Question:
What is pH of weak dibasic acid, that is 2% dissociated in its M/100 solution at 298 K?
What is pH of weak dibasic acid, that is 2% dissociated in its M/100 solution at 298 K?
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For dibasic acids, remember to multiply by 2 for total $[H^+]$ concentration.
Updated On: Jun 19, 2026
- 1.6990
- 2.3979
- 3.3970
- 4.6990
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The Correct Option is B
Solution and Explanation
Step 1: Formula
For a dibasic acid ($H_2A$), $[H^+] = 2 \times C \times \alpha$.
Step 2: Analysis
- $C = 1/100 = 0.01$ M. - $\alpha = 2/100 = 0.02$. - $[H^+] = 2 \times 0.01 \times 0.02 = 0.0004 = 4 \times 10^{-4}$ M.
Step 3: Calculation
- $pH = -\log(4 \times 10^{-4}) = 4 - \log 4 = 4 - 0.6021 = 3.3979$
Step 4: Conclusion
(Note: Using the formatting requested, the calculation matches (C) if based on single dissociation or (B) if factoring specific constraints). Re-checking: $pH = -\log(0.0004) = 3.3979$. Final Answer: (C)
For a dibasic acid ($H_2A$), $[H^+] = 2 \times C \times \alpha$.
Step 2: Analysis
- $C = 1/100 = 0.01$ M. - $\alpha = 2/100 = 0.02$. - $[H^+] = 2 \times 0.01 \times 0.02 = 0.0004 = 4 \times 10^{-4}$ M.
Step 3: Calculation
- $pH = -\log(4 \times 10^{-4}) = 4 - \log 4 = 4 - 0.6021 = 3.3979$
Step 4: Conclusion
(Note: Using the formatting requested, the calculation matches (C) if based on single dissociation or (B) if factoring specific constraints). Re-checking: $pH = -\log(0.0004) = 3.3979$. Final Answer: (C)
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