Question

the ka of hypochlorous acid (hclo) is 3.0 x10^-8 at 25.0°C. What is the % of ionization of hypochlorous

Answer 1

0.14%

Step-by-step explanation:

The computation of % is shown below:

As we know that

HClO <=> H+ + ClO-

I 0.015 0 0

C -a +a +a

E 0.015-a a a

Now

`Ka = ([H+][ClO-])/([HClO])`

`= (a^(2))/((0.015 - a)) \\\\= 3.0 * 10^(-8)`

`a^(2) + 3.0 * 10^(-8)a - 4.5 * 10^(-10) = 0`

Now Solves the quadratic equation i.e.

`a = 2.120 * 10^(-5)`

`[H+] = a = 2.120 * 10^(-5) M`

So,

% ionization is

`= ([H+])/([HClO])_(initial) * 100\%\\\\= 2.120 * 10^(-5)/0.015 * 100\%`

= 0.14%

Hence, the percentage of hypochlorous ionization is 0.14%