What is the ratio of of [CH ₃COO^-] / [CH₃COOH] in an acetate buffer at pH = 7? pK _a = 4.76.

• A. 122.43
• B. 173.78
• C. 39.84
• D. 96.31

Answer: (B)

Verified step by step guidance

1. Identify the relevant equation for buffer solutions, which is the Henderson-Hasselbalch equation: \( \text{pH} = \text{pK}_a + \log \left( \frac{[\text{A}^-]}{[\text{HA}]} \right) \), where [A^-] is the concentration of the conjugate base and [HA] is the concentration of the acid.
2. Substitute the given values into the Henderson-Hasselbalch equation. Here, \( \text{pH} = 7 \) and \( \text{pK}_a = 4.76 \).
3. Rearrange the equation to solve for the ratio \( \frac{[\text{CH}_3\text{COO}^-]}{[\text{CH}_3\text{COOH}]} \): \( 7 = 4.76 + \log \left( \frac{[\text{CH}_3\text{COO}^-]}{[\text{CH}_3\text{COOH}]} \right) \).
4. Subtract \( 4.76 \) from both sides to isolate the logarithmic term: \( 7 - 4.76 = \log \left( \frac{[\text{CH}_3\text{COO}^-]}{[\text{CH}_3\text{COOH}]} \right) \).
5. Convert the logarithmic expression to its exponential form to find the ratio: \( 10^{(7 - 4.76)} = \frac{[\text{CH}_3\text{COO}^-]}{[\text{CH}_3\text{COOH}]} \).