n (HCl) = C * V
n (HCl) = 0.08 * 0.02
n (HCl) = 1.6*10^-3mols
n (NaOH) = C * V
n (NaOH) = 0.05 * 0.03
n (NaOH) = 1.5*10^-3mols
n (excess) = n (HCl) - n (NaOH)
n (excess) = 1.6*10^-3 - 1.5*10^-3
n (excess) = 1*10^-4 mols of HCl
C = n / V
C = 1*10^-4 / 0.05
C = 2*10^-3mols
pH = - log [H+]
pH =...