...............CH3COOH ==> H^+ + CH3COO^-
I................0.20 M..............0...............0
C......................-x...............x...............x
E..................0.20-x.............x...............x
Ka = 1.8E-5 = (x)(x)/(0.20-x)
Solve for x = (H^+), then pH = -log(H^+)
Post your work if you get stuck.
What would be the PH of a 0.20 M CH3COOH solution if no salt were present
1 answer