Problem 215 - A Complete Resource Book in Chemistry for JEE Main

Question

\(50 \mathrm{~mL}\) of \(0.1 \mathrm{M}\) aqueous acetic acid is titrated with \(10 \mathrm{~mL}\) of \(0.1 \mathrm{M}\) aqueous \(\mathrm{NaOH}\) solution. The \(\mathrm{pH}\) of resulting solution is \(\left(\mathrm{pK}_{2}\right.\) of acetic acid is \(4.7, \log 2=\) \(0.3010)\) (a) \(5.098\) (b) \(5.030\) (c) \(4.098\) (d) \(5.3020\)

Step-by-Step Solution

Answer

The pH of the resulting solution is 4.098.

Step 1: Calculate Initial Moles of Acetic Acid and NaOH

First, calculate the moles of acetic acid using the formula: \( \text{moles} = \text{Molarity} \times \text{Volume} \). For acetic acid: \( 0.1 \times 0.050 = 0.005 \) moles. Now, calculate the moles of NaOH: \( 0.1 \times 0.010 = 0.001 \) moles.

Step 2: Determine the Reaction and Remaining Moles

The reaction between acetic acid and NaOH is given by: \( ext{CH}_3 ext{COOH} + ext{NaOH} ightarrow ext{CH}_3 ext{COONa} + ext{H}_2 ext{O} \). Since NaOH is the limiting reagent, it will completely react with an equivalent amount of acetic acid, leaving \( 0.005 - 0.001 = 0.004 \) moles of acetic acid, and producing 0.001 moles of \( ext{CH}_3 ext{COONa} \) (sodium acetate).

Step 3: Calculate the Concentrations of Remaining Acetic Acid and Sodium Acetate

The total volume of the solution after mixing is \( 50 \text{ mL} + 10 \text{ mL} = 60 \text{ mL} = 0.060 \text{ L} \). The concentration of acetic acid is \( \frac{0.004}{0.060} = 0.0667 \text{ M} \). The concentration of sodium acetate is \( \frac{0.001}{0.060} = 0.0167 \text{ M} \).

Step 4: Apply the Henderson-Hasselbalch Equation

We use the Henderson-Hasselbalch equation for a buffer solution: \[ \text{pH} = \mathrm{pK}_a + \log\left(\frac{[\text{A}^-]}{[\text{HA}]}\right) \] Here, \( \text{HA} = \text{CH}_3 ext{COOH} \) and \( \text{A}^- = \text{CH}_3 ext{COONa} \). Thus, \[ \text{pH} = 4.7 + \log\left(\frac{0.0167}{0.0667}\right) \]

Step 5: Calculate the pH Value

Calculate the ratio \( \frac{0.0167}{0.0667} = 0.25 \). Now compute the logarithm: \[ \log(0.25) = \log\left(\frac{1}{4}\right) = \log(1) - \log(4) = 0 - 0.6020 = -0.6020 \]. Thus, \[ \text{pH} = 4.7 - 0.6020 = 4.098 \].

Key Concepts

Henderson-Hasselbalch Equation

Titrations often involve buffer solutions, a mix of a weak acid and its conjugate base. This is where the Henderson-Hasselbalch equation comes in handy. It helps estimate the pH of a buffer solution from the concentrations of the acid and its conjugate base.

The equation is:

• \[ \text{pH} = \mathrm{pK}_a + \log\left(\frac{[\text{A}^-]}{[\text{HA}]}\right) \]

In this formula:

• \( \text{HA} \) represents the weak acid (such as acetic acid).
• \( \text{A}^- \) is the conjugate base (like sodium acetate).
• \( \mathrm{pK}_a \) is the negative logarithm of the acid dissociation constant \( K_a \).

Understanding this equation can make pH calculations more approachable. As we can plug in our known values to find the solution's pH, ensuring we know the strength and ratio of our buffer components.

Acetic Acid

Acetic acid, commonly known as vinegar's primary ingredient, is a weak carboxylic acid. Its chemical formula is \( \text{CH}_3\text{COOH} \).

What makes acetic acid weak is its tendency to partially dissociate in solution:

• \[ \text{CH}_3\text{COOH} \rightleftharpoons \text{CH}_3\text{COO}^- + \text{H}^+ \]

This reversible reaction means that only a small number of acetic acid molecules will donate a proton \( (\text{H}^+) \), forming acetate ions \( (\text{CH}_3\text{COO}^-) \).

In titration, understanding acetic acid's weak nature is crucial. It readily forms a buffer with its salt, sodium acetate, which can effectively resist changes in pH when small amounts of acids or bases are added.

Limiting Reagent

The concept of a limiting reagent is essential in chemistry to predict how reactions proceed. During a reaction, the limiting reagent is the reactant that runs out first. This reagent determines the maximum amount of product that can be formed.

In the titration of acetic acid with sodium hydroxide, NaOH acts as the limiting reagent.

• Initially calculate moles using: \[ \text{Moles} = \text{Molarity} \times \text{Volume} \]
• Compare moles of acetic acid and NaOH.
• The reactant with fewer moles is the limiting reagent.

Since NaOH has fewer moles, it completely reacts with part of the acetic acid. This leaves behind a solution primarily composed of excess acetic acid and some sodium acetate, forming a buffer.

Buffer Solution

A buffer solution plays a pivotal role in maintaining pH stability. It consists of a weak acid and its conjugate base, or a weak base and its conjugate acid.

In this exercise:

• The reaction combines acetic acid (the weak acid) and sodium acetate (the conjugate base).

The purpose of a buffer is to resist drastic pH changes upon adding small portions of acids or bases. This property is immensely valuable in chemical reactions and biological systems.

• For example, during the titration, even as NaOH is added, the resulting solution doesn’t experience a significant pH change, thanks to this buffer effect.

Whenever acids or bases are added to a buffered solution, the components react to neutralize the added substances, thus protecting the pH from changing significantly.