Question

Calculate the pH of 0.0095M solution of acetic acid give that Ka = 1.76*10^-5.


I started the question by using -log(0.0095) but I'm not sure if that's all I need to do.

Any help is much appreciated!

Answer 1

Ka of acetic acid is 1.7 * 10^-5
it should be provided in the appendix of your chem textbook

.....CH3COOH <===> H+ + CH3COO-
I.....0.0095M...........0...........0
C.... -x......................+x..........+x
E...0.0095M-x...........x............x

Ka = [H+] [CH3COO-] / [CH3COOH]
= x * x / (0.0095M - x)
= x^2 / (0.0095M - x) = 1.7*10^-5

rearrange into quadratic equation
x^2 = (0.0095M -x) * 1.7*10^-5
x^2 +(1.7*10^-5)*X -(0.0096M)*(1.7*10^-5) = 0

x = 0.00040M

but [H+] = x
[H+] = 0.00040M

pH = -log[H+] = 3.40

Answer 2

ohh that makes so much more sense now! thanks again chmvijay you have been a great help! :)

Answer 3

ur welcome :)