Question

Please help I’m timed !!!

Answer 1

Answer:

b. $\% diss =4.4x10^{-2}\%$

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:

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

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

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

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:

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

Which is choice b.

Best regards!