The values of Delta G_f^o (in kJ mol^{-1}) f | vedclass

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(A) The dissociation reaction is: $Mg(OH)_{2(s)} 	o Mg^{2+}_{(aq)} + 2OH^-_{(aq)}$Calculate the standard Gibbs free energy change $(\Delta G^o)$:$\De

Vedclass · Accepted Answer

$8.4 	imes 10^{-12} \ M^3$

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(A) The dissociation reaction is: $Mg(OH)_{2(s)} 	o Mg^{2+}_{(aq)} + 2OH^-_{(aq)}$Calculate the standard Gibbs free energy change $(\Delta G^o)$:$\Delta G^o = [\Delta G_f^o(Mg^{2+}) + 2 	imes \Delta G_f^o(OH^-)] - [\Delta G_f^o(Mg(OH)_2)]$$\Delta G^o = [-456.0 + 2(-157.3)] - (-833.9)$$\Delta G^o = [-456.0 - 314.6] + 833.9 = 63.3 \ kJ \ mol^{-1} = 63300 \ J \ mol^{-1}$Using the relation $\Delta G^o = -RT \ln K_{sp}$ or $\Delta G^o = -2.303 RT \log K_{sp}$:$63300 = -2.303 	imes 8.314 	imes 298 	imes \log K_{sp}$$63300 = -5705.85 	imes \log K_{sp}$$\log K_{sp} = -11.0938$$K_{sp} = 10^{-11.0938} \approx 8.06 	imes 10^{-12} \ M^3$Rounding to the nearest provided option,the correct answer is $8.4 	imes 10^{-12} \ M^3$.

The values of $\Delta G_f^o$ (in $kJ \ mol^{-1}$) for $Mg(OH)_2$ at $25 \ ^oC$ are given below. Calculate the solubility product constant $(K_{sp})$.
$Mg^{2+}_{(aq)} = -456.0; OH^{-}_{(aq)} = -157.3; Mg(OH)_{2(s)} = -833.9$

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The values of $\Delta G_f^o$ (in $kJ \ mol^{-1}$) for $Mg(OH)_2$ at $25 \ ^oC$ are given below. Calculate the solubility product constant $(K_{sp})$.$Mg^{2+}_{(aq)} = -456.0; OH^{-}_{(aq)} = -157.3; Mg(OH)_{2(s)} = -833.9$