PH chemistry question?

Question

Calculate the change in pH (to 3 sig figures) that occurs when 1.00 mol of sodium acetate (CH3COONa) is added to 1.00M acetic acid (CH3COOH). Assume no change in volume. Ka = 1.80x10-5

show all steps and working please!

Answer 1

Initial concentration acetic acid = 1.00 M

CH3COOH <-----> CH3COO- + H+

1.80 x 10^-5 = x^2 / 1.00 -x

x = [H+] = 0.00424 M

pH = - log 0.00424 =2.37

When acetate is added we get the same concentration of acetic acid and acetate

pH = pKa + log [CH3COO-] / [CH3COOH]

pH = 4.74 + log 1 = 4.74

pH change = 4.74 - 2.37 = 2.37

I'm a chemist