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Salt of Strong Acid and Weak Base

Aniline behav...

Question

Aniline behaves as a weak base. When $0.1 M, 50 m L$ solution of aniline was mixed with $0.1 M, 25 m L$ solution of $H C l$ the $p H$ of resulting solution was 8. The $p H$ of $0.01 M$ solution of anilinium chloride will be:

5

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4.5

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Solution

The correct option is A $5$
Given:
Milimoles of Aniline $= 0.1 \times 50 = 5$
Milimoles of $H C l = 0.1 \times 25 = 2.5$
Reaction taking place will be
$C_{6} H_{5} N H_{2} + H C l ⇌ C_{6} H_{5} N H_{3}^{+} C l^{-}$
$5 + 2.5 ⇌ 0$
$2.5 + 0 ⇌ 2.5$

Given $p H = 8$
$\Rightarrow p O H = 14 - 8 = 6 = p K_{b}$

$p H$ of $0.01 M$ solution of anilinium chloride:

Using the formula,
$p H = 7 - \frac{1}{2} P K_{b} - \frac{1}{2} l o g C$

putting the values,

$p H = 7 - \frac{1}{2} (6) - \frac{1}{2} l o g (0.01)$

$p H = 5$

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Salt of Strong Acid and Weak Base

Standard XIII Chemistry

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Aniline behaves as a weak base. When 0.1M, 50mL solution of aniline was mixed with 0.1M, 25mL solution of HCl the pH of resulting solution was 8. The pH of 0.01 M solution of anilinium chloride will be:

Aniline behaves as a weak base. When $0.1 M, 50 m L$ solution of aniline was mixed with $0.1 M, 25 m L$ solution of $H C l$ the $p H$ of resulting solution was 8. The $p H$ of $0.01 M$ solution of anilinium chloride will be: