The $p H$ of a $0.02 M$ ammonia solution which is $5 %$ ionized will be:

2

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5

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7

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11

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Solution

The correct option is C $11$
$C = 0.02 M N H_{3} = B a s e$
$α = 5$ % $= 0.05$

$({O H}^{-}) = C \times α$

$({O H}^{-}) = 0.02 \times 0.05$

$= 10^{- 3}$

$∴ p O H = - l o g (O H) = - l o g (10^{- 3})$
$= - (- 3)$
$= 3$
$w e k n o w, p H + p O H = 14$
$p H = 14 - p O H$
$= 14 - 3 = 11$

Here answer is option D.

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