Acid–base Equilibria

Question

A buffer is made by mixing 0.10 M acetic acid and 0.20 M sodium acetate. (Ka for acetic acid = 1.8×10⁻⁵). The pH of the buffer is (25°C):

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1. **Buffer Type**: Acidic buffer (Weak acid + Salt with strong base).
2. **Henderson–Hasselbalch Equation**: $pH = pK_a + \log \frac{[Salt]}{[Acid]}$.
3. **$pK_a$ Calculation**:
   - $pK_a = -\log(1.8 	imes 10^{-5})$
   - $pK_a = 5 - \log 1.8 = 5 - 0.255 = 4.745$.
4. **Substituting Values**:
   - $[Salt] = 0.20$ M, $[Acid] = 0.10$ M.
   - $pH = 4.745 + \log(0.20 / 0.10)$
   - $pH = 4.745 + \log 2$
   - $pH = 4.745 + 0.301 = 5.046$.
5. **Result**: The buffer $pH$ is approximately **$5.04$**.

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Acid–base Equilibria

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