Using the Gibbs energy change, `Delta G^(@)=+ 63.3 kJ`, for the following reaction,
`Ag_(2)CO_(3)hArr2Ag^(+)(aq)+CO_(3)^(2-)`
the `K_(sp)` of `Ag_(2)CO_(3)(s)` in water at `25^(@)C` is
`(R=8.314 JK^(-1)mol^(-1))`

Options

A. `7.9xx10^-2`
B. `8.0xx10^-12`
C. `2.9xx10^-3`
D. `3.2xx10^-26`

Answer

To find the solubility product constant \( K_{sp} \) of \( Ag_2CO_3 \) using the given Gibbs energy change \( \Delta G^\circ = +63.3 \, \text{kJ} \), we can follow these steps:

### Step-by-Step Solution:

1. **Convert Gibbs Energy to Joules**:
\[ \Delta G^\circ = 63.3 \, \text{kJ} = 63.3 \times 10^3 \, \text{J} = 63300 \, \text{J} \]

2. **Determine the Temperature in Kelvin**:
\[ T = 25^\circ C = 273 + 25 = 298 \, \text{K} \]

3. **Use the Gibbs Free Energy Equation**:
The relationship between Gibbs free energy and the equilibrium constant is given by:
\[ \Delta G^\circ = -RT \ln K_{sp} \]
Rearranging this gives:
\[ \ln K_{sp} = -\frac{\Delta G^\circ}{RT} \]

4. **Substitute the Values**:
Using \( R = 8.314 \, \text{J K}^{-1} \text{mol}^{-1} \):
\[ \ln K_{sp} = -\frac{63300}{8.314 \times 298} \]

5. **Calculate the Denominator**:
\[ 8.314 \times 298 \approx 2477.572 \]

6. **Calculate \( \ln K_{sp} \)**:
\[ \ln K_{sp} = -\frac{63300}{2477.572} \approx -25.53 \]

7. **Convert from Natural Logarithm to Base 10**:
To convert from natural logarithm to base 10, we use the conversion:
\[ \ln K_{sp} = 2.303 \log K_{sp} \]
Thus,
\[ \log K_{sp} = \frac{\ln K_{sp}}{2.303} \approx \frac{-25.53}{2.303} \approx -11.09 \]

8. **Calculate \( K_{sp} \)**:
\[ K_{sp} = 10^{\log K_{sp}} = 10^{-11.09} \]

9. **Final Calculation**:
To express \( 10^{-11.09} \):
\[ K_{sp} \approx 10^{-12} \times 10^{0.91} \approx 10^{-12} \times 8.13 \approx 8.13 \times 10^{-12} \]

### Final Answer:
\[ K_{sp} \approx 8 \times 10^{-12} \]