The relationship between standard Gibbs free energy change ($\Delta G^\circ$) and the equilibrium constant (here, solubility product $K_{sp}$) is given by: $\Delta G^\circ = -RT \ln K_{sp} = -2.303 RT \log K_{sp}$ Given: $\Delta G^\circ = +63.3 \text{ kJ} = 63.3 \times 10^3 \text{ J}$ $R = 8.314 \text{ J K}^{-1} \text{mol}^{-1}$ $T = 25^\circ\text{C} = 298 \text{ K}$ Substituting the values into the equation: $63.3 \times 10^3 = -2.303 \times 8.314 \times 298 \times \log K_{sp}$ $\log K_{sp} = -\frac{63300}{2.303 \times 8.314 \times 298}$ $\log K_{sp} = -11.09$ $K_{sp} = \text{antilog}(-11.09) = 10^{-11.09} = 10^{0.91} \times 10^{-12} \approx 8.128 \times 10^{-12} \approx 8.0 \times 10^{-12}$.