Question

The $\Delta G^{\circ}$ for the reaction $=+57 \,kJ$
$M X_{(s)} \rightleftharpoons M_{(a q)}^{+}+X_{(a q)}^{-}$
Hence $K_{s p}$ for the reaction is

• A. $10^{-5}$
• B. $10^{-6}$
• C. $10^{-10}$
• D. $10^{-16}$

Answer 1

Answer: (C)

$\Delta G^{\circ}=-2.303 \times 8.31 \times 298 \log K_{s p}$

$\Rightarrow \log K_{s p}=\frac{-57000}{2.303 \times 8.31 \times 298}$

On solving we get $K_{s p} \simeq 10^{-10}$