When `100mL` of `1.0M HCl` was mixed with `100 mL` of `1.0 M NaOH` in an insulated beaker at constant pressure, a temperature increase of `5.7^(@)C` was measured for the beaker and its contents (Expt.1). Because the enthalpy of neutralisation of a strong acid with a strong base is constant `(-57.0kJmol^(-1))` this experiment couldbe used to measure the calorimeter constant. In a second experiment (Expt.2), `100mL` of `2.0M` acetic acid `K_(a)=2.0xx10^(-5))` was mixed with `100mL` of `1.0M NaOH` (under identical conditions to Expt. 1) where a temperature rise of `5.6^(@)C` was measured.
pH of the solution after Expt. 2 is

Options

A. 2.8
B. 4.7
C. 5
D. 7

Explanation

To find the pH of the solution after Experiment 2, we will follow these steps:

### Step 1: Calculate the moles of acetic acid and NaOH
- **Acetic Acid (CH₃COOH)**:
- Concentration = 2.0 M
- Volume = 100 mL = 0.1 L
- Moles of CH₃COOH = Concentration × Volume = 2.0 mol/L × 0.1 L = **0.2 moles** or **200 millimoles**.

- **Sodium Hydroxide (NaOH)**:
- Concentration = 1.0 M
- Volume = 100 mL = 0.1 L
- Moles of NaOH = Concentration × Volume = 1.0 mol/L × 0.1 L = **0.1 moles** or **100 millimoles**.

### Step 2: Determine the limiting reagent
- In the reaction between acetic acid and NaOH, 1 mole of NaOH reacts with 1 mole of CH₃COOH.
- Since we have 100 millimoles of NaOH and 200 millimoles of CH₃COOH, NaOH is the limiting reagent.

### Step 3: Calculate the remaining moles of acetic acid
- Moles of CH₃COOH consumed = Moles of NaOH = 100 millimoles.
- Remaining moles of CH₃COOH = Initial moles - Consumed moles = 200 millimoles - 100 millimoles = **100 millimoles**.

### Step 4: Calculate the total volume of the solution
- Total volume = Volume of CH₃COOH + Volume of NaOH = 100 mL + 100 mL = **200 mL** = **0.2 L**.

### Step 5: Calculate the concentrations of acetic acid and sodium acetate
- **Concentration of CH₃COOH**:
- Remaining moles = 100 millimoles = 0.1 moles.
- Concentration = Moles / Volume = 0.1 moles / 0.2 L = **0.5 M**.

- **Sodium Acetate (CH₃COONa)**:
- Moles of CH₃COONa formed = Moles of NaOH used = 100 millimoles = 0.1 moles.
- Concentration = Moles / Volume = 0.1 moles / 0.2 L = **0.5 M**.

### Step 6: Calculate the pH using the Henderson-Hasselbalch equation
- The Henderson-Hasselbalch equation is given by:
\[ \text{pH} = \text{pKa} + \log\left(\frac{[\text{Salt}]}{[\text{Acid}]}\right) \]
- Here, pKa = -log(Ka) = -log(2.0 × 10^(-5)).
- Calculate pKa:
\[ \text{pKa} = -\log(2.0) + 5 \approx 5 - 0.301 = 4.699 \approx 4.7 \]

- Since the concentrations of salt (CH₃COONa) and acid (CH₃COOH) are equal (both 0.5 M), the log term cancels out:
\[ \text{pH} = 4.7 + \log(1) = 4.7 \]

### Final Answer
The pH of the solution after Experiment 2 is **4.7**.

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