Question
At what minimum pH will 10–3 M – Al(OH)3 go into solution (V = 1 L) as Al(OH)4− and at what maximum pH, it will be dissolved as Al3+? Given: log 2 = 0.3
Al(OH)4− $\rightleftharpoons$Al3+ + 4OH–; Keq = 1.6 × 10–34
Al(OH)3 $\rightleftharpoons$Al3+ + 3OH–; Keq = 8.0 × 10–33
Options
Solution
$Al(OH)_{3}(s) + OH^{-} \rightleftharpoons Al(OH)_{4}^{-};$
From question ? 10-3 M
$K = \frac{8 \times 10^{- 33}}{1.6 \times 10^{- 34}} = 50$
$50 = \frac{10^{- 3}}{\left\lbrack OH^{-} \right\rbrack} \Rightarrow \left\lbrack OH^{-} \right\rbrack = 2 \times 10^{- 5}\text{ M} \Rightarrow P^{H} = 9.30$
As the calculated [OH–] is minimum OH–, PH is minimum.
$Al(OH)_{3}(s) \rightleftharpoons \underset{10^{- 3}M}{Al^{3 +}} + 3OH^{-};K = 8 \times 10^{- 33} $$${8 \times 10^{- 33} = 10^{- 3} \times \left\lbrack OH^{-} \right\rbrack \Rightarrow \left\lbrack OH^{-} \right\rbrack = 2 \times 10^{- 10}\text{ M} }{\Rightarrow P^{H} = 4.30}$$
As the calculated [OH–] is maximum OH–, PH is maximum.
Create a free account to view solution
View Solution Free