Answer:
0.14%
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
[tex]Ka = \frac{[H+][ClO-]}{[HClO]}[/tex]
[tex]= \frac{a^{2}}{(0.015 - a)} \\\\= 3.0 \times 10^{-8}[/tex]
[tex]a^{2} + 3.0 \times 10^{-8}a - 4.5 \times 10^{-10} = 0[/tex]
Now Solves the quadratic equation i.e.
[tex]a = 2.120 \times 10^{-5}[/tex]
[tex][H+] = a = 2.120 \times 10^{-5} M[/tex]
So,
% ionization is
[tex]= \frac{[H+]}{[HClO]}_{initial} \times 100\%\\\\= 2.120 \times 10^{-5}\div0.015 \times 100\%[/tex]
= 0.14%
Hence, the percentage of hypochlorous ionization is 0.14%
