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Solubility Product

The pH of a 0...

Question

The pH of a 0.001 M solution of $N H_{4} O H$ (having dissociation constant $K_{b}$ $= 1.0 \times 10^{- 5}$ ) is equal to:

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Solution

The correct option is D 10
$N H_{4} O H \to N H_{4}^{+} + O H^{-}$

Conc. of $[O H^{-}] = c α = \sqrt{C \times K_{b}} = \sqrt{0.001 \times 10^{- 5}} = 10^{- 4}$

So $p O H = - l o g [O H^{-}] = 4$

$p H = 14 - p O H = 10$

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