What are the \(\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]\) and the \(\mathrm{pH}\) of a propanoic acidpropanoate buffer that consists of \(0.35 M \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COONa}\) and \(0.15 M \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COOH}\left(K_{\mathrm{a}}\right.\) of propanoic acid \(\left.=1.3 \times 10^{-5}\right) ?\)

The Henderson-Hasselbalch equation is key for understanding buffer solutions. Buffers are solutions that resist changes in pH when small amounts of acid or base are added. The equation is \( \text{pH} = \text{pKa} + \text{log} \frac{[\text{A}^-]}{[\text{HA}]} \). Here, \( \text{pH} \) is the measure of acidity, \(\text{pKa}\) is the acid dissociation constant, \([\text{A}^-]\) is the concentration of the conjugate base, and \([\text{HA}]\) is the concentration of the acid. This equation relates the pH of a solution to the pKa and the ratio of the concentrations of the acid and its conjugate base. It helps you predict how the pH will change when acid or base is added to the buffer solution.

Understanding the relationship between \( \text{pKa} \) and \( \text{Ka} \) is essential for buffer calculations. The \( \text{Ka} \) value, or acid dissociation constant, measures how completely an acid dissociates in water. A small \( \text{Ka} \) value indicates a weak acid that doesn't release many ions in water. The \( \text{pKa} \) is the negative logarithm of \( \text{Ka} \): \( \text{pKa} = -\text{log}(\text{Ka}) \). For example, if \( \text{Ka} \) for propanoic acid is \( 1.3 \times 10^{-5} \), then \(\text{pKa} = -\text{log}(1.3 \times 10^{-5}) \), which gives \(\text{pKa} \thickapprox 4.89 \). The \(\text{pKa} \) value makes it easier to understand the acidity of a solution, especially when dealing with logarithmic scales in pH calculations.

Calculating the pH is a crucial step in buffer solution problems. Using the Henderson-Hasselbalch equation, you can find the pH if you know the \(\text{pKa}\) and the concentration ratio of the conjugate base (A-) and the acid (HA). For example, with a \(\text{pKa}\) of 4.89 and a concentration ratio of 2.33 (from \(\frac{0.35}{0.15}\)), the pH is: \( \text{pH} = 4.89 + \text{log} (2.33) \). Using a calculator, this gives \( \text{pH} \thickapprox 4.89 + 0.368 = 5.26 \). This method shows how to connect the dots between \(\text{pKa}\), concentration ratios, and pH.

Determining the concentration ratio of the acid and conjugate base ( \( \frac{[\text{A}^-]}{[\text{HA}]} \) ) is fundamental in buffer calculations. This ratio is part of the Henderson-Hasselbalch equation and directly influences the pH. For example, given \( [\text{CH}_3\text{CH}_2\text{COO}^-] = 0.35 \text{ M} \) and \( [\text{CH}_3\text{CH}_2\text{COOH}] = 0.15 \text{ M} \), the ratio is \( \frac{0.35}{0.15} \thickapprox 2.33 \). Knowing this ratio helps in substituting the correct values into the Henderson-Hasselbalch equation and achieving an accurate pH calculation.