Problem 170 - A Complete Resource Book in Chemistry for JEE Main

Question

A buffer solution is prepared by mixing \(10 \mathrm{~mL}\) of \(1.0\) M acetic and \(20 \mathrm{~mL}\) of \(0.5 \mathrm{M}\) sodium acetate and then diluted to \(100 \mathrm{~mL}\) with distilled water. If the pKa of \(\mathrm{CH}_{3} \mathrm{COOH}\) is \(4.76\), what is the \(\mathrm{pH}\) of the buffer solution prepared? (a) \(5.21\) (b) \(4.76\) (c) \(4.34\) (d) \(5.35\)

Step-by-Step Solution

Answer

The pH of the buffer solution is (b) 4.76.

Step 1: Calculate Moles of Acetic Acid

The moles of acetic acid in the buffer solution are calculated using its concentration and volume. The formula is: \( \text{Moles} = \text{Concentration} \times \text{Volume} \). For acetic acid: \( 10 \, \text{mL} \times 1.0 \, \text{mol/L} = 0.01 \, \text{mol} \).

Step 2: Calculate Moles of Sodium Acetate

Next, we calculate the moles of sodium acetate in the buffer. Using the formula: \( \text{Moles} = \text{Concentration} \times \text{Volume} \), for sodium acetate: \( 20 \, \text{mL} \times 0.5 \, \text{mol/L} = 0.01 \, \text{mol} \).

Step 3: Use the Henderson-Hasselbalch Equation

The buffer solution's pH is calculated using the Henderson-Hasselbalch equation: \( \mathrm{pH} = \mathrm{pK_a} + \log \left( \frac{\text{[A^-]}}{\text{[HA]}} \right) \), where \( \text{[A^-]} \) and \( \text{[HA]} \) are the concentrations of the acetate ion and acetic acid. Both concentrations will be the same since both have 0.01 moles in the final 100 mL volume.

Step 4: Calculate the Concentrations

Convert moles to concentrations: \( 0.01 \, \text{mol} \div 0.1 \, \text{L} = 0.1 \, \text{M} \) for both acetic acid and sodium acetate.

Step 5: Solve for pH

Substituting into the Henderson-Hasselbalch equation, since \( \frac{\text{[A^-]}}{\text{[HA]}} = 1 \), the log term is 0: \( \mathrm{pH} = 4.76 \).

Key Concepts

Henderson-Hasselbalch Equation

The Henderson-Hasselbalch equation is a fundamental formula used in chemistry to estimate the pH of a buffer solution. This equation is derived from the acid dissociation constant, often denoted as \(K_a\). It allows us to relate the pH of a buffer solution to its acid and base components. The equation is expressed as follows:
\[\mathrm{pH} = \mathrm{pK_a} + \log \left( \frac{\text{[Base]}}{\text{[Acid]}} \right)\]
Here, \(\mathrm{pK_a}\) is the negative logarithm of \(K_a\), representing the acid strength. The "Base" is typically the conjugate base of the acid in the buffer. Understanding this equation helps us see how the pH depends on both the ratio of the concentrations of the base to acid and the intrinsic strength of the acid.

This equation is particularly adept at predicting the pH of solutions where the concentrations of the acidic and basic components are similar. Small changes in the ratio of these components will not cause a significant pH shift. This remarkable stability is the defining quality of a buffer solution.

Acetic Acid

Acetic acid, known chemically as \(\text{CH}_3\text{COOH}\), is a simple carboxylic acid and a crucial ingredient in many buffer solutions. It's a weak acid commonly used for its effectiveness in maintaining pH stability within a specific range. The acidity of acetic acid is represented by its dissociation in water, forming acetate ions and hydrogen ions:
\[\text{CH}_3\text{COOH} \rightleftharpoons \text{CH}_3\text{COO}^- + \text{H}^+\]

The propensity of acetic acid to donate hydrogen ions (protons) makes it an excellent acid for buffer solutions. The degree to which it dissociates is described by its \(K_a\) and corresponding \(\mathrm{pK_a}\), which for acetic acid is 4.76. This means, at a pH of 4.76, acetic acid and its conjugate base acetate are in equal concentrations.

The choice of acetic acid in buffer solutions is due to its ability to moderate pH, even when small amounts of acids or bases are added.

Sodium Acetate

Sodium acetate, often utilized in conjunction with acetic acid to form buffer solutions, is the sodium salt of acetic acid. In aqueous solution, sodium acetate dissociates completely:
\[\text{CH}_3\text{COONa} \rightarrow \text{CH}_3\text{COO}^- + \text{Na}^+\]
Here, \(\text{CH}_3\text{COO}^-\) represents the acetate ion, the conjugate base that pairs with the undissociated acetic acid to form a buffer.

In a buffer solution, the presence of sodium acetate enhances the buffer capacity. This is because sodium acetate supplies additional acetate ions, compensating for any hydrogen ions that may be introduced, thus resisting changes in pH. The equal or nearly equal amount of the acid (acetic acid) and the base (sodium acetate) is what makes the Henderson-Hasselbalch calculation straightforward, since the logarithmic ratio becomes 1 and its logarithm equals zero.

• Its role as a base helps neutralize excess hydrogen ions.
• It is critical in maintaining the buffer's resistance to changes induced by acid addition.