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(A) The equilibrium constant expression for the reaction $Fe(OH)_{3(s)} \rightleftharpoons Fe^{3+}_{(aq)} + 3OH^{-}_{(aq)}$ is given by:$K_{sp} = [Fe^

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$64$

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(A) The equilibrium constant expression for the reaction $Fe(OH)_{3(s)} ightleftharpoons Fe^{3+}_{(aq)} + 3OH^{-}_{(aq)}$ is given by:$K_{sp} = [Fe^{3+}] [OH^{-}]^3$Let the initial concentration of $Fe^{3+}$ be $[Fe^{3+}]_1$ and $OH^{-}$ be $[OH^{-}]_1$.$K_{sp} = [Fe^{3+}]_1 [OH^{-}]_1^3$When the concentration of $OH^{-}$ is decreased by $1/4$ times,let the new concentration be $[OH^{-}]_2 = \frac{1}{4} [OH^{-}]_1$.Let the new concentration of $Fe^{3+}$ be $[Fe^{3+}]_2 = x [Fe^{3+}]_1$.Since $K_{sp}$ remains constant at a constant temperature:$K_{sp} = [Fe^{3+}]_2 [OH^{-}]_2^3$$[Fe^{3+}]_1 [OH^{-}]_1^3 = (x [Fe^{3+}]_1) 	imes (\frac{1}{4} [OH^{-}]_1)^3$$1 = x 	imes (\frac{1}{4})^3$$1 = x 	imes \frac{1}{64}$$x = 64$Thus,the concentration of $Fe^{3+}$ will increase by $64$ times.

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

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