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Calculate the pH at which the following reaction will be at equilibrium in the basic medium $A_{2} (s) ⇌ A^{-} (a q) + A O_{3}^{-} (a q)$ . The equilibrium concentrations at $300 k$ are
$[A^{-}] = 0.1 M$
$[A O_{3}^{-}] = 0.1 M$
Given $Δ G_{f}^{0}$ of $O H^{-} (a q) = - 150 K J / m o l e$
$Δ G_{f}^{0} o f H_{2} O (l) = - 233 K J / m o l e$
$Δ G_{f}^{0} o f A^{-} (a q) = - 50 K J / m o l e$
$Δ G_{f}^{0} o f A O_{3}^{-} (a q) = - 123.5 K J / m o l e$
$R = \frac{25}{3} J / m o l e - k \ln 10 = 2.3$

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Detailed Solution

The balanced equation $3 A_{2} (s) + 6 O H^{-} \to 5 A^{-} + A O_{3} + 3 H_{2} O$

$Δ G_{R \times n}^{0} = [5 \times (- 50) + (- 123.5) + 3 \times (- 233)] - [3 \times 0 + 6 \times (- 150)] = - 172.5 K J Δ G_{R \times n}^{0} = - R T \ln K k = \frac{{(0.1)}^{5} \times (0.1)}{{[O H^{-}]}^{6}} K = 10^{30}$

On solving

$[O H] = 10^{- 6} P^{O H} = 6 P^{H} = 8$

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