SATHEE: Equilibrium - Result Question 72

Equilibrium - Result Question 72

####76. Using the Gibbs energy change, $Δ G^{\circ} = + 63.3 k J$ , for the following reaction, [2014]

$A g_{2} C O_{3} \to 2 A g^{+} (a q) + C O_{3}^{2 -} (a q)$

the $K_{s p}$ of $A g_{2} C O_{3} (s)$ in water at $25^{\circ} C$ is:( $R = 8.314 J K^{- 1} m o l^{- 1}$ )

(a) $3.2 \times 10^{- 26}$

(b) $8.0 \times 10^{- 12}$

(c) $2.9 \times 10^{- 3}$

(d) $7.9 \times 10^{- 2}$

Show Answer

Solution:

(b) $Δ G = - 2.303 \log K$

here $K = [A g^{+}]^{2} [C O_{3}^{2 -}] = K_{s p}$

$∴ 63.3 \times 10^{+ 3} = - 2.303 \times 8.314 \times 298 \log K_{s p}$

$∴ \log K_{s p} = - \frac{63.3 \times 10^{+ 3}}{5705.8} = - 11.09$

$∴ K_{s p} = A n t i l o g (- 11.09) = 8 \times 10^{- 12}$

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