (D) 8 × 10^-13 M

Answer 1

The Correct Option is B

Explanation:
${Ag}_2{CO}_3(s)\rightleftharpoons 2{Ag}^{+}(aq.)+{CO}_3^{-2}(aq)$$Ksp={\left[A{g}^{+}\right]}^2\left[C{O}_3^{-2}\right]$$8\times {10}^{-12}=(0.1+2S{)}^2(S)$As $K_{\text{sp }}$ is so small, so $0.1+2S=0.1$$S=8\times {10}^{-10}M$Hence, the correct option is (B).