Question:
What is the molar ratio [salt]/[acid] required to prepare an acetate buffer of pH 5.0 ($pKa=4.76$)?
What is the molar ratio [salt]/[acid] required to prepare an acetate buffer of pH 5.0 ($pKa=4.76$)?
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If $pH > pKa$, the ratio of [salt]/[acid] must be greater than 1.
Updated On: May 28, 2026
- 1.74
- 2.74
- 0.74
- 1.85
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The Correct Option is A
Solution and Explanation
Step 1: Concept
The pH of a buffer solution can be calculated using the Henderson-Hasselbalch equation: $pH = pKa + \log\frac{[Salt]}{[Acid]}$.
Step 2: Meaning
We need to find the ratio $\frac{[Salt]}{[Acid]}$ when the desired $pH$ is 5.0 and the $pKa$ is 4.76.
Step 3: Analysis
Substitute the values into the equation: $5.0 = 4.76 + \log\frac{[Salt]}{[Acid]}$ $5.0 - 4.76 = \log\frac{[Salt]}{[Acid]}$ $0.24 = \log\frac{[Salt]}{[Acid]}$ Taking antilog on both sides: $\frac{[Salt]}{[Acid]} = 10^{0.24} \approx 1.7378$.
Step 4: Conclusion
The calculated molar ratio is approximately 1.74.
Final Answer: (A)
The pH of a buffer solution can be calculated using the Henderson-Hasselbalch equation: $pH = pKa + \log\frac{[Salt]}{[Acid]}$.
Step 2: Meaning
We need to find the ratio $\frac{[Salt]}{[Acid]}$ when the desired $pH$ is 5.0 and the $pKa$ is 4.76.
Step 3: Analysis
Substitute the values into the equation: $5.0 = 4.76 + \log\frac{[Salt]}{[Acid]}$ $5.0 - 4.76 = \log\frac{[Salt]}{[Acid]}$ $0.24 = \log\frac{[Salt]}{[Acid]}$ Taking antilog on both sides: $\frac{[Salt]}{[Acid]} = 10^{0.24} \approx 1.7378$.
Step 4: Conclusion
The calculated molar ratio is approximately 1.74.
Final Answer: (A)
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