Question

Calculate the pH of 0.02 M monobasic acid having 2% dissociation.

• A. 3.4
• B. 4.5
• C. 5.1
• D. 5.8

Hint:

Use the shortcut: $\text{pH} = n - \log m$ when $[H^+] = m \times 10^{-n}$ (here $4 \times 10^{-4}$, so pH $= 4 - \log 4$).

Answer 1

The Correct Option is A

Step 1: Concept
For a weak monobasic acid, $[H^+] = C \times \alpha$, where $C$ is the molar concentration and $\alpha$ is the degree of dissociation.

Step 2: Meaning
Concentration $C = 0.02$ M and percentage dissociation is 2%, so $\alpha = \dfrac{2}{100} = 0.02$.

Step 3: Analysis
Calculate $[H^+]$: \[[H^+] = 0.02 \times 0.02 = 4 \times 10^{-4}\ \text{M}.\] \[\text{pH} = -\log(4 \times 10^{-4}) = 4 - \log 4 = 4 - 0.602 = 3.398 \approx 3.4.\]

Step 4: Conclusion
The pH is approximately 3.4. Final Answer: (A)