Question
The solubility product of AgC2O4 at 25°C is 2.3 × 10–11 M3. A solution of K2C2O4 containing 0.15 moles in 500 ml water is shaken at 25°C with excess of Ag2CO3 till the following equilibrium is reached.
Ag2CO3 + K2C2O4 $\rightleftharpoons$ Ag2C2O4 + K2CO3
At equilibrium, the solution contains 0.035 mole of K2CO3. Assuming the degree of dissociation of K2C2O4 and K2CO3 to be equal, calculate the solubility product of Ag2CO3.
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Solution
$Ag_{2}CO_{3}(s) + C_{2}O_{4}^{2 -} \rightleftharpoons Ag_{2}C_{2}O_{4}(s) + CO_{3}^{2 -}$
$\text{Initial}\frac{0.15}{500} \times 1000 = 0.3\text{ M}$ 0
$\text{Final}0.3 - x = 0.23\text{ M}x = \frac{0.035}{500} \times 1000 = 0.07\text{ M}$
Now, $K_{eq} = \frac{\left\lbrack CO_{3}^{2 -} \right\rbrack}{\left\lbrack C_{2}O_{4}^{2 -} \right\rbrack} \times \frac{\left\lbrack Ag^{+} \right\rbrack^{2}}{\left\lbrack Ag^{+} \right\rbrack^{2}} = \frac{K_{sp}\left( Ag_{2}CO_{3} \right)}{K_{sp}\left( Ag_{2}C_{2}O_{4} \right)}$
Or, $\frac{0.07}{0.23} = \frac{K_{sp}\left( Ag_{2}CO_{3} \right)}{2.3 \times 10^{- 11}} \Rightarrow K_{sp}\left( Ag_{2}CO_{3} \right) = 7 \times 10^{- 12}$
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