26. Calculate the pH of a buffer solution that is 0.050 M in benzoic acid \( \left(\mathrm{HC}_{7} \mathrm{H}_{5} \mathrm{O}_{2}, \mathrm{~K}_{\mathrm{a}}=6.5 \times 10^{-5}\right) \) and 0.150 M in sodium benzoate \( \left(\mathrm{NaC}_{7} \mathrm{H}_{5} \mathrm{O}_{2}\right) \). \[ \mathrm{HC}_{7} \mathrm{H}_{5} \mathrm{O}_{2}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\mathrm{I}) \rightleftharpoons \mathrm{NaC}_{7} \mathrm{H}_{5} \mathrm{O}_{2}(\mathrm{aq})+\mathrm{OH}-(\mathrm{aq}) \] A. 4.19 B. 4.66 C. 3.71 D. 9.81

To calculate the pH of the buffer solution containing benzoic acid and sodium benzoate, we can utilize the Henderson-Hasselbalch equation: \[ \text{pH} = \text{p}K_a + \log\left(\frac{[\text{A}^-]}{[\text{HA}]}\right) \] Here, \( [\text{A}^-] \) is the concentration of the base (sodium benzoate) and \( [\text{HA}] \) is the concentration of the acid (benzoic acid). First, we need to find \( pK_a \): \[ pK_a = -\log(K_a) = -\log(6.5 \times 10^{-5}) \approx 4.187 \] Next, plug in the concentrations of sodium benzoate and benzoic acid into the Henderson-Hasselbalch equation: \[ \text{pH} = 4.187 + \log\left(\frac{0.150}{0.050}\right) = 4.187 + \log(3) \approx 4.187 + 0.477 \approx 4.66 \] Thus, the pH of the buffer solution is approximately **4.66** (Option B). ### Fun Fact: Buffers are nature's way of keeping things balanced! Imagine your body—like a tightrope walker trying to maintain equilibrium. Buffers help your blood maintain a stable pH, ensuring you stay healthy and functioning well. ### Real-World Application: Have you ever added vinegar to a baking soda mix? That fizzing reaction is a bit like a buffer at work! The acidity of acetic acid in vinegar interacts with the basic sodium bicarbonate, creating a fun, bubbly explosion while helping keep the pH more stable during baking. Buffers are everywhere, from your kitchen to your biology, and they keep things in check!