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Q.

Percentage of ionization of 0.1 M NH₄OH having p^H=9 at 298 K will be

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a

$10^{- 1}$

b

$10^{- 3}$

c

$10$

d

$10^{- 2}$

answer is D.

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Detailed Solution

$p^{H} = 9 ∴ p^{O H} = 5$

$\begin{array}{l} [O H^{-}] = C α \\ 10^{- 5} = 0.1 α \\ α = 10^{- 4} \end{array}$

% of ionization = $10^{- 4} \times 100 = 10^{- 2}$

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Percentage of ionization of 0.1 M NH4OH having pH=9 at 298 K will be