Question

Use the free energy change, $\Delta G^o=+63.3kJ$, for this reaction, to calculate the $K_{sp}$ of ${Ag}_2{CO}_3\left(s\right)$ in water at ${25}^{o}C$:

${Ag}_2{CO}_3\left(s\right)\leftrightharpoons {Ag}^{+}(aq.)+{CO}_3^{2-}(aq.)$

• A. $3.2\times {10}^{-26}$
• B. $8\times {10}^{-12}$
• C. $2.9\times {10}^{-3}$
• D. $7.9\times {10}^{-2}$

Answer 1

The correct option is B $8\times {10}^{-12}$

We know that, $\Delta G^o=-RT$ In $K_{sp}$

$\Rightarrow +63.3\times {10}^{3}Jmo{l}^{-1}=-8.314Jmo{l}^{-1}{K}^{-1}\times 298K$ In $K_{sp}$

$\Rightarrow$ In $K_{sp}=-\frac{63.3\times {10}^3}{8.314\times 298}$

$\Rightarrow K_{sp}=8\times {10}^{-12}$ .