​ 13.​Calculate the pH of a solution of 0.10M HOCl if 0.10M NaOCl is added. ​Ka HOCl = 3.5 x 10-8 (a) 7.46 (b) 8.0 (c) 6.54 (d) 8.54 (e) None

Answer 1

Henderson Hasselbalch equation:

pH = pKa + log [salt]/[acid]

pKa = -log Ka = -log 3.5x10-8 = 7.46

Since [salt] = [acid] (1/2 the equivalence point), the pH = pKa since log 1 = 0

So, pH = 7.46 (ANSWER a)

Answer 2

Hi Asked (a tutor doesn't know this??),

you have a buffer, at the equivalence point, of a weak acid. So pH =p(Ka) = -log(3.5*10^(-8)). You should be able to pick the correct choice in your head! It is one of the first 4 options ....

-- Cheers, --Mr. d.