a) pH = -log 0.10 M = 1.00

b) So, the equilibrium is CH3COOH + H2O <--> H3O+ + CH3COO-.

Ka = 1.8X10^-5 = [H3O+][CH3COO-]/[CH3COOH]

Let x = [H3O+] = [CH3COO-]. [CH3COOH] = 0.10 M. Then:

1.8X10^-5 = x^2 / 0.10

x = [H3O+] = 1.34X10^-3

pH = -log (1.34X10^-3) = 2.87