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 5.05.

Step 1: Determine the Moles of Acetic Acid

Calculate the moles of acetic acid (CH₃COOH) neutralized by NaOH using the formula: \( \text{moles} = \text{molarity} \times \text{volume in L} \). \[ \text{Moles of acetic acid} = 0.1 \text{ M} \times 6 \times 10^{-3} \text{ L} = 0.0006 \text{ moles} \]

Step 2: Determine the Moles of Sodium Acetate

Calculate the moles of sodium acetate (CH₃COONa) that reacted with HCl using the same formula: \( \text{moles} = \text{molarity} \times \text{volume in L} \). \[ \text{Moles of sodium acetate} = 0.1 \text{ M} \times 12 \times 10^{-3} \text{ L} = 0.0012 \text{ moles} \]

Step 3: Use Henderson-Hasselbalch Equation

The pH of a buffer solution can be calculated using the Henderson-Hasselbalch equation: \[ \text{pH} = \text{pK}_a + \log \left( \frac{[\text{A}^-]}{[\text{HA}]} \right) \] where \([\text{A}^-]\) is the concentration of the sodium acetate, and \([\text{HA}]\) is the concentration of the acetic acid. Use the moles from previous steps to find the concentrations: \( [\text{A}^-] = \frac{0.0012}{0.018}, [\text{HA}] = \frac{0.0006}{0.018} \).

Step 4: Calculate the pH

Substitute the values into the Henderson-Hasselbalch equation: \[ \text{pH} = 4.75 + \log \left( \frac{0.0012/0.018}{0.0006/0.018} \right) \] Simplifying gives: \[ \text{pH} = 4.75 + \log(2) \] As \( \log(2) \approx 0.3010 \): \[ \text{pH} = 4.75 + 0.3010 = 5.051 \] The pH, rounded to two decimal places, is approximately 5.05.

Key Concepts

Buffer Solution

A buffer solution is a unique type of solution that helps maintain a consistent pH, even when small quantities of an acid or a base are added to it. Buffer solutions are composed of a weak acid and its conjugate base, or a weak base and its conjugate acid. For example, in our exercise, we have acetic acid (\(\text{CH}_3\text{COOH}\)) and sodium acetate (\(\text{CH}_3\text{COONa}\)), which form an acetate buffer system.
This type of solution is essential because it resists drastic changes in pH. Here’s why:

• When an acid (\(\text{H}^+\)) is added to the buffer, the conjugate base (\(\text{A}^-\)) present in the solution neutralizes it.
• When a base (\(\text{OH}^-\)) is added, the weak acid (\(\text{HA}\)) in the buffer neutralizes the base.

This buffering action keeps the pH within a narrow range, crucial for processes that require a stable pH, such as enzymatic reactions in biological systems.

pKa of Acetic Acid

The pKa value of an acid is a vital parameter that tells us about the acid's strength. Specifically, it measures the tendency of a compound to donate protons (\(\text{H}^+\)). Essentially, the lower the pKa value, the stronger the acid, because a low pKa indicates a high ability to donate protons.
The pKa of acetic acid is 4.75, which means acetic acid is a weak acid. It partially dissociates in a solution to form acetate ions and hydrogen ions. In the context of buffer calculations, the pKa is crucial because it helps determine the pH of the solution when used in the Henderson-Hasselbalch equation. The closer the pH value of the solution is to the pKa, the more effective the buffer is. Acetic acid and sodium acetate form an effective buffer system around this pKa value, thus allowing stability in the pH of the solution.

Neutralization Reaction

A neutralization reaction occurs when an acid and a base react to form water and a salt, effectively canceling each other's effect. In the given exercise:

• Acetic acid (\(\text{CH}_3\text{COOH}\)) was neutralized by sodium hydroxide (\(\text{NaOH}\)), forming water and sodium acetate (\(\text{CH}_3\text{COONa}\)).
• Sodium acetate reacted with hydrochloric acid (\(\text{HCl}\)), yielding acetic acid and sodium chloride (\(\text{NaCl}\)).

These neutralization reactions play a pivotal role in determining the concentrations of the acid and base components needed for subsequent pH calculation using the Henderson-Hasselbalch equation.

pH Calculation

Calculating pH, especially in buffer solutions, can be accomplished using the Henderson-Hasselbalch equation, which is pivotal in buffering capacity analysis. The equation is expressed as:\[\text{pH} = \text{pK}_a + \log \left( \frac{[\text{A}^-]}{[\text{HA}]} \right)\]
This equation takes into account the acid dissociation constant (pKa) and the ratio of the concentration of the conjugate base \(([\text{A}^-])\) to that of the acid \(([\text{HA}])\) present in the solution. In our example:

• The moles of sodium acetate (conjugate base) calculated is 0.0012 moles.
• Acetic acid (weak acid) amounted to 0.0006 moles.

The solution’s total volume, 18 ml, allows us to compute both concentrations. Substituting these into the equation provides the calculated pH, affirming the buffer's efficacy. The computation in our exercise resulted in a pH of approximately 5.05, closely maintaining the expected buffer range based on the initial pKa of acetic acid.