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If the concentration of $\mathrm{OH}^{-}$ions in the reaction $\mathrm{Fe}(\mathrm{OH})_3(s) \rightleftharpoons \mathrm{Fe}^{3+}(a q)+3 \mathrm{OH}(a q)$ is decreased by $\frac{1}{4}$ times, then equilibrium concentration of $\mathrm{Fe}^{3+}$ will increase by
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The correct answer is:
64 times
Key Idea : The concentration of solids taken to be unity.
$\begin{aligned}
& \mathrm{Fe}(\mathrm{OH})_3(s) \rightleftharpoons \mathrm{Fe}^{3+}(a q)+3 \mathrm{OH}^{-}(a q) \\
& \therefore \quad K=\left[\mathrm{Fe}^{3+}\right]\left[\mathrm{OH}^{-}\right]^3
\end{aligned}$
Hence, if $\mathrm{OH}^{-}$ion concentration is decreased by $\frac{1}{4}$ times, then equilibrium concentration of $\mathrm{Fe}^{3+}$ will increase by 64 times.
$\begin{aligned}
& \mathrm{Fe}(\mathrm{OH})_3(s) \rightleftharpoons \mathrm{Fe}^{3+}(a q)+3 \mathrm{OH}^{-}(a q) \\
& \therefore \quad K=\left[\mathrm{Fe}^{3+}\right]\left[\mathrm{OH}^{-}\right]^3
\end{aligned}$
Hence, if $\mathrm{OH}^{-}$ion concentration is decreased by $\frac{1}{4}$ times, then equilibrium concentration of $\mathrm{Fe}^{3+}$ will increase by 64 times.
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