Question

Calculate the pH of 0.01 M sulphuric acid.

• A. 1.699
• B. 2.00
• C. 0.699
• D. 3.398

Hint:

Never forget the basicity! For $H_2SO_4$ (dibasic), multiply molarity by 2 before taking the log to get $[H^+]$.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
Sulphuric acid (\( H_2SO_4 \)) is a strong dibasic acid. It dissociates completely in water.
Because it is dibasic, one mole of \( H_2SO_4 \) produces two moles of \( H^+ \) ions.
Key Formula or Approach:
1. \( [H^+] = 2 \times [Acid] \) for strong dibasic acids.
2. \( pH = -\log_{10}[H^+] \).
Step 2: Detailed Explanation:
Given molarity of \( H_2SO_4 = 0.01 \) M.
Since it fully dissociates:
\[ [H^+] = 2 \times 0.01 = 0.02 \text{ M} = 2 \times 10^{-2} \text{ M} \] Now, calculate the \( pH \):
\[ pH = -\log(2 \times 10^{-2}) = 2 - \log 2 \] Using the value \( \log 2 \approx 0.301 \):
\[ pH = 2 - 0.301 = 1.699 \] Step 3: Final Answer:
The pH is 1.699.
This matches Option (A).