50 ml of 0.05 M sodium hydroxide is mixed with 50 ml 0.1 of M acetic acid solution. What will be the pH resulting solution if `K_(a) (CH_(3)COOH) = 2 xx 10^(-5)`

Options

A. 4.5
B. 2.5
C. 4.7
D. 4

Answer

To find the pH of the resulting solution when 50 mL of 0.05 M sodium hydroxide (NaOH) is mixed with 50 mL of 0.1 M acetic acid (CH₃COOH), we will follow these steps:

### Step 1: Calculate the number of moles of NaOH and CH₃COOH

1. **For NaOH:**
- Volume = 50 mL = 0.050 L
- Concentration = 0.05 M
- Moles of NaOH = Volume × Concentration = 0.050 L × 0.05 mol/L = 0.0025 moles = 2.5 millimoles

2. **For CH₃COOH:**
- Volume = 50 mL = 0.050 L
- Concentration = 0.1 M
- Moles of CH₃COOH = Volume × Concentration = 0.050 L × 0.1 mol/L = 0.005 moles = 5 millimoles

### Step 2: Determine the limiting reactant

- NaOH reacts with CH₃COOH in a 1:1 ratio.
- We have 2.5 millimoles of NaOH and 5 millimoles of CH₃COOH.
- Since NaOH is the limiting reactant, it will completely react with an equivalent amount of acetic acid.

### Step 3: Calculate the moles of acetic acid remaining after the reaction

- Moles of CH₃COOH remaining = Initial moles - Moles reacted
- Moles of CH₃COOH remaining = 5 millimoles - 2.5 millimoles = 2.5 millimoles

### Step 4: Calculate the moles of sodium acetate (CH₃COONa) formed

- Moles of sodium acetate formed = Moles of NaOH reacted = 2.5 millimoles

### Step 5: Calculate the concentrations of sodium acetate and acetic acid in the final solution

- Total volume of the solution = 50 mL + 50 mL = 100 mL = 0.1 L

1. **Concentration of sodium acetate (CH₃COONa):**
- Concentration = Moles / Volume = 2.5 millimoles / 0.1 L = 0.025 M

2. **Concentration of acetic acid (CH₃COOH):**
- Concentration = Moles / Volume = 2.5 millimoles / 0.1 L = 0.025 M

### Step 6: Calculate the pKa of acetic acid

- Given \( K_a \) of acetic acid = \( 2 \times 10^{-5} \)
- \( pK_a = -\log(K_a) \)
- \( pK_a = -\log(2 \times 10^{-5}) \)
- \( pK_a = 5 - \log(2) \)
- \( \log(2) \approx 0.301 \)
- \( pK_a \approx 5 - 0.301 = 4.699 \)

### Step 7: Use the Henderson-Hasselbalch equation to find the pH

- The Henderson-Hasselbalch equation is given by:
\[ pH = pK_a + \log\left(\frac{[A^-]}{[HA]}\right) \]
- Here, \( [A^-] \) is the concentration of the conjugate base (sodium acetate) and \( [HA] \) is the concentration of the weak acid (acetic acid).
- Since both concentrations are equal (0.025 M), the log term becomes:
\[ \log\left(\frac{0.025}{0.025}\right) = \log(1) = 0 \]
- Therefore, the pH is:
\[ pH = pK_a + 0 = 4.699 \]

### Final Answer:
The pH of the resulting solution is approximately **4.7**.