Acetic acid = HAc
.........HAc ==> H^+ + Ac^-
I........0.05....0......0
C........-x......x......x
E......0.05-x...x......x
You know x = 0.05 x 0.02 = ? = (H^+).
Then pH = -log(H^+) = ?
What is the pH of a 0.05 M solution of acetic acid knowing that it was 2% ionized
3 answers
2.8
2.8
4.1
4.1