Question

$\underset{(0.09 - x)\text{ M}}{en + H_{2}O} \rightleftharpoons \underset{(x - y)\text{ M}}{enH^{+}} + \underset{(x + y)\text{ M}}{OH^{-}};K_{b_{1}} = 8.1 \times 10^{- 5}$

$\underset{(x - y)\text{ M}}{enH^{+} + H_{2}O} \rightleftharpoons \underset{y\text{ M}}{enH_{2}^{2 +}} + \underset{(x + y)\text{ M}}{OH^{-}};K_{b_{2}} = 7.0 \times 10^{- 8}$

Now, $8.1 \times 10^{- 5} = \frac{(x - y)(x + y)}{(0.09 - x)}K \approx \frac{x.x}{0.09} \Rightarrow x = 2.7 \times 10^{- 3}$

and $7.0 \times 10^{- 8} = \frac{y.(x + y)}{(x - y)} \approx \frac{y.x}{x} \Rightarrow y = 7.0 \times 10^{- 8}$

∴ [en H⁺] = (x – y) ≈ x M = 2.7 × 10^–3 M

[enH₂²⁺ ]= y = 7.0 × 10⁻⁸ M

∴ [OH^–] = (x + y) = x = 2.7 × 10^–3 M

and P^OH = – log (2.7 × 10^–3) = 2.56 ⇒ PH = 11.44