Problem 164 - A Complete Resource Book in Chemistry for JEE Main

Question

\(18 \mathrm{ml}\) of mixture of acetic acid and sodium acetate required \(6 \mathrm{ml}\) of \(0.1 \mathrm{M} \mathrm{NaOH}\) for neutralization of the acid and \(12 \mathrm{ml}\) of \(0.1 \mathrm{M} \mathrm{HCl}\) for reaction with salt separately. If \(\mathrm{pK}_{\mathrm{a}}\) of the acid is \(4.75\), what is the \(\mathrm{pH}\) of the mixture? (a) \(5.05\) (b) \(4.75\) (c) \(4.5\) (d) \(4.6\)

Step-by-Step Solution

Answer

The pH of the mixture is approximately 5.05.

Step 1: Determine Moles of NaOH Required

To find the moles of acetic acid (CH\(_3\)COOH) present, use the volume and concentration of NaOH. The moles of NaOH used for neutralization is calculated as:\[\text{Moles of NaOH} = 6 \, ext{ml} \times 0.1 \, \text{M} = 0.6 \, \text{mmol}\]This amount of NaOH is used to neutralize the acetate ion from acetic acid.

Step 2: Calculate Moles of HCl Required

The HCl reacts with sodium acetate (CH\(_3\)COONa) in the solution. We find the moles of HCl as:\[\text{Moles of HCl} = 12 \, ext{ml} \times 0.1 \, \text{M} = 1.2 \, \text{mmol}\]This amount of HCl reacts with the sodium acetate to release acetic acid.

Step 3: Apply Henderson-Hasselbalch Equation

Using the Henderson-Hasselbalch equation to find the pH:\[\mathrm{pH} = ext{pK}_a + \log\left(\frac{\text{[A]^−}}{\text{[HA]}}\right)\]Where \([A^-]\) is acetate and \([HA]\) is acetic acid. From previous calculations, \([A^-] = 1.2\,\text{mmol}\) and \([HA] = 0.6\,\text{mmol}\), so substitute these into the equation:\[\mathrm{pH} = 4.75 + \log\left(\frac{1.2}{0.6}\right) = 4.75 + \log(2)\]

Step 4: Calculate Logarithmic Value

Calculate \(\log(2)\) which is approximately \(0.3010\), so:\[\mathrm{pH} = 4.75 + 0.3010 = 5.051\]Round this result to match the closest answer option.

Key Concepts

Henderson-Hasselbalch Equation

The Henderson-Hasselbalch equation is a vital tool in chemistry for calculating the pH of a buffer solution. In simple terms, this equation relates the pH, pKa (the acid dissociation constant), and the ratio of the concentrations of the conjugate base and the weak acid of a buffer solution. The equation is expressed as follows:\[\mathrm{pH} = \mathrm{pK}_a + \log\left(\frac{[A^-]}{[HA]}\right)\]Here,

• \(\mathrm{pH}\) is what we want to find.
• \(\mathrm{pK}_a\) represents the strength of the acid, with smaller values indicating stronger acids.
• \([A^-]\) is the concentration of the conjugate base, and \([HA]\) is the concentration of the undissociated weak acid.

The equation derives from the equilibrium expression for weak acids and bases, and it requires the solution to be at equilibrium for accurate calculations. It’s perfect for buffer solutions where both the weak acid and its conjugate base are present.

Calculating pH

Understanding how to calculate pH is essential in chemistry, particularly when dealing with buffer solutions or weak acids and bases. pH is a measure of the acidity or basicity of a solution. When you want to calculate the pH of a buffer solution, you often use the Henderson-Hasselbalch equation.In the given problem, the solution involves determining the concentrations of acetic acid and its conjugate base (acetate ions) first:

• From the neutralization process with NaOH, you find the moles of acetic acid, which got neutralized to form acetate ions.
• From the acidic reaction with HCl, additional moles of acetate are accounted for by releasing acetic acid back into the mixture.

Once the concentrations are determined, the values are substituted back into the Henderson-Hasselbalch equation to compute the \(\mathrm{pH}\). This pH value is pivotal as it describes how the mixture responds to additional acid or base. Note that accurate pH calculation reflects how well the buffer can resist pH changes, which is crucial for maintaining consistent chemical environments in many biological and chemical processes.

Neutralization Reaction

A neutralization reaction involves an acid and a base reacting to form water and a salt. This type of reaction is essential for determining the composition and the concentration of each component in buffer solutions. In the exercise provided, we see two different reactions: 1. **Acid Neutralization**: The addition of NaOH to the acetic acid results in a typical neutralization reaction where acetic acid gives up protons to NaOH, converting to acetate ions. 2. **Conjugate Base Reaction**: HCl was used to overpower the acetate ions back into acetic acid. This restores the equilibrium of the buffer, which is critical for calculating the actual pH. These reactions are fundamental in understanding how buffers work by providing or absorbing hydrogen ions to maintain essential pH levels. An understanding of neutralization reactions helps in the analysis and calculation of the buffer solutions, which helps anticipate the solution's behavior when additional materials are introduced.

Buffer Solutions

Buffer solutions play a crucial role in stabilizing the pH in various chemical and biological processes. These solutions consist of a weak acid and its corresponding conjugate base (or a weak base and its conjugate acid) present in equilibrium. The primary functionality of a buffer is its ability to resist changes in pH by neutralizing small quantities of added acids or bases. For instance, in the exercise, the solution was a mixture of acetic acid (a weak acid) and sodium acetate (the salt providing the conjugate base). This combination creates an equilibrium where:

• If acid is added (in the form of HCl), the acetate ions react to form acetic acid, thereby minimizing pH change.
• If a base like NaOH is added, it reacts with the acetic acid to form more acetate, stabilizing the pH.

This remarkable property of buffers is invaluable, especially in biological systems where even slight pH changes can be harmful. Understanding how buffer solutions work, allows chemists to predict and control the effects of pH in various environments, ensuring stability and desired chemical behaviors.