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

The Correct Option is C

To solve this problem, we need to understand the equilibrium involved in the reaction:

$Fe(OH)_3(s) \rightleftharpoons Fe^{3+}_{(aq)} + 3 OH^-_{(aq)}$

The equilibrium constant expression for this reaction is given by:

$K_{sp} = [Fe^{3+}][OH^-]^3$

Here, $K_{sp}$ is the solubility product constant, which remains constant at a given temperature.

If the concentration of $OH^-$ ions is decreased by $\\frac{1}{4}$, the new concentration of $OH^-$ becomes:

$[OH^-]_{new} = \\frac{1}{4} [OH^-]_{initial}$

Substituting the new concentration in the equilibrium expression, we have:

$K_{sp} = [Fe^{3+}]_{new} (\\frac{1}{4} [OH^-]_{initial})^3$

Comparing the new and initial scenarios, we get:

$K_{sp} = [Fe^{3+}]_{initial} [OH^-]_{initial}^3 = [Fe^{3+}]_{new} \\frac{[OH^-]_{initial}^3}{4^3}$

This simplifies to:

$[Fe^{3+}]_{new} = 4^3 \times [Fe^{3+}]_{initial}$

Thus, $[Fe^{3+}]_{new}$ is $64$ times the initial concentration of $Fe^{3+}$.

By reducing the $OH^-$ concentration to a quarter, the equilibrium shifts to increase the concentration of $Fe^{3+}$ by 64 times to maintain the constant $K_{sp}$.

Therefore, the correct answer is:

64 times