In the reaction Fe(OH)_{3(s)} rightleftharpoo | vedclass

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Question

(C) For the equilibrium: $Fe(OH)_{3(s)} \rightleftharpoons Fe^{3+}_{(aq)} + 3OH^{-}_{(aq)}$The solubility product expression is: $K_{sp} = [Fe^{3+}][O

Vedclass · Accepted Answer

$64$

Vedclass · Answer

(C) For the equilibrium: $Fe(OH)_{3(s)} ightleftharpoons Fe^{3+}_{(aq)} + 3OH^{-}_{(aq)}$The solubility product expression is: $K_{sp} = [Fe^{3+}][OH^{-}]^3$Let the initial concentrations be $[Fe^{3+}]_1$ and $[OH^{-}]_1$.$K_{sp} = [Fe^{3+}]_1 [OH^{-}]_1^3$ ... $(i)$If the new concentration $[OH^{-}]_2 = \frac{1}{4} [OH^{-}]_1$, then:$K_{sp} = [Fe^{3+}]_2 [OH^{-}]_2^3 = [Fe^{3+}]_2 (\frac{1}{4} [OH^{-}]_1)^3$$K_{sp} = [Fe^{3+}]_2 	imes \frac{1}{64} [OH^{-}]_1^3$ ... $(ii)$From $(i)$ and $(ii)$:$[Fe^{3+}]_1 [OH^{-}]_1^3 = [Fe^{3+}]_2 	imes \frac{1}{64} [OH^{-}]_1^3$$[Fe^{3+}]_2 = 64 	imes [Fe^{3+}]_1$Therefore, the concentration of $Fe^{3+}$ increases by $64$ times.

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

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