Question

20 mL of a weak monobasic acid (HA) requires 20 mL 0.2 M NaOH for complete titration. If pH of solution upon addition of 10 mL of this alkali to 25 mL of the above solution of HA is 5.8. The pKa of the weak acid is

• A. 6.1
• B. 5.8
• C. 5.98
• D. 5.58

Answer 1

Answer: (C)

5.98

Answer 2

meq. of acid = meq of base $\Rightarrow$ 20 × M = 20 × 0.2 = 4

Molarity of HA = 0.2 M

HA + OH $\longrightarrow$A– + H2O 5 2

m. mole 3 – 2

m. mole

$\therefore$ pH = pKa + log $\frac{\lbrack A^{-}\rbrack}{\lbrack HA\rbrack}$ $\Rightarrow$ 5.8 = pKa + log $\left( \frac{2}{3} \right)$

$\Rightarrow$ pKa = 5.98