An aqueous solution of $0.24 M$ aniline ( $K_{b} = 4.166 \times 10^{- 10}$ ) is mixed with $N a O H$ solution to maintain anilinium ion concentration to $1 \times 10^{- 8} M$ .
The $p O H$ of $N a O H$ solution used was:

Open in App

Solution

When NaOH is added,
$C_{6} H_{5} N H_{2} + H O H ⇌ C_{6} H_{5} N H_{3}^{+} + O H^{-}$
0.24 $10^{- 8}$ y
$\frac{[C_{6} H_{5} N H_{3}^{+}] [O H^{-}]}{[C_{6} H_{5} N H_{2}]} = K_{b}$

$\frac{y \times 10^{- 8}}{0.24} = 4.166 \times 10^{- 10}$ $⟹ y = 1 \times 10^{- 2} = [O H^{-}]$

$p O H = - l o g [O H^{-}] = - l o g (1 \times 10^{- 2}) = 2$

Suggest Corrections

Similar questions

0.24 M

1 \times 10^{- 8} M

K_{a}

C_{6} H_{5} {N H}_{3}^{+} = 2.4 \times 10^{- 5} M

3 \times 10^{- 5} m o l L^{- 1}

0.01

R N H_{2}

K_{b} = 2 \times 10^{- 6}

10^{- 4}

O H^{-}

H C l

p H

K_{w} = 10^{- 14}

p O H

=

10^{- 13}

Join BYJU'S Learning Program