Question

Using the Gibbs energy change, $\Delta G^{\circ}=+63.3 \,kJ$ for the following reaction, $Ag _{2} CO _{3}(s) \rightleftharpoons 2 Ag ^{+}(a q)+ CO _{3}^{2-}(a q)$ the $K_{s p}$ of $Ag _{2} CO _{3}(s)$ in water at $25^{\circ} C$ is $\left(R=8.314 \,JK ^{-1} \,mol ^{-1}\right)$

• A. $ 3.2 \times 10^{-26} $
• B. $ 8.0 \times 10^{-12} $
• C. $ 2.9 \times 10^{-3} $
• D. $ 7.9 \times 10^{-2} $

Answer 1

Answer: (B)

$\Delta G^{\circ}$ is related to $K_{s p}$ by the equation
$\Delta G^{\circ}=-2.303 R T \log K_{s p}$
Given, $\Delta G^{\circ}=+63.3 \,K J$
$=63.3 \times 10^{3} \,J$
Thus, substitute $\Delta G^{\circ}=63.3 \times 10^{3} \,J$
$R=8.314 JK ^{-1} mol ^{-1}$ and $T =298 K [25+273 \, K ]$
from the above equation to get
$63.3 \times 10^{3}=-2.303 \times 8.314 \times 298 \log K_{ sp }$
$\therefore \log K_{ sp }=-11.09$
$\Rightarrow K_{ sp }=\operatorname{antilog}(-11.09)$
$K_{ sp }=8.0 \times 10^{-12}$