Chemistry Question

Question

Calculate the pH of a solution involving the addition of 35 mL of 0.1 M acetic acid, CH3COOH, to 350 mL of water. Ka for acetic acid is 1.8 x 10^(-5). Thank you!

Answer 1

[CH3COOH] = (35 ml)(0.1 M) = (350 ml)(x M) and x = 0.01M 

CH3COOH ==> CH3COO- + H+

Ka = [H+][CH3COO-]/[CH3COOH]

1.8x10^-5 = (x)(x)/0.01

x² = 1.8x10^-7 

x = 4.2x10^-4 M = [H+]

pH = -log 4.2x10^-4

pH = 3.37