In order to prepare a buffer solution of pH 5.74, sodium acetate is added to acetic acid. If the con — Ionic Equilibrium Chemistry Question
Question
In order to prepare a buffer solution of pH 5.74, sodium acetate is added to acetic acid. If the concentration of acetic acid in the buffer is 1.0 M, the concentration of sodium acetate in the buffer is ..................... M. (Round off to the Nearest Integer). [Given: pKa (acetic acid) = 4.74]
💡 Solution & Explanation
**Step 1: Apply the Henderson-Hasselbalch equation** For a buffer solution: $$\text{pH} = \text{pK}_a + \log\frac{[\text{A}^-]}{[\text{HA}]}$$ Where: - pH = 5.74 - pKa = 4.74 - [A⁻] = concentration of sodium acetate (unknown) - [HA] = concentration of acetic acid = 1.0 M **Step 2: Substitute the known values** $$5.74 = 4.74 + \log\frac{[\text{CH}_3\text{COO}^-]}{1.0}$$ **Step 3: Solve for the logarithm term** $$5.74 - 4.74 = \log\frac{[\text{CH}_3\text{COO}^-]}{1.0}$$ $$1.0 = \log\frac{[\text{CH}_3\text{COO}^-]}{1.0}$$ **Step 4: Convert from logarithmic to exponential form** $$10^{1.0} = \frac{[\text{CH}_3\text{COO}^-]}{1.0}$$ **Step 5: Calculate the concentration** $$[\text{CH}_3\text{COO}^-] = 10^{1.0} = 10.0 \text{ M}$$ Therefore, the answer is **10.00**.