Calculate the pH of the following.
$0.002$ N $N H_{4} O H$ having $2.3$ % dissociation.

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Solution

$N H_{4} O H$ is weak base and partially dissociated.
$N H_{4} O H ⇌ N H_{4}^{\oplus} + ⊝ O H$
Cone before dissociation 1 0 0
Cone after dissociation $1 - α$ $α$ $α$
$∴ [⊝ O H] = C α = 2 \times 10^{- 3} \times \frac{23}{100} = 4.6 \times 106 - 5 M$
$∵ p O H = l o g [⊝ O H] = - l o g (4.6 \times 10^{- 5})$
$∴ p O H = - 433.72 ∴ p H = 14 - 4.3372$
$p H = 9.6628$

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