Question

In the reaction; $Fe ( OH )_{3( s )} \rightleftharpoons Fe ^{3+}{ }_{( aq )}+3 OH _{\text {(aq) }}^{-}$, if the concentration of $OH ^{-}$ ions is decreased by $\frac{1}{4}$ times, then the equilibrium concentration of $Fe ^{3+}$ will increase by,

• A. 8 times
• B. 16 times
• C. 64 times
• D. 4 times

Answer 1

Answer: (C)

64 times

Answer 2

The correct answer is C:64 times
Given relation,
$Fe ( OH )_{3}(s) \rightleftharpoons Fe ^{3+}(a q)+3 OH ^{-}(a q)$

$K_{ eq }=\frac{\left[ Fe ^{3+}\right]\left[ OH ^{-}\right]^{3}}{\left[ Fe ( OH )_{3}\right]}=\frac{\left[ Fe ^{3+}\right]\left[ OH ^{-}\right]^{3}}{1}$

When $[OH^{-} ]$ is decreased by $\frac{1}{4}$ times, $K$ becomes

$K_{ eq } =\left[ Fe ^{3+}\right]\left(\frac{1}{4} \cdot[\overrightarrow{ O } H ]\right)^{3}$
$=\left[ Fe ^{3+}\right] \frac{1}{64}\left[ OH ^{-}\right]^{3}$

To keep the value of $K_{\text {eq }}$ constant, the $\left[ Fe ^{3+}\right]$ must be increased by 64 times so that $K_{ eq }$ remains as such.

$K_{ eq }=64 \times \frac{1}{64}\left[ Fe ^{3+}\right]\left[ OH ^{-}\right]^{3}$
$K_{ eq }=\left[ Fe ^{3+}\right]\left[ OH ^{-}\right]^{3}$