100 mL of 0.02 M benzoic acid pKa = 4.2 is titrated using 0.02 M NaOH. pH values after 50 mL and 100 mL of NaOH have been added are
3.70, 7
4.2, 7
4.2, 8.1
4..2, 8.25
On 50ml NaOH addition C6H5COOH+OH- → C6H5COO-+H2O t = 0 2 1 0 teq 1 1 1 pH = pKa = 4.2 On 100 ml NaOH addition C6H5COOH+OH-→ C6H5COO-+ H2O t = 0; 2 2 0 teq 2 2 2 C6H5COO- = 2200 = 0.01 M ∴ pH = 7 + pKa + 12 log C = 7 + 4.22 + 12 log 0.01 = 8.1