Ionic EquilibriumHard
Question
What is the pH of 6.67 × 10−3 M aqueous solution of Al(OH)3 if its first dissociation is 100%, second dissociation is 50% and the third dissociation is negligible.
Options
A.2
B.12
C.11
D.3
Solution
$\left\lbrack OH^{-} \right\rbrack = 6.67 \times 10^{- 3} + \frac{6.67 \times 10^{- 3}}{2} + 0 \approx 10^{- 2}\text{ M}$
$\therefore P^{OH} = - \log\left( 10^{- 2} \right) = 2.0 \Rightarrow P^{H} = 12.0$
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