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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

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

Vedclass · Answer

(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 $(K_{sp})$.
$Mg^{2+}_{(aq)} = -456.0; OH^{-}_{(aq)} = -157.3; Mg(OH)_{2(s)} = -833.9$

Similar Questions

The solubility of salt $AX_2$ is $6 \times 10^{-12} \text{ mol dm}^{-3}$. What is the solubility product $(K_{sp})$ of the salt?

$PbCl_2$ has a maximum concentration of $1.0 \times 10^{-3} \ M$ in its saturated aqueous solution at $25 \ ^oC$. Its solubility in a $0.1 \ M \ NaCl$ solution will be:

Equal volumes of $0.002 \, M$ solutions of sodium iodate and cupric chlorate are mixed together. Will it lead to precipitation of copper iodate?
(For cupric iodate $K_{sp} = 7.4 \times 10^{-8}$)

If $[S^{2-}] = 0.6 \times 10^{-2} \ mol \ m^{-3}$ and $[Hg^{2+}] = [Mn^{2+}] = [Fe^{2+}] = [Zn^{2+}] = 1 \times 10^{-16} \ mol \ m^{-3}$,which of the following will precipitate first in an aqueous solution?

For a compound $AB$,the degree of ionization is $80\%$. What is the solubility $(s)$ of $AB$ if its solubility product $(K_{sp})$ is $6.4 \times 10^{-9}$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module

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 $(K_{sp})$.$Mg^{2+}_{(aq)} = -456.0; OH^{-}_{(aq)} = -157.3; Mg(OH)_{2(s)} = -833.9$