Q. Ionisation of weak acid can be calculated by the formula

189 209 Equilibrium Report Error

A

$100 \frac{K _{a}}{C}$

B

$\frac{100}{1 + 1 0 ^{(p K_{a} - p H)}}$

C

Both (a) and (b)

D

None of these

Solution:

For weak acid dissociation equilibria, degree of dissociation $α$ is given as :
$α = \frac{K _{a}}{C} ∴ % α = 100 \frac{K _{a}}{C}$
Also, $K_{a} = \frac{[ H ^{+} ] [ A ^{-} ]}{[ H A ]} = \frac{[ H ^{+} ] c α}{c ( 1 - α )} = \frac{[ H ^{+} ] α}{( 1 - α )}$
$lo g K_{a} = lo g H^{+} + lo g \frac{α}{1 - α}$
or $p K_{a} = p H + lo g \frac{1 - α}{α}$
$p K_{a} - p H = lo g \frac{1 - α}{α}$
$\frac{1 - α}{α} = 1 0^{p K_{a} - p H}$
or $\frac{1}{α} = 1 0^{p K_{a} - p H} + 1$
$∴ α = \frac{1}{[ 1 + 1 0 ^{p K_{a} - p H} ]}$
or $% α = \frac{100}{[ 1 + 1 0 ^{p K_{a} - p H} ]}$