Calculate the $H^{+}$ ion concentration of a $0.01 N$ weak monobasic acid. The value of dissociation constant is $4.0 \times 10^{- 10}$ .

Open in App

Solution

$H A ⇌ H^{+} + A^{-}$
According to ostwald's dilution law for weak monobasic acid
$K_{a} = C α^{2}$
It is given that $K_{a} = 4.0 \times 10^{- 10}, C = 0.01$
So, $α = \sqrt{\frac{K_{a}}{C}} = 2 \times 10^{- 4}$ .
Concentration of Hydrogen ion = $C α = 2 \times 10^{- 6} m o l / l$

Suggest Corrections

Similar questions

H^{+}

4.0 \times 10^{- 10}

4 \times 10^{- 10} . p H

0.01 N

p H

0.05 M

8.0 \times 10^{- 9}

H^{+}

0.10 M

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

10^{- 6}

P^{H}

0.01 N

Join BYJU'S Learning Program