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If the concentration of $OH^-$ ions in the reaction
$Fe(OH)_3(s) \rightleftharpoons Fe^{3+}(aq.) + 3OH^-(aq.)$
is decreased by $\dfrac{1}{4}$ times, then equilibrium concentration of $Fe^{3+}$ will increase by -
$Fe(OH)_3(s) \rightleftharpoons Fe^{3+}(aq.) + 3OH^-(aq.)$
is decreased by $\dfrac{1}{4}$ times, then equilibrium concentration of $Fe^{3+}$ will increase by -
A
64 times
B
4 times
C
8 times
D
16 times
Detailed Solution
$Fe(OH)_3(s) \rightleftharpoons Fe^{3+}(aq.) + 3OH^-(aq.)$
$K = \dfrac{[Fe^{3+}][OH^-]^3}{[Fe(OH)_3]}$
The activity of a pure solid is taken as unity, so $K = [Fe^{3+}][OH^-]^3$.
$K$ is constant at a given temperature.
When the concentration of $OH^-$ is decreased to $\dfrac{1}{4}$ of its value: $K = [Fe^{3+}]_{new}\left(\dfrac{[OH^-]}{4}\right)^3 = \dfrac{[Fe^{3+}]_{new}[OH^-]^3}{64}$
Equating with the original expression: $[Fe^{3+}]_{new} = 64\,[Fe^{3+}]$
So the equilibrium concentration of $Fe^{3+}$ will be increased by 64 times in order to keep the value of $K$ constant.
$K = \dfrac{[Fe^{3+}][OH^-]^3}{[Fe(OH)_3]}$
The activity of a pure solid is taken as unity, so $K = [Fe^{3+}][OH^-]^3$.
$K$ is constant at a given temperature.
When the concentration of $OH^-$ is decreased to $\dfrac{1}{4}$ of its value: $K = [Fe^{3+}]_{new}\left(\dfrac{[OH^-]}{4}\right)^3 = \dfrac{[Fe^{3+}]_{new}[OH^-]^3}{64}$
Equating with the original expression: $[Fe^{3+}]_{new} = 64\,[Fe^{3+}]$
So the equilibrium concentration of $Fe^{3+}$ will be increased by 64 times in order to keep the value of $K$ constant.