Question

What is the pH of a 0.1 M NaOCl solution (Ka of HOCl is 2.95 x10-8)

I did:
x squared/.10= 2.95x10-8
x squared= 2.95x10-9
x= 5.43x10-5
-log(5.43x10-5)=
4.27 pH
However, the options I'm given are either
a.) 3.74 b.) 6.47 c.) 8.23 d.) 10.27

What steps am I doing wrong?

Answer 1

You are treating this as an acid perhaps. NaOCl is a salt and the pH is determined by the hydrolysis of the salt; i.e., Kb of OCl^-.
........OCl^- + HOH ==> HOCl + OH^-
I.......0.1M..............0.....0
C.......-x................x.....x
E......0.1-x..............x.....x

Kb for OCl^- = (Kw/Ka for HOCL) = (x)(x)/(0.1-x) = ?
Solve for x = (OH^-), convert to pOH then to pH. My calculator is on the blink and I can't calculate that; however, it appears that c and d are in the right ballpark.