Ionic EquilibriumHard

Question

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

Options

A.3.50, 7
B.4.2, 7
C.4.2, 8.1
D.4.2, 8.25

Solution

       C6H5COOH + OH- → C6H5COO-   +  H2O
t = 0  2               1
teq    1               -          1
        pH = pKa = 4.2
    C6H5COOH + OH-   C6H5COO- + H2O
t=0    2               2
teq    -               -        2
[C6H5COO-] = = 0.01 M
∴ pH = 7 + pKa + log C = 7 + log (0.01) = 8.1 

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