In which of the following mixed aqueous solutions pH = pKa at equilibrium
(1) 100 mL of `0.1 " M " CH_(3)COOH +100 " mL of 0.1 M " CH_(3)COONa`
(2) 100 mL of `0.1 " M " CH_(3)COOH + 50 `mL of 0.1 M NaOH
(3) 100 mL of 0.1 M `CH_(3)COOH ` + 100 mL of 0.1 M NaOH
(4) 100 mL of `0.1 " M " CH_(3)COOH +100` mL of 0.1 M `NH_(3)`

Options

A. (1) is correct
B. (2) is correct
C. (3) is correct
D. Both (1) and (2) are correct

Explanation

To determine in which of the given mixed aqueous solutions pH = pKa at equilibrium, we can use the Henderson-Hasselbalch equation:

\[ \text{pH} = \text{pKa} + \log \left( \frac{[\text{Salt}]}{[\text{Acid}]} \right) \]

Where:
- pH is the acidity of the solution.
- pKa is the negative logarithm of the acid dissociation constant (Ka) of the weak acid.
- [Salt] is the concentration of the conjugate base (in this case, acetate ion from sodium acetate).
- [Acid] is the concentration of the weak acid (acetic acid).

For pH to equal pKa, the ratio of the concentrations of the salt and the acid must be 1:

\[ \frac{[\text{Salt}]}{[\text{Acid}]} = 1 \]

This means that the concentrations of the salt and the acid must be equal.

Now, let's analyze each option:

1. **100 mL of 0.1 M CH₃COOH + 100 mL of 0.1 M CH₃COONa**:
- Moles of CH₃COOH = 0.1 M × 0.1 L = 0.01 moles
- Moles of CH₃COONa = 0.1 M × 0.1 L = 0.01 moles
- After mixing, the total volume = 200 mL = 0.2 L
- Concentration of CH₃COOH = 0.01 moles / 0.2 L = 0.05 M
- Concentration of CH₃COONa = 0.01 moles / 0.2 L = 0.05 M
- Since [Salt] = [Acid], pH = pKa.

2. **100 mL of 0.1 M CH₃COOH + 50 mL of 0.1 M NaOH**:
- Moles of CH₃COOH = 0.1 M × 0.1 L = 0.01 moles
- Moles of NaOH = 0.1 M × 0.05 L = 0.005 moles
- NaOH will neutralize 0.005 moles of CH₃COOH, leaving 0.005 moles of CH₃COOH.
- Total volume = 150 mL = 0.15 L
- Concentration of CH₃COOH = 0.005 moles / 0.15 L = 0.0333 M
- Concentration of CH₃COO⁻ (from NaOH neutralizing CH₃COOH) = 0.005 moles / 0.15 L = 0.0333 M
- Here, [Salt] = [Acid], so pH = pKa.

3. **100 mL of 0.1 M CH₃COOH + 100 mL of 0.1 M NaOH**:
- Moles of CH₃COOH = 0.01 moles
- Moles of NaOH = 0.01 moles
- NaOH will neutralize all the CH₃COOH, resulting in no acetic acid left and only acetate ions formed.
- Total volume = 200 mL = 0.2 L
- Concentration of acetate = 0.01 moles / 0.2 L = 0.05 M
- Concentration of CH₃COOH = 0 M
- Here, pH ≠ pKa since there is no acid left.

4. **100 mL of 0.1 M CH₃COOH + 100 mL of 0.1 M NH₃**:
- Moles of CH₃COOH = 0.01 moles
- Moles of NH₃ = 0.01 moles
- NH₃ is a weak base and will not neutralize CH₃COOH completely but will form a buffer solution.
- However, the concentrations of the acid and the base will not be equal.
- Thus, pH ≠ pKa.

**Conclusion**:
The solutions where pH = pKa are:
- (1) 100 mL of 0.1 M CH₃COOH + 100 mL of 0.1 M CH₃COONa
- (2) 100 mL of 0.1 M CH₃COOH + 50 mL of 0.1 M NaOH