Problem 27 - Chemistry: The Central Science in SI Units, Global Edition

Question

A buffer contains 0.20 mol of acetic acid and 0.25 mol of sodium acetate in \(2.50 \mathrm{~L}\). (a) What is the pH of this buffer? (b) What is the pH of the buffer after the addition of 0.05 mol of \(\mathrm{NaOH}\) ? (c) What is the pH of the buffer after the addition of \(0.05 \mathrm{~mol}\) of \(\mathrm{HCl}\) ?

Step-by-Step Solution

Answer

The initial pH of the buffer solution is 4.85. After the addition of 0.05 mol of NaOH, the pH increases to 5.44. When 0.05 mol of HCl is added to the buffer, the pH decreases to 4.63.

Step 1: Find initial pH (a) using the Henderson-Hasselbalch equation

The Henderson-Hasselbalch equation is: \[pH = pKa + \log \frac{[A^-]}{[HA]}\] Where: - pH is the power of hydrogen ion activity in a solution - pKa is the negative logarithm of the acid ionization constant (Ka) - [A-] is the concentration of the conjugate base (acetate ion) in moles per liter - [HA] is the concentration of the weak acid (acetic acid) in moles per liter The acid dissociation constant, Ka, for acetic acid is \(1.8 \times 10^{-5}\). First, we find the pKa value for acetic acid: \(pKa = -\log(Ka) = -\log(1.8 \times 10^{-5}) = 4.74\) Now calculate the initial concentrations of acetate ions and acetic acid: Initial concentration of sodium acetate, \([A^-] = \frac{0.25\:mol}{2.50\:L} = 0.10\:M\) Initial concentration of acetic acid, \([HA] = \frac{0.20\:mol}{2.50\:L} = 0.080\:M\) Now using the Henderson-Hasselbalch equation to find the pH: \(pH = 4.74 + \log \frac{0.10}{0.080} = 4.74 + 0.11 = 4.85\) So, the initial pH of the buffer is 4.85.

Step 2: Calculate the pH after the addition of NaOH (b)

When NaOH is added to the buffer solution, it reacts with the acetic acid: \(\mathrm{NaOH + CH_3COOH \rightarrow CH_3COONa + H_2O}\) Since 0.05 mol of NaOH is added, it neutralizes 0.05 mol of acetic acid and produces 0.05 mol of sodium acetate: New concentration of sodium acetate, \([A^-] = \frac{0.25\:mol + 0.05\:mol}{2.50\:L} = 0.12\:M\) New concentration of acetic acid, \([HA] = \frac{0.20\:mol - 0.05\:mol}{2.50\:L} = 0.060\:M\) Using the Henderson-Hasselbalch equation to find the new pH: \(pH = 4.74 + \log \frac{0.12}{0.060} = 4.74 + 0.70 = 5.44\) So, the pH of the buffer after the addition of 0.05 mol of NaOH is 5.44.

Step 3: Calculate the pH after the addition of HCl (c)

When HCl is added to the buffer solution, it reacts with the acetate ions: \(\mathrm{HCl + CH_3COO^- \rightarrow CH_3COOH + Cl^-}\) Since 0.05 mol of HCl is added, it reacts with 0.05 mol of acetate ions and produces 0.05 mol of acetic acid: New concentration of sodium acetate, \([A^-] = \frac{0.25\:mol - 0.05\:mol}{2.50\:L} = 0.080\:M\) New concentration of acetic acid, \([HA] = \frac{0.20\:mol + 0.05\:mol}{2.50\:L} = 0.10\:M\) Using the Henderson-Hasselbalch equation to find the new pH: \(pH = 4.74 + \log \frac{0.080}{0.10} = 4.74 - 0.11 = 4.63\) So, the pH of the buffer after the addition of 0.05 mol of HCl is 4.63.

Key Concepts

Henderson-Hasselbalch equation

The Henderson-Hasselbalch equation is a crucial tool in acid-base chemistry, particularly for calculating the pH of buffer solutions. It predicts the pH of a solution containing a weak acid and its conjugate base. This can be represented as: \[pH = pKa + \log \frac{[A^-]}{[HA]}\]- **pH** is the measure of acidity or basicity of a solution. It indicates how acidic or alkaline the solution is.- **pKa** is the logarithmic measure of the acid dissociation constant (Ka) of a weak acid. It signifies how easily the acid donates a proton to the base. - **[A^-]** denotes the concentration of the conjugate base of the acid in moles per liter.- **[HA]** represents the concentration of the weak acid in moles per liter.This equation simplifies the process of calculating pH by relating the concentrations of an acid and its conjugate base to its dissociation constant. It serves as a bridge between the concentrations and pH, helping us understand how changes in the amounts of [A^-] or [HA] alter the pH.

Acid-base chemistry

Understanding acid-base chemistry is fundamental in determining how substances interact in solutions, particularly in buffer systems. A buffer consists of a mixture of a weak acid and its corresponding conjugate base, or vice versa. This combination helps maintain a stable pH in a solution, even when an acid or base is added. • **Buffers** act by neutralizing added acids or bases, preventing drastic pH changes. They are essential in biological systems, industrial processes, and chemical research where pH stability is crucial. • **Weak acids**, like acetic acid ( CH₃COOH ), partially dissociate in a solution, creating equilibrium between the acid and its ionized form. • **Conjugate bases** are formed when the acid donates a proton. For acetic acid, it forms acetate ions ( CH₃COO^- ). In the context of the exercise mentioned, the buffer system comprises acetic acid and sodium acetate, demonstrating how buffers help sustain pH stability even with additions of either NaOH, a strong base, or HCl, a strong acid.

pH calculation

pH calculation is vital to understand how the acidity of a solution is determined and adjusted. The pH is calculated using the Henderson-Hasselbalch equation, especially in buffer systems, allowing us to predict and control the pH in various chemical reactions and solutions. When calculating pH:

• Utilize the molarity (concentration) of the acid and its conjugate base.
• Apply the pKa value, derived from the Ka of the acid, to find stability areas of the pH scale.

The initial pH of a solution gives a snapshot of its current state, while changes in pH after adding acids or bases indicate the buffer action. For instance, in the exercise, the initial pH was found to be 4.85, providing a baseline from which the effect of adding either 0.05 mol of NaOH or HCl could be measured. Each addition alters the concentrations of acetic acid and acetate, showing the buffer’s response and its capacity to maintain near-constant pH.