91Ó°ÊÓ

The Henderson-Hasselbalch equation is an essential formula in chemistry, specifically for understanding buffer solutions. It connects the pH of a solution with the pKa (negative logarithm of the acid dissociation constant) and the concentrations of the acid and its conjugate base. The equation is:
\[ \text{pH} = \text{p}K_a + \text{log} \frac{[\text{A}^-]}{[\text{HA}]} \]
Where:

• \text{pH\text} \text{ is the measure of the acidity of the solution.}


• \text{p}K_a\text is the logarithmic form of the acid dissociation constant (\text{K_a}).


• [\text{A}^-]\text is the concentration of the conjugate base.


• [\text{HA}]\text is the concentration of the weak acid.

This equation is particularly useful in calculating the pH of buffer solutions because it directly relates the pH to the ratio of the concentrations of the components of the buffer. It also helps in understanding how the pH changes with the addition of more acid or base.

The acid dissociation constant, denoted as \( K_a \), is a quantitative measure of the strength of an acid in solution. It represents the equilibrium constant for the dissociation of a weak acid into its ion and proton. The formula for \( K_a \) is:
\[ K_a = \frac{[\text{H}^+][\text{A}^-]}{[\text{HA}]} \]
Where:

• [\text{H}^+] is the concentration of hydrogen ions (protons).


• [\text{A}^-] is the concentration of the conjugate base.


• [\text{HA}] is the concentration of the weak acid.

The pKa is simply the negative logarithm of the \( K_a \). It's a more convenient way to express the strength of an acid. So
\( \text{p}K_a = -\text{log} \text{K}_a \).
For the exercise provided, with \( K_a = 1.3 \times 10^{-5} \), we find the pKa using the formula:
\( \text{p}K_a = -\text{log} (1.3 \times 10^{-5}) \ \text{ which is approximately } 4.89 \). This value is important to plug into the Henderson-Hasselbalch equation.

Calculating the pH of a buffer solution involves using the Henderson-Hasselbalch equation. Here’s a step-by-step overview:

Given the pH and the \( K_a \), first determine the pKa value using the relation \ \( \text{p}K_a = -\text{log K}_a \).

With the pKa value and the known pH, use the Henderson-Hasselbalch equation \ \( \text{pH} = \text{p}K_a + \text{log} \frac{[\text{A}^-]}{[\text{HA}]} \).

Rearrange the equation to solve for the concentration ratio \ \( \frac{[\text{A}^-]}{[\text{HA}]} \).

Substitute the known values into the equation. For the given exercise, the pH is 5.44 and the pKa is approximately 4.89.

Solve for the concentration ratio:\ \ \( 5.44 = 4.89 + \text{log} \frac{[\text{Pr}^{-}] }{[\text{HPr}] } \).

Simplify the equation to find:\ \( 0.55 = \text{log} \frac{[\text{Pr}^{-}] }{[\text{HPr}] } \).

Finally, we find the ratio by calculating the antilog (base 10): \ \( \frac{[\text{Pr}^{-}] }{[\text{HPr}] } = 10^{0.55} \) \ \approx 3.55.

Hence, the concentration ratio of \ \( \frac{[\text{Pr}^{-}] }{[\text{HPr}] } \) in the buffer solution is approximately 3.55.