What are the \(\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]\) and the \(\mathrm{pH}\) of a benzoic acid-benzoate buffer that consists of \(0.33 \mathrm{MC}_{6} \mathrm{H}_{5} \mathrm{COOH}\) and \(0.28 \mathrm{MC}_{6} \mathrm{H}_{5} \mathrm{COONa}\) \(\left(K_{\mathrm{a}}\right.\) of benzoic acid \(\left.=6.3 \times 10^{-5}\right) ?\)

The Henderson-Hasselbalch equation is commonly used to estimate the pH of a buffer solution. This equation relates the pH, pKa (the negative logarithm of the acid dissociation constant, Ka), and the ratio of the concentrations of the conjugate base (\text{A}^-) to the acid (\text{HA}). It is expressed as:
\[\text{pH} = \text{p}K_a + \text{log} \frac{[\text{A}^-]}{[\text{HA}]}\]
Here, \text{p}K_a represents the strength of the acid. A lower \text{p}K_a value indicates a stronger acid, as it dissociates more in water. The ratio \(\frac{[\text{A}^-]}{[\text{HA}]}\) helps determine whether the solution is more acidic or basic. If the ratio is greater than 1, the solution is more basic; if less than 1, the solution is more acidic. To use this equation properly, focus on:

• \text{Knowing the concentration of both the acid and the conjugate base.
• \text{Accurately calculating pKa from Ka.

The relationship between pKa and Ka is crucial for buffer solution calculations. The \text{K}_a (acid dissociation constant) quantifies the strength of an acid in water and is calculated from the concentrations of the products and reactants at equilibrium. For benzoic acid, \text{K}_a is given as 6.3 x 10^-5.
To find pKa, use the formula:
\[\text{p}K_a = -\text{log}(K_a) \] So, \[\text{p}K_a = -\text{log}(6.3 \times 10^{-5}) = 4.2\]
The lower the pKa value, the stronger the acid (and thus the higher the Ka value). In essence, pKa makes it easier to express very small numbers (like Ka) that are common in acid-base chemistry.

Calculating the pH of a buffer solution involves several steps. Let's apply these steps to a benzoic acid-benzoate buffer consisting of 0.33 M benzoic acid and 0.28 M sodium benzoate.
First, we use the Henderson-Hasselbalch equation:
\[\text{pH} = \text{p}K_a + \text{log} \frac{[\text{A}^-]}{[\text{HA}]}\]
Given pKa = 4.2, \text{[A}^{-}\text{]} = 0.28 M, and \text{[HA]} = 0.33 M, we substitute these values into the equation:
\[\text{pH} = 4.2 + \text{log} \frac{0.28}{0.33}\]
Next, calculate the ratio: \[\frac{0.28}{0.33} = 0.848\] Then find the logarithm of the ratio: \[\text{log } 0.848 = -0.071\]
Finally, solve for pH: \[\text{pH} = 4.2 - 0.071 = 4.129\]
To find \text{[H}_3\text{O}^+], use the relationship:
\[ \text{[H}_3\text{O}^+] = 10^{-\text{pH}}\] So, \( [\text{H}_3\text{O}^+] = 10^{-4.129} = 7.47 \times 10^{-5} M\)
Understanding each step ensures accurate buffer pH calculations and better handling of buffer systems in practical applications.