Question

Calculate the ionisation constant of 0.08 moldm$^{-3}$ of a monobasic acid having pH = 2.

• A. $3531 \times 10^{-7}$
• B. $2.081 \times 10^{-6}$
• C. $3.456 \times 10^{-8}$
• D. $1.25 \times 10^{-3}$

Hint:

$K_a = C\alpha^2$ is used when $\alpha$ is very small; otherwise, use $K_a = \frac{[H^+]^2}{C}$.

Answer 1

The Correct Option is D

Step 1: Find $[H^+]$
$pH = 2 \implies [H^+] = 10^{-2} = 0.01$ M.

Step 2: Formula
For a weak monobasic acid, $K_a = \frac{[H^+]^2}{C - [H^+]}$. Since $[H^+]$ is significant compared to $C$, we check: $K_a \approx \frac{[H^+]^2}{C}$.

Step 3: Calculation
$K_a = \frac{(0.01)^2}{0.08} = \frac{0.0001}{0.08} = \frac{1}{800} = 0.00125$.
$K_a = 1.25 \times 10^{-3}$.

Step 4: Conclusion
The ionisation constant is $1.25 \times 10^{-3}$.
Final Answer: (D)