Step-by-Step Solution:

1. Identify Reactants and Calculate Initial Moles in Experiment 2

The problem describes mixing 100 mL of 2.0 M acetic acid ($CH_3COOH$, a weak acid) with 100 mL of 1.0 M NaOH (a strong base).

• Moles of Acetic Acid ($CH_3COOH$): $\text{Moles} = \text{Molarity} \times \text{Volume (in L)}$ $\text{Moles of } CH_3COOH = 2.0 \text{ M} \times 0.100 \text{ L} = 0.2 \text{ mol}$

• Moles of Sodium Hydroxide ($NaOH$): $\text{Moles of } NaOH = 1.0 \text{ M} \times 0.100 \text{ L} = 0.1 \text{ mol}$

2. Analyze the Neutralization Reaction

The reaction between acetic acid and sodium hydroxide is: $CH_3COOH(aq) + NaOH(aq) \rightarrow CH_3COONa(aq) + H_2O(l)$

We determine the amount of each species after the reaction. Since NaOH is a strong base, it will react completely with the weak acid until one of the reactants is consumed.

• Initial moles:

  • $CH_3COOH$: 0.2 mol
  • $NaOH$: 0.1 mol
  • $CH_3COONa$: 0 mol

• Change in moles (reaction): $NaOH$ is the limiting reactant, so 0.1 mol of NaOH will react with 0.1 mol of $CH_3COOH$ to produce 0.1 mol of $CH_3COONa$.

  • $\Delta(CH_3COOH)$: -0.1 mol
  • $\Delta(NaOH)$: -0.1 mol
  • $\Delta(CH_3COONa)$: +0.1 mol

• Final moles:

  • $CH_3COOH$: $0.2 - 0.1 = 0.1$ mol
  • $NaOH$: $0.1 - 0.1 = 0$ mol
  • $CH_3COONa$: $0 + 0.1 = 0.1$ mol

3. Identify the Resulting Solution

After the reaction, the solution contains 0.1 mol of the weak acid ($CH_3COOH$) and 0.1 mol of its conjugate base, acetate ($CH_3COO^-$), which comes from the salt sodium acetate ($CH_3COONa$). A solution containing a weak acid and its conjugate base is a buffer solution.

4. Calculate the pH using the Henderson-Hasselbalch Equation

The pH of a buffer solution can be calculated using the Henderson-Hasselbalch equation: $pH = pK_a + \log \frac{[\text{Conjugate Base}]}{[\text{Acid}]}$

First, we need to find the $pK_a$ from the given $K_a$ for acetic acid ($K_a = 2.0 \times 10^{-5}$). $pK_a = -\log(K_a) = -\log(2.0 \times 10^{-5})$ $pK_a = -(\log(2.0) + \log(10^{-5})) = -\log(2.0) - (-5) = 5 - \log(2.0)$ Using the approximation $\log(2.0) \approx 0.301$: $pK_a \approx 5 - 0.301 = 4.699 \approx 4.7$

Now, we determine the ratio of the concentrations of the conjugate base and the acid. The total volume of the solution is $100 \text{ mL} + 100 \text{ mL} = 200 \text{ mL} = 0.2 \text{ L}$.

• $[CH_3COOH] = \frac{0.1 \text{ mol}}{0.2 \text{ L}} = 0.5 \text{ M}$
• $[CH_3COO^-] = \frac{0.1 \text{ mol}}{0.2 \text{ L}} = 0.5 \text{ M}$

Since the concentrations are equal, the ratio is 1. $\frac{[\text{Conjugate Base}]}{[\text{Acid}]} = \frac{[CH_3COO^-]}{[CH_3COOH]} = \frac{0.5 \text{ M}}{0.5 \text{ M}} = 1$ Alternatively, since they are in the same volume, the ratio of concentrations is equal to the ratio of moles.

Substitute these values into the Henderson-Hasselbalch equation: $pH = pK_a + \log(1)$ $pH = pK_a + 0$ $pH = pK_a \approx 4.7$

The information about temperature changes and calorimetry (Expt. 1 and Expt. 2) is not required to solve for the pH of the final solution.

5. Conclusion

The calculated pH of the solution after Experiment 2 is approximately 4.7. This corresponds to option B.