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If the concentration of $\mathrm{Ag}^{+}$ions in the saturated solution of $\mathrm{Ag}_2 \mathrm{CO}_3$ is $1.20 \times 10^{-14} \mathrm{~mol} \mathrm{~L}^{-1}$, then find the solubility product of $\mathrm{Ag}_2 \mathrm{CO}_3$.
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The correct answer is:
$6.90 \times 10^{-12}$
$\mathrm{Ag}_2 \mathrm{CO}_3 \longrightarrow 2 \mathrm{Ag}^{+}+\mathrm{CO}_3^{2-}$
Let solubility be $S$.
The solubility of silver will be $2 S$ as two moles ions are dissociated.
$$
\begin{aligned}
K_{\text {sp }} & =\left[\mathrm{Ag}^{+}\right]^2\left[\mathrm{CO}_3^{2-}\right] \\
K_{\text {sp }} & =(2 S)^2 \times S=4 S^3 \\
S & =1.20 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1} \text { (given) } \\
K_{\text {sp }} & =4 \times\left(1.20 \times 10^{-4}\right)^3=6.90 \times 10^{-12}
\end{aligned}
$$
Let solubility be $S$.
The solubility of silver will be $2 S$ as two moles ions are dissociated.
$$
\begin{aligned}
K_{\text {sp }} & =\left[\mathrm{Ag}^{+}\right]^2\left[\mathrm{CO}_3^{2-}\right] \\
K_{\text {sp }} & =(2 S)^2 \times S=4 S^3 \\
S & =1.20 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1} \text { (given) } \\
K_{\text {sp }} & =4 \times\left(1.20 \times 10^{-4}\right)^3=6.90 \times 10^{-12}
\end{aligned}
$$
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