100 mL of 0.02 M benzoic acid (pKa = 4.2) is titrated by using 0.02 M NaOH. pH after 50 mL and 100 mL of NaOH have been added are

Millimoles of benzoic acid = 2Millimole of NaOH in first addition = 1Millimoles of NaOH in second addition = 2                     C6H5COOH+NaOH⇌C6H5COONa+H2O Initial                      2                   0                   0                0 After first  addition of NaOH  (2-2)=0           (2-2) = 0        2               After first addition, mixture is a buffer containing 1 millimole each of C6H5COONa and C6H5COOH∴ pH=pKa∗+log⁡C6H5COONaC6H5COOH⇒pKa=4.2After second addition, there are only 2 millimoles of, C6H5COONa in 200 mL solution and due to hydrolysis, solution is basic.C6H5COO−=2×10−302=1×10−2MC6H5COO−+H2O⇌C6H5COOH+OH− ∴ pH=7+pKa2+logC2 =7+2.1-1=8.1.