Question

Please help I’m timed !!!

Determine the percent dissociation of a 0.18 M solution of hypochlorous acid, HClO. The Ka for the acid is

3.5 x 10-8

Оа

Oь

Ос

Od

7.9' 10-3 %

4.4'10-2%

6.3' 10-9%

3.5' 10-6 %

Answer 1

b.
`\% diss =4.4x10^(-2)\%`

Step-by-step explanation:

Hello there!

In this case, given the ionization reaction of HClO as weak acid:

`HClO\rightleftharpoons H^++ClO^-`

We can write the equilibrium expression as shown below:

`Ka=3.5x10^(-8)=([H^+][ClO^-])/([HClO])`

In such a way, via the definition of x as the reaction extent, we can write:

`3.5x10^(-8)=(x^2)/([HClO])`

As long as Ka<<<<1 so that the x on the bottom can be neglected. Thus, we solve for x as shown below:

`x=\sqrt{3.5x10^(-8)*0.18} =\\\\x=7.94x10^(-5)M`

And finally the percent dissociation:

`\% diss=(x)/([HClO]) *100\%\\\\\% diss=(7.94x10^(-5)M)/(0.18)*100\% \\\\\% diss =0.044\%=4.4x10^(-2)\%`

Which is choice b.

Best regards!