Chemical and Ionic Equilibrium 2 Question 58
58. A sample of $\mathrm{AgCl}$ was treated with $5.00 \mathrm{~mL}$ of $1.5 \mathrm{M}$ $\mathrm{Na}{2} \mathrm{CO}{3}$ solution to give $\mathrm{Ag}{2} \mathrm{CO}{3}$. The remaining solution contained $0.0026 \mathrm{~g}$ of $\mathrm{Cl}^{-}$ions per litre. Calculate the solubility product of $\mathrm{AgCl}$. $\left[K_{\mathrm{sp}}\left(\mathrm{Ag}{2} \mathrm{CO}{3}\right)=8.2 \times 10^{-12}\right]$
$(1997,5 M)$
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Solution:
- $2 \mathrm{AgCl}(s)+\mathrm{CO}{3}^{2-} \rightleftharpoons \mathrm{Ag}{2} \mathrm{CO}_{3}(s)+2 \mathrm{Cl}^{-}$
$$ \begin{aligned} K & =\frac{\left[\mathrm{Cl}^{-}\right]^{2}}{\left[\mathrm{CO}{3}^{2-}\right]}=\frac{\left[\mathrm{Cl}^{-}\right]^{2}}{\left[\mathrm{CO}{3}^{2-}\right]} \times \frac{\left[\mathrm{Ag}^{+}\right]^{2}}{\left[\mathrm{Ag}^{+}\right]^{2}}=\frac{\left[K_{\text {sp }}(\mathrm{AgCl})\right]^{2}}{K_{\text {sp }}\left(\mathrm{Ag}{2} \mathrm{CO}{3}\right)} \ {\left[\mathrm{Cl}^{-}\right] } & =\frac{0.0026}{35.5} \mathrm{M}=7.3 \times 10^{-5} \mathrm{M} \end{aligned} $$
The above concentration of $\mathrm{Cl}^{-}$indicates that $\left[\mathrm{CO}_{3}^{2-}\right]$ remains almost unchanged.
$$ \begin{aligned} & \frac{7.3 \times 10^{-5}}{1.5}=\frac{\left[K_{\text {sp }}(\mathrm{AgCl})\right]^{2}}{8.2 \times 10^{-12}} \ & K_{\text {sp }}(\mathrm{AgCl})=2 \times 10^{-8} \end{aligned} $$
