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

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

Answer 1

Answer: (C)

4.2, 8.1

Answer 2

C6H5COOH + OH– $\longrightarrow$ C6H5COO– + H2O

t=0 2 1

teq 1 – 1

pH = pKa = 4.2

C6H5COOH + OH– $\longrightarrow$ C6H5COO– + H2O

t=0 2 2

teq – – 2

[C6H5COO–] = $\frac{2}{200}$= 0.01 M

\ pH = 7 + pKa + $\frac{1}{2}$ log C = 7 + $\frac{4.2}{2}$ + $\frac{1}{2}$ log (0.01)

= 8.1