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



Ос

Od

7.9' 10-3 %

4.4'10-2%

6.3' 10-9%

3.5' 10-6 %

Answers

Answer 1

Answer:

b. [tex]\% diss =4.4x10^{-2}\%[/tex]

Explanation:

Hello there!

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

[tex]HClO\rightleftharpoons H^++ClO^-[/tex]

We can write the equilibrium expression as shown below:

[tex]Ka=3.5x10^{-8}=\frac{[H^+][ClO^-]}{[HClO]}[/tex]

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

[tex]3.5x10^{-8}=\frac{x^2}{[HClO]}[/tex]

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

[tex]x=\sqrt{3.5x10^{-8}*0.18} =\\\\x=7.94x10^{-5}M[/tex]

And finally the percent dissociation:

[tex]\% diss=\frac{x}{[HClO]} *100\%\\\\\% diss=\frac{7.94x10^{-5}M}{0.18}*100\% \\\\\% diss =0.044\%=4.4x10^{-2}\%[/tex]

Which is choice b.

Best regards!


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