Question

A weak acid (HA) after treatment with 12 mL of 0.1 M strong base (BOH) has a pH of 5. At the end point, the volume of same base required is 27 mL. Ka of acid is (log2 = 0.3)

• A. 1.8 × 10–5
• B. 8 × 10–6
• C. 1.8 × 10–6
• D. 8 × 10–5

Answer 1

Answer: (B)

8 × 10–6

Answer 2

m. moles of HA taken = 27 × 0.1 = 2.7

HA + OH– $\longrightarrow$ A– + H2O

t=0 2.7 1.2

teq 1.5 – 1.2

pH = pKa + log a

$\Rightarrow$ 5 = pKa + log $\left( \frac { 1.2 } { 1.5 } \right)$ = pKa + log $\frac { 4 } { 5 }$