The pH of $10^{- 3} M N H_{4} O H$ ( $K_{b} = 10^{- 5}$ ) is:

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Solution

The correct option is C 10
For weak base ammonium hydroxide,
$[O H^{-}] = \sqrt{k_{b} \times c}$
Substitute values in the above expression,
$[O H^{-}] = \sqrt{10^{- 5} \times 10^{- 3}} = 10^{- 4}$ M
$p O H = - l o g [O H^{-}] = - l o g 10^{- 4} = 4$
$p H = 14 - p O H = 14 - 4 = 10$

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