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: Concept
Sulphuric acid ($H_2SO_4$) is a strong dibasic acid that fully dissociates: $H_2SO_4 \to 2H^+ + SO_4^{2-}$.

Step 2: Meaning
Because it releases two protons per molecule, $[H^+] = 2 \times [H_2SO_4]$.

Step 3: Analysis
Given $[H_2SO_4] = 0.01\ \text{M} = 10^{-2}\ \text{M}$: \[[H^+] = 2 \times 10^{-2}\ \text{M}.\] \[\text{pH} = -\log(2 \times 10^{-2}) = 2 - \log 2 = 2 - 0.301 = 1.699.\]

Step 4: Conclusion
The pH of 0.01 M $H_2SO_4$ is 1.699. Final Answer: (A)