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Salt of Strong Acid and Weak Base

1.2 mol of ...

Question

$1.2$ mol of $C H_{3} N H_{2} (p K_{b} = 3.3)$ is added to $0.8$ moles of $H C l$ and the solution is diluted to one litre, resulting $p H$ of solution is:

10.7

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3.6

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10.4

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11.3

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Solution

The correct option is B $10.4$
As we know,
$p O H = p K_{b} + l o g \frac{s a l t}{b a s e}$

$p O H = 3.3 + l o g \frac{0.8}{0.4} = 3.6$

$p H = 14 - p O H = 10.4$

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