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: Understanding the Concept:
A monobasic acid is one that yields one hydrogen ion (\( H^+ \)) per molecule upon dissociation.
For weak acids, the concentration of \( H^+ \) ions depends on the total concentration (\( C \)) and the degree of dissociation (\( \alpha \)).
Key Formula or Approach:
1. \( [H^+] = C \cdot \alpha \).
2. \( pH = -\log_{10}[H^+] \).
Step 2: Detailed Explanation:
Given:
Concentration \( C = 0.02 \) M.
Degree of dissociation \( \alpha = 2% = 0.02 \).
First, we find the hydrogen ion concentration:
\[ [H^+] = C \cdot \alpha = 0.02 \times 0.02 = 0.0004 \text{ M} \] \[ [H^+] = 4 \times 10^{-4} \text{ M} \] Now, calculate the \( pH \):
\[ pH = -\log(4 \times 10^{-4}) = -(\log 4 + \log 10^{-4}) \] \[ pH = -(0.602 - 4) = 4 - 0.602 = 3.398 \] Rounding to one decimal place, we get approximately 3.4.
Step 3: Final Answer:
The pH is 3.4.
This matches Option (A).