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NEET CHEMISTRYMedium

What is the value of KspK_{sp} of Ag2CO3(s)\text{Ag}_2\text{CO}_3\text{(s)} in water at 25∘C25^\circ\text{C} for the following reaction: Ag2CO3(s)→2Ag+(aq)+CO32−(aq)\text{Ag}_2\text{CO}_3\text{(s)} \rightarrow 2\text{Ag}^+\text{(aq)} + \text{CO}_3^{2-}\text{(aq)}? [Given: R=8.314 J K–1 mol–1R=8.314 \text{ J K}^{–1}\text{ mol}^{–1}; ΔG∘=+63.3 kJ\Delta G^\circ=+63.3 \text{ kJ}]

A

3.2×10263.2 \times 10^{26}

B

8.0×10−128.0 \times 10^{-12}

C

2.9×10−32.9 \times 10^{-3}

D

7.9×10−27.9 \times 10^{-2}

Step-by-Step Solution

The relationship between standard Gibbs free energy change (ΔG∘\Delta G^\circ) and the equilibrium constant (here, solubility product KspK_{sp}) is given by: ΔG∘=−RTln⁡Ksp=−2.303RTlog⁡Ksp\Delta G^\circ = -RT \ln K_{sp} = -2.303 RT \log K_{sp} Given: ΔG∘=+63.3 kJ=63.3×103 J\Delta G^\circ = +63.3 \text{ kJ} = 63.3 \times 10^3 \text{ J} R=8.314 J K−1mol−1R = 8.314 \text{ J K}^{-1} \text{mol}^{-1} T=25∘C=298 KT = 25^\circ\text{C} = 298 \text{ K} Substituting the values into the equation: 63.3×103=−2.303×8.314×298×log⁡Ksp63.3 \times 10^3 = -2.303 \times 8.314 \times 298 \times \log K_{sp} log⁡Ksp=−633002.303×8.314×298\log K_{sp} = -\frac{63300}{2.303 \times 8.314 \times 298} log⁡Ksp=−11.09\log K_{sp} = -11.09 Ksp=antilog(−11.09)=10−11.09=100.91×10−12≈8.128×10−12≈8.0×10−12K_{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}.

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