Calculate the pH at which the following reaction will be at equilibrium in basic medium:
$A_{2} (s) ⇌ A^{-} (a q) + {AO}_{3}^{-} (a q)$
when the equilibrium concentration at 300 K are :
$[A^{-}] = 0.1 M, [{AO}_{3}^{-}] = 0.1 M$
Given : $Δ G_{f} of {OH}^{-} (a q) = - 150 kJ / mole$
$Δ G_{f}^{\circ}$ of $H_{2} O (l) = - 233 kJ / mole$
$Δ G_{f}^{\circ}$ of $A^{-} (a q) = - 50 kJ / mole$
$Δ G_{f}^{\circ}$ of $A^{-} (a q) = - 50 kJ / mole$
$R = \frac{25}{3} J {mole}^{- 1} K^{- 1}, \log_{e} 10 = 2.3$

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answer is 8.

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Detailed Solution

$\begin{array}{l} 3 A_{2} (s) + 6 {OH}^{-} \overset{↓}{⟵} {AA}^{-} (a q) + {AO}_{3}^{-} (α q) + 3 H_{2} O \\ K^{\circ} = \frac{(0.1)^{5} \times 0.1}{{[{OH}^{-}]}^{6}} \end{array}$

$\begin{array}{l} Δ G_{R \times n}^{\circ} = 5 \times (- 50) - 123.5 - 3 \times 233 + 6 \times 150 \\ = - 172.5 kJ / mol = - 2.3 \times \frac{25}{3} \times 300 \log k^{\circ} \\ \Rightarrow K^{\circ} ≅ 10^{30} = \frac{10^{- 6}}{{[{OH}^{-}]}^{6}} \\ \Rightarrow [{OH}^{-}] = 10^{- 6} \\ \Rightarrow pH = 8 \end{array}$