Q27 / 180 · KCET 2016 · Chemistry
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,
decreased by \( \frac{1}{4} \) times, then the equilibrium concentration of \( F e^{3+} \) will increase by,
- A\( 8 \) times
- B\( 16 \) times
- Correct answerC\( 64 \) times
- D\( 4 \) times
Solution
\[
\begin{array}{l}
\mathrm{Fe}(\mathrm{OH})_{3}(\mathrm{~s}) \rightleftharpoons \mathrm{Fe}^{3+}(a q)+3 \mathrm{OH}^{-}(a q) \\
K_{\text {eq }}=\left[\mathrm{Fe}^{3+}\right]\left[\mathrm{OH}^{-}\right]^{3} \\
K_{\text {eq }}=\left[\mathrm{Fe}^{3+}\right]\left[\frac{1}{4} \mathrm{OH}^{-}\right]^{3}=\frac{\left[\mathrm{Fe}^{3+}\right]\left[\mathrm{OH}^{-}\right]^{3}}{64}
\end{array}
\]
Therefore, to maintain the equilibrium, the equilibrium concentration of \( F e^{3+} \) will increase by \( 64 \) times
Therefore, to maintain the equilibrium, the equilibrium concentration of \( F e^{3+} \) will increase by \( 64 \) times
Hence, the correct answer is option: 'C'.