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In the reaction; \( \mathrm{F} e(\mathrm{OH})_{3}(\mathrm{~s}) \rightleftharpoons \mathrm{F} e^{3+}(a q)+3 \mathrm{OH}^{-}(a q) \), if the concentration of \( \mathrm{OH}^{-} \)ions is
decreased by \( \frac{1}{4} \) times, then the equilibrium concentration of \( F e^{3+} \) will increase by,
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decreased by \( \frac{1}{4} \) times, then the equilibrium concentration of \( F e^{3+} \) will increase by,
Solution:
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Verified Answer
The correct answer is:
\( 64 \) times
\[
\begin{array}{l}
\mathrm{Fe}(\mathrm{OH})_{3}(\mathrm{~s}) \rightleftharpoons \mathrm{Fe}^{3+}(a q)+3 \mathrm{OH}^{-}(a q) \\
K_{\text {eq }}=\left[\mathrm{Fe}^{3+}\right]\left[\mathrm{OH}^{-}\right]^{3} \\
K_{\text {eq }}=\left[\mathrm{Fe}^{3+}\right]\left[\frac{1}{4} \mathrm{OH}^{-}\right]^{3}=\frac{\left[\mathrm{Fe}^{3+}\right]\left[\mathrm{OH}^{-}\right]^{3}}{64}
\end{array}
\]
Therefore, to maintain the equilibrium, the equilibrium concentration of \( F e^{3+} \) will increase by \( 64 \) times
\begin{array}{l}
\mathrm{Fe}(\mathrm{OH})_{3}(\mathrm{~s}) \rightleftharpoons \mathrm{Fe}^{3+}(a q)+3 \mathrm{OH}^{-}(a q) \\
K_{\text {eq }}=\left[\mathrm{Fe}^{3+}\right]\left[\mathrm{OH}^{-}\right]^{3} \\
K_{\text {eq }}=\left[\mathrm{Fe}^{3+}\right]\left[\frac{1}{4} \mathrm{OH}^{-}\right]^{3}=\frac{\left[\mathrm{Fe}^{3+}\right]\left[\mathrm{OH}^{-}\right]^{3}}{64}
\end{array}
\]
Therefore, to maintain the equilibrium, the equilibrium concentration of \( F e^{3+} \) will increase by \( 64 \) times
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