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Factors Affecting Alpha

0.1 mole of C...

Question

0.1 mole of $C H_{3} N H_{2} (K_{b} = 5 \times 10^{- 4}$ is mixed with 0.08 mole of $H C l$ and

diluted to one litre. What will be the $H^{+}$ concentration in the solution?

A

8 x 10^-2 M

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B

8 x 10^-11 M

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C

1.6 x 10^-11 M

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D

8 x 10^-5 M

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Solution

The correct option is B
8 x 10^-11 M

$C H_{3} N H_{2} + H C l \to C H_{3} N H_{3}^{+} C l^{-}$

Initially,
$C H_{3} N H_{2}$ = $0.1$ , $H C l$ = $0.08$ , $C H_{3} N H_{3}^{+} C l^{-}$ = $0$
At Equilibrium,
$C H_{3} N H_{2}$ = $0.02$ , $H C l$ = $0$ , $C H_{3} N H_{3}^{+} C l^{-}$ = $0.08$

$p O H = p K + l o g \frac{0.08}{0.02}$
= $p K + 0.602$
= $3.30 + 0.602$ = $3.902$

$∴ p H = 10.1$
$[H^{+}] = 7.99 \times 10^{- 11}$

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