Question

Calculate the degree of ionization of 0.04 M HOCL
solution having ionization constant
1.25 x 104 ?

Answer 1

The degree of ionization of 0.04 M HOCL  solution having ionization constant  $1.25\times 10^4$ is 99.9%

Explanation:

The ionization of the acid is represented by:

$HOCl\rightarrow ClO^-+H^+$

 cM              0             0

$c-c\alpha$        $c\alpha$          $c\alpha$  

So ionization constant will be:

$K_a=\frac{(c\alpha)^{2}}{c-c\alpha}$

Give c= concentration = 0.04 M and

$\alpha$ = ionization constant = ?

$1.25\times 10^4=\frac{(0.04\times \alpha)^2}{(0.04-0.04\times \alpha)}$

$\alpha=0.999$

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