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.
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