Problem 32 - Advanced Problems in Physical Chemistry for Competitive Examinations

Question

Two buffers, \(X\) and \(Y\) of \(p H 4.0\) and \(6.0\) respectively are prepared from acid HA and the salt NaA. Both the buffers are \(0.50 \mathrm{M}\) in HA. What would be the pH of the solution obtained by mixing equal volumes of the two buffers? \(K_{\mathrm{a}}\) of \(\mathrm{HA}=1.0 \times 10^{-5} \cdot(\log 5.05=0.7)\) (a) \(5.0\) (b) \(4.3\) (c) \(4.7\) (d) \(5.7\)

Step-by-Step Solution

Answer

The pH of the solution obtained by mixing equal volumes of the two buffers is (c) 4.7.

Step 1: Calculate the concentration of A- in each buffer

Use the Henderson-Hasselbalch equation to find the concentration of A- in buffer X and buffer Y. The equation is \[ pH = pK_a + \log\frac{[A^-]}{[HA]} \] For buffer X: \[4.0 = \log5 + \log\frac{[A^-_X]}{0.50} \] Solve for [A^-_X]. For buffer Y: \[6.0 = \log5 + \log\frac{[A^-_Y]}{0.50} \] Solve for [A^-_Y].

Step 2: Calculate the new concentrations after mixing

When equal volumes of buffer X and buffer Y are mixed, the concentrations of HA and A- are halved because the volume doubles. \[ [HA]_{new} = \frac{[HA]}{2} \] \[ [A^-_{X_{new}}] = \frac{[A^-_X]}{2} \] \[ [A^-_{Y_{new}}] = \frac{[A^-_Y]}{2} \] Also, calculate the total concentration of A- in the mixture by adding \[ [A^-_{X_{new}}] \] and \[ [A^-_{Y_{new}}] \].

Step 3: Calculate the pH of the resulting mixture

Use the Henderson-Hasselbalch equation again with the new concentrations of HA and A- to find the pH of the mixture. \[ pH = pK_a + \log\frac{[A^-]_{combined}}{[HA]_{new}} \] Substitute \[ [A^-]_{combined} \] and \[ [HA]_{new} \] into the equation and solve for the pH.

Key Concepts

pH Calculation

Understanding how to calculate the pH of a solution is an essential skill in chemistry, especially when studying acid-base reactions and buffer systems. The pH is a measure of the acidity or basicity of an aqueous solution. It is defined as the negative logarithm to base 10 of the hydrogen ion concentration, or \[ pH = -\text{log} [H^+] \].
A pH value of less than 7 indicates an acidic solution, while a value greater than 7 indicates a basic solution. A pH of 7 is neutral, like pure water. The pH can be affected by adding acids, bases, or by dilution. In the case of buffer solutions, such as those used in the exercise, the pH remains relatively stable despite the addition of small amounts of acids or bases. The Henderson-Hasselbalch equation reflects this relationship and is used to calculate the pH of buffer solutions, which we will explore more in the next section.

Buffer Solution

A buffer solution is a special type of solution that resists changes in pH when small quantities of an acid or a base are added. It is composed of a weak acid (HA) and its conjugate base (A-), or a weak base and its conjugate acid. The key to its buffering action lies in the equilibrium that exists between these two species.
The Henderson-Hasselbalch equation provides a way to relate pH, pKa (the negative logarithm of the acid dissociation constant, Ka), and the ratio of the concentrations of the conjugate base and the weak acid: \[ \text{pH} = \text{pKa} + \text{log} \frac{[A^-]}{[HA]} \].
This equation essentially describes how the pH of the buffer changes with the concentration ratio of the conjugate acid-base pair. In the textbook exercise, buffer solutions X and Y are created from a weak acid HA and its salt NaA. By providing the pKa of HA and the pH of the buffers, one can determine the ratio of the conjugate base to the acid and hence the ability of the buffer to maintain its pH.

Acid Dissociation Constant (Ka)

The acid dissociation constant, Ka, is a quantitative measure of the strength of an acid in solution. It is the equilibrium constant for the dissociation of the acid into its ions in water. For a general weak acid, HA, that dissociates as follows: \[ HA \rightleftharpoons H^+ + A^- \],
the expression for the Ka is the product of the concentrations of the hydrogen ions ([H+]) and the conjugate base ([A-]) divided by the concentration of the undissociated acid ([HA]), or: \[ K_a = \frac{[H^+][A^-]}{[HA]} \].
The larger the Ka, the stronger the acid because a greater proportion of the acid is dissociated into its ions at equilibrium. The pKa, which is simply \(-\text{log} K_a\), is often used because it is directly related to the pH scale. In solving our problem about buffer solutions, understanding the concept of Ka allows us to appreciate the significance of the equilibrium between the weak acid and its conjugate base and how it contributes to the buffering action of the solution.