For the equilibrium reaction: Fe(OH)_3(s) rig | vedclass

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Question

(C) The equilibrium constant expression for the reaction is $K_{sp} = [Fe^{3+}][OH^-]^3$. Since $K_{sp}$ is constant at a given temperature,we have $[

Vedclass · Accepted Answer

$64$

Vedclass · Answer

(C) The equilibrium constant expression for the reaction is $K_{sp} = [Fe^{3+}][OH^-]^3$. Since $K_{sp}$ is constant at a given temperature,we have $[Fe^{3+}]_1 [OH^-]_1^3 = [Fe^{3+}]_2 [OH^-]_2^3$. Given that $[OH^-]_2 = \frac{1}{4} [OH^-]_1$,we substitute this into the equation: $[Fe^{3+}]_1 [OH^-]_1^3 = [Fe^{3+}]_2 (\frac{1}{4} [OH^-]_1)^3$. $[Fe^{3+}]_1 [OH^-]_1^3 = [Fe^{3+}]_2 	imes \frac{1}{64} [OH^-]_1^3$. $[Fe^{3+}]_2 = 64 [Fe^{3+}]_1$. Thus,the concentration of $Fe^{3+}$ increases by a factor of $64$.

For the equilibrium reaction: $Fe(OH)_3(s) \rightleftharpoons Fe^{3+}(aq) + 3OH^-(aq)$,if the concentration of $OH^-$ is reduced to $\frac{1}{4}$ of its initial value,by what factor will the concentration of $Fe^{3+}$ increase at equilibrium?

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