What is the molar solubility of Fe(OH)_2 (K_{ | vedclass

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(D) The solubility equilibrium for $Fe(OH)_2$ is: $Fe(OH)_2(s) \rightleftharpoons Fe^{2+}(aq) + 2OH^-(aq)$.Given $pH = 13.0$,we calculate $pOH = 14.0

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$8.0 	imes 10^{-14}$

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(D) The solubility equilibrium for $Fe(OH)_2$ is: $Fe(OH)_2(s) ightleftharpoons Fe^{2+}(aq) + 2OH^-(aq)$.Given $pH = 13.0$,we calculate $pOH = 14.0 - 13.0 = 1.0$.Thus,$[OH^-] = 10^{-pOH} = 10^{-1.0} = 0.1 \ M$.The solubility product expression is $K_{sp} = [Fe^{2+}][OH^-]^2$.Substituting the values: $8.0 	imes 10^{-16} = [Fe^{2+}](0.1)^2$.$[Fe^{2+}] = \frac{8.0 	imes 10^{-16}}{0.01} = 8.0 	imes 10^{-14} \ M$.Since the molar solubility $S = [Fe^{2+}]$,the solubility is $8.0 	imes 10^{-14} \ M$.

What is the molar solubility of $Fe(OH)_2$ $(K_{sp} = 8.0 \times 10^{-16})$ at $pH$ $13.0$ ?

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