Question:
A solution is prepared by adding 0.5 L of 0.5 M NaOH solution to 0.5 L of x M HCOOH solution. The pH of resultant solution is 4.74. What is x in mol $L^{-1}$? (pKa (HCOOH) = 3.74)
A solution is prepared by adding 0.5 L of 0.5 M NaOH solution to 0.5 L of x M HCOOH solution. The pH of resultant solution is 4.74. What is x in mol $L^{-1}$? (pKa (HCOOH) = 3.74)
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Henderson-Hasselbalch equation is perfect for buffer pH problems.
Updated On: Jun 6, 2026
- 0.45
- 0.5
- 0.55
- 0.75
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The Correct Option is D
Solution and Explanation
Step 1: Concept
Buffer solution: $pH = pKa + \log([Salt]/[Acid])$.
Step 2: Meaning
Moles of NaOH $= 0.5 \times 0.5 = 0.25$ mol. Moles of HCOOH $= 0.5 \times x = 0.5x$ mol.
Step 3: Analysis
After reaction, salt (HCOONa) $= 0.25$ mol. Remaining Acid $= 0.5x - 0.25$ mol. $4.74 = 3.74 + \log(0.25 / (0.5x - 0.25))$. $1 = \log(0.25 / (0.5x - 0.25)) \implies 10 = 0.25 / (0.5x - 0.25)$. $5x - 2.5 = 0.25 \implies 5x = 2.75 \implies x = 0.55$. Wait, adjusting calculation: If $pH=4.74$, $pK_a=3.74$, ratio is 10. $0.25/(0.5x-0.25) = 10 \implies 5x-2.5=0.25 \implies 5x=2.75 \implies x=0.55$. Check option C vs D.
Step 4: Conclusion
Recalculating: $x = 0.55$.
Final Answer: (C)
Buffer solution: $pH = pKa + \log([Salt]/[Acid])$.
Step 2: Meaning
Moles of NaOH $= 0.5 \times 0.5 = 0.25$ mol. Moles of HCOOH $= 0.5 \times x = 0.5x$ mol.
Step 3: Analysis
After reaction, salt (HCOONa) $= 0.25$ mol. Remaining Acid $= 0.5x - 0.25$ mol. $4.74 = 3.74 + \log(0.25 / (0.5x - 0.25))$. $1 = \log(0.25 / (0.5x - 0.25)) \implies 10 = 0.25 / (0.5x - 0.25)$. $5x - 2.5 = 0.25 \implies 5x = 2.75 \implies x = 0.55$. Wait, adjusting calculation: If $pH=4.74$, $pK_a=3.74$, ratio is 10. $0.25/(0.5x-0.25) = 10 \implies 5x-2.5=0.25 \implies 5x=2.75 \implies x=0.55$. Check option C vs D.
Step 4: Conclusion
Recalculating: $x = 0.55$.
Final Answer: (C)
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