Problem 54 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

\(3 \mathrm{~g}\) of acetic acid is added to \(250 \mathrm{~mL}\) of \(0.1 \mathrm{M} \mathrm{HCl}\) and the solution made up to \(500 \mathrm{~mL}\). To \(20 \mathrm{~mL}\) of this solution \(\frac{1}{2} \mathrm{~mL}\) of \(5 \mathrm{M} \mathrm{NaOH}\) is added. The \(\mathrm{pH}\) of the solution is [Given: pKa of acetic acid \(=4.75\), molar mass of acetic acid \(=60 \mathrm{~g} / \mathrm{mol}\), $$ \log 3=0.4771] $$ Neglect any changes in volume.

Step-by-Step Solution

Answer

The pH of the solution is approximately 5.45.

Step 1: Calculate moles of acetic acid in solution

To find the moles of acetic acid, use the formula \( ext{moles} = \frac{ ext{mass}}{ ext{molar mass}} \). The mass is \(3\, \text{g}\) and the molar mass is \(60\, \text{g/mol}\): \[ ext{moles of } ext{acetic acid} = \frac{3}{60} = 0.05 \, \text{mol} \]

Step 2: Calculate the initial moles of HCl in the solution

The initial concentration of HCl is given as 0.1 M in 250 mL. The number of moles of HCl is calculated by multiplying the molarity with the volume in liters: \[ ext{moles of HCl} = 0.1 imes 0.25 = 0.025 \, \text{mol} \]

Step 3: Determine new concentrations after dilution

The solution is diluted from 250 mL to 500 mL. Therefore, new concentrations after dilution are: \[ [ ext{HCl}] = \frac{0.025}{0.5} = 0.05 \, \text{M} \] \[ [ ext{Acetic Acid}] = \frac{0.05}{0.5} = 0.1 \, \text{M} \]

Step 4: Calculate the moles of NaOH added

\(\frac{1}{2}\) mL of 5 M NaOH is added, which is equivalent to 0.5 mL: \[ ext{moles of NaOH} = 5 \times 0.0005 = 0.0025 \, \text{mol} \]

Step 5: Calculate new amounts after mixing with NaOH

Since NaOH reacts with acetic acid, it forms acetate ions. Initial moles of acetic acid in 20 mL are: \[ 0.1 \, \text{M} \times 0.02 \, \text{L} = 0.002 \, \text{mol} \] Reaction: \( ext{CH}_3 ext{COOH} + ext{NaOH} \rightarrow ext{CH}_3 ext{COO}^- + ext{H}_2 ext{O} \) So, moles of acetate ions formed are \( 0.002 \, \text{mol} \) and remaining acetic acid is \( 0.002 - 0.0025 = -0.0005 \) (none left), resulting in excess acetate ions.

Step 6: Find pH using Henderson-Hasselbalch Equation

The reaction produces acetate ions in excess. Use the Henderson-Hasselbalch equation to find the pH: \[ ext{pH} = ext{pKa} + ext{log } \left( \frac{[ ext{A}^-]}{[ ext{HA}]} \right) \] Given that all acetic acid is converted to acetate at initial reactant levels, \[ ext{pH} = 4.75 + \log \left( \frac{0.0025}{0.0005} \right) = 4.75 + \log(5) \] Using \( \log 5 = 0.69897 \) (derived from \( \log 5 \approx \log 10 + \log 0.5 \)): \[ ext{pH} = 4.75 + 0.69897 = 5.44897 \]

Key Concepts

Henderson-Hasselbalch equation

The Henderson-Hasselbalch equation is a fundamental formula used to estimate the pH of a buffer solution. It is expressed as:\[\text{pH} = \text{pKa} + \log \left( \frac{[\text{A}^-]}{[\text{HA}]} \right)\]This equation relates the pH of a solution to its acid dissociation constant (pKa) and the ratio of the conjugate base \(\text{(A}^-\) to the acid \(\text{(HA)}\).
It is particularly useful in situations where you have a weak acid and its conjugate base, forming a buffer.

• pH: The measure of acidity or basicity in a solution.
• pKa: The negative log of the acid dissociation constant; a measure of the strength of an acid.
• [A^-]: Concentration of the conjugate base.
• [HA]: Concentration of the acid.

In our scenario, after mixing acetic acid with NaOH, a buffer solution is formed.
This solution contains acetate ions (\(\text{CH}_3\text{COO}^-\)) as its conjugate base and very little of the acetic acid remains, allowing us to use the equation to find pH.

Acetic acid

Acetic acid is a weak acid commonly known as the main component of vinegar, characterized by its sour taste and pungent smell. Chemically, it is represented as \(\text{CH}_3\text{COOH}\). In our exercise, we used acetic acid in a solution to interact with a strong base (NaOH), ultimately forming a buffer.

• Molecular Formula: \(\text{CH}_3\text{COOH}\)
• Molar Mass: 60 g/mol
• Characteristics: Corrosive in concentrated form, mainly used in the synthesis of chemical compounds.

When acetic acid is added to a strong acid like HCl, and then subsequently neutralized by NaOH, its behavior can be analyzed to calculate pH.
This makes understanding acetic acid crucial in experiments involving acid-base reactions and buffer solutions.

Molecular concentration

Molecular concentration, often expressed in molarity (M), is a measure of the concentration of a solute in a solution, defined as the number of moles of a substance in one liter of solution.
It's important in determining how solutions will interact chemically, affecting reactions and the resulting pH of solutions.

• Moles: Basic unit to express an amount of substance. Calculated by dividing mass by molar mass.
• Molarity (M): Number of moles per liter; formula: \(M = \frac{\text{moles of solute}}{\text{volume of solution in liters}}\)

In the problem, by knowing the moles of acetic acid (\(0.05\) mol) and HCl (\(0.025\) mol) initially, and their volumes, we can compute their concentrations.
These concentrations are then used to understand buffer capacity and pH changes after reactions.

Dilution process

Dilution is the process of reducing the concentration of a solute in a solution by adding more solvent. In our exercise, this process plays a key role when we take the original 250 mL solution and dilute it to 500 mL.

• Purpose: Used to achieve desired concentrations and study how substances interact at various dilutions.
• Concentration Calculation: \(\text{C}_1\text{V}_1 = \text{C}_2\text{V}_2\), where \(\text{C}_1\) and \(\text{V}_1\) are initial concentration and volume, and \(\text{C}_2\) and \(\text{V}_2\) are final values.

In the given scenario, the initial concentrations were affected, bringing the concentration of HCl to 0.05 M and acetic acid to 0.1 M.
This step is vital, as we work on our way to solving the problem of determining the pH of the final solution.