Calculate the...

Question

Calculate the degree of ionisation and hydroxyl ion concentration in $0.2 M$ ${N H}_{3}$ solution.
( $K_{b} = 1.85 \times 10^{- 5}$ for $298 K$ )

Open in App

Solution

$N H_{3} + H_{2} O ⇌ N H_{4}^{+} + O H^{-}$
$0.2$
$0.2 (1 - α)$ $0.2 α$ $0.2 α$
$K_{b} = \frac{[N H_{4}^{+}] [{O H}^{-}]}{[N H_{3}]} = \frac{(C α) (C α)}{C (1 - α)}$
Assume $α < 0.01$ so $1 - α = 1$
$1.85 \times 10^{- 5} = 0.2 (α)^{2}$
$α = 0.962 \times 10^{- 2}, [{O H}^{-}] = 0.2 α = 1.924 \times 10^{- 3} M$

Suggest Corrections

Similar questions

K_{a}

2 \times 10^{- 5}

0.2 M

0.15

500 {c m}^{3}

0.2 M

K_{b}

= 1.5 \times 10^{- 9} M

0.4 M

{C H}_{3} C O O H

0.2 M

{C H}_{3} C O O N a

H^{+}

1.8 \times 10^{- 5}

H C N

4 \times 10^{- 10}

0.2 M

H C N

1 m o l L^{- 1}

K C N

K_{a}

2.0 \times 10^{- 5}

0.2 M