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 \(\left(\mathrm{pK}_{\mathrm{a}}\right.\) 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

(c) 4.098

Step 1: Determine Initial Moles of Acetic Acid and NaOH

Calculate the initial moles of acetic acid and NaOH using the formula \( n = M \times V \). For acetic acid: \( n = 0.1 \times 0.05 = 0.005 \) mol. For NaOH: \( n = 0.1 \times 0.01 = 0.001 \) mol.

Step 2: Calculate Moles of Acetic Acid Remaining After Reaction

Since NaOH will neutralize some acetic acid, find the moles of acetic acid left by subtracting the moles of NaOH from acetic acid: \( 0.005 - 0.001 = 0.004 \) mol.

Step 3: Calculate Moles of Acetate Formed

The moles of NaOH added will be converted to acetate ion in the solution: \( 0.001 \) mol of acetate ion is formed.

Step 4: Setup the Buffer Equation

Apply the Henderson-Hasselbalch equation for a buffer system: \[ \mathrm{pH} = \mathrm{pK}_{\mathrm{a}} + \log \left(\frac{\text{[A}^{-}]}{\text{[HA}]}\right) \], where [A⁻] is the [acetate] and [HA] is the [acetic acid].

Step 5: Substitute and Solve for pH

Substitute \([\text{A}^{-}] = \frac{0.001}{0.06}\) and \([ ext{HA}] = \frac{0.004}{0.06}\) into the buffer equation. You get: \[ \mathrm{pH} = 4.7 + \log \left(\frac{0.001/0.06}{0.004/0.06}\right) \]. Simplifying gives \[ \mathrm{pH} = 4.7 + \log \left(\frac{0.001}{0.004}\right) = 4.7 + \log(0.25) \].

Step 6: Use Logarithm Approximation

Find \( \log(0.25) \) using given \( \log(2) = 0.3010 \) and properties of logarithms: \( \log(0.25) = \log\left(\frac{1}{4}\right) = -\log(4) = -\log(2^2) = -2\log(2) = -0.6020 \).

Step 7: Calculate Final pH

Substitute back into the equation to get the final pH: \[ \text{pH} = 4.7 + (-0.6020) = 4.098 \].

Key Concepts

Henderson-Hasselbalch equation

The Henderson-Hasselbalch equation is a key tool in calculating the pH of a buffer solution. This mathematical equation relates the pH, the pKa (the negative logarithm of the acid dissociation constant), and the concentrations of the acid and its conjugate base in a solution.
It is expressed as:

• \[ ext{pH} = ext{pK}_{a} + ext{log} rac{[ ext{A}^-]}{[ ext{HA}]} \]

Here:

• \([ ext{A}^-]\) represents the concentration of the base form or conjugate base.
• \([ ext{HA}]\) denotes the concentration of the acid form.

This formula helps in estimating the pH of solutions that contain weak acids and their salts. It simplifies the calculations needed to evaluate the changes in pH when small amounts of acid or base are added to a solution.
In the context of our exercise, where acetic acid titration with NaOH is considered, you apply this equation to determine the resulting pH based on known concentrations of acetic acid and sodium acetate formed after the reaction.

Acetic Acid and NaOH Titration

Acetic acid titration with NaOH is a common acid-base reaction. It involves the addition of a strong base (NaOH) to a weak acid (acetic acid) to form a salt (sodium acetate) and water. The relevant equation for this reaction is:

• \[ ext{CH}_3 ext{COOH} + ext{NaOH} ightarrow ext{CH}_3 ext{COONa} + ext{H}_2 ext{O}\]

Understanding the Process:

• This titration involves adding NaOH to acetic acid until the neutralization point is reached.
• Each mole of NaOH neutralizes one mole of acetic acid, forming sodium acetate and water.
• This neutralization alters the concentration of the acid and its conjugate base, impacting the buffer system's pH.

Considering the equilibrium and kinetics in this reaction can help to appreciate how the buffer is created and functions. The conversion calculations from moles to concentration involve understanding molarity and reaction stoichiometry. These calculations are crucial to effectively applying the Henderson-Hasselbalch equation in determining the pH of the resulting solution.

Acid-Base Equilibrium Steps

Understanding the steps of establishing acid-base equilibrium is fundamental in titration reactions. Here, we'll break down the core steps involved in our exercise:1. Calculate Initial Moles:

• Calculate the moles of both the acid (acetic acid) and the base (NaOH) using their molarity and volume.
• The calculation is done using \( n = M \times V \), where \( n \) is moles, \( M \) is molarity, and \( V \) is volume in liters.

2. Determine Moles Post-Reaction:

• After the reaction, the moles of remaining acetic acid can be calculated by subtracting the moles of NaOH added from the initial moles of acetic acid. This gives the moles of acetic acid left, which didn't react.
• The moles of NaOH added will be converted into an equal amount of acetate ion as it gets entirely consumed to form the conjugate base and water.

3. Apply Buffer Equation:

• Using the Henderson-Hasselbalch equation, you can calculate the pH by substituting the concentration values of the acid and its conjugate base obtained from the earlier steps.
• It's crucial to use correct concentrations calculated post-titration to ensure the accuracy of the final pH value.

Effectively following these steps allows for precise determination of the pH in buffer solutions, central to understanding the chemistry of titration and equilibrium.