The pH of a 0.55 mol L-1 solution of a weak acid, HZ, is 3.76.

Calculate Ka for HZ?

Please show workings.

Cheers.

Hi!!

Assuming that the dissociation equation is:
HZ(aq) <---> H+ + Z- (aq)

Then:
Ka = [H+]^2/[HZ]   ([H+] = [Z-])


[H+] = 10^(-pH) = 10^(-3.76) = 1.7378*10^-4

Therefore:
Ka = (1.7378*10^-4)^2/0.55 = 5.491*10^-8


At the following Powerpoint file you will find an excellent source for
a further reading and explanations on this topic:
"Acids and Bases" by Tsun-Mei Chang at University of Wisconsin-Parkside site:
http://uwp.edu/~chang/C102/ch16/chang-16.ppt


I hope this helps you. Feel free to request for a clarification if you need it.


Best regards,
livioflores-ga

Excellent answer thanks for your help :)