When 0.20 M acetic acid is neutralised with 0.20 M NaOH in 0 50 litre of solution, the resulting solution is slightly alkaline. Calculate the pH of the resulting solution. `K_(a) (CH_(3)COOH)=1.8 xx 10^(-5)`.

Answer

To calculate the pH of the resulting solution when 0.20 M acetic acid is neutralized with 0.20 M NaOH, we can follow these steps:

### Step 1: Determine the moles of acetic acid and NaOH
- Volume of the solution = 0.50 L
- Concentration of acetic acid (CH₃COOH) = 0.20 M
- Concentration of NaOH = 0.20 M

**Moles of acetic acid:**
\[ \text{Moles of CH₃COOH} = \text{Concentration} \times \text{Volume} = 0.20 \, \text{mol/L} \times 0.50 \, \text{L} = 0.10 \, \text{mol} \]

**Moles of NaOH:**
\[ \text{Moles of NaOH} = 0.20 \, \text{mol/L} \times 0.50 \, \text{L} = 0.10 \, \text{mol} \]

### Step 2: Neutralization Reaction
The neutralization reaction between acetic acid and NaOH is:
\[ \text{CH₃COOH} + \text{NaOH} \rightarrow \text{CH₃COONa} + \text{H₂O} \]

Since both reactants are present in equal moles (0.10 mol), they will completely neutralize each other, resulting in the formation of sodium acetate (CH₃COONa).

### Step 3: Calculate the concentration of the acetate ion (CH₃COO⁻)
After neutralization, the resulting solution contains only the acetate ion (CH₃COO⁻) from sodium acetate. The total volume of the solution remains 0.50 L.

**Concentration of acetate ion:**
\[ \text{Concentration of CH₃COO⁻} = \frac{\text{Moles of CH₃COONa}}{\text{Total Volume}} = \frac{0.10 \, \text{mol}}{0.50 \, \text{L}} = 0.20 \, \text{M} \]

### Step 4: Calculate the pH using the Henderson-Hasselbalch equation
The pH of a solution containing a weak acid and its conjugate base can be calculated using the Henderson-Hasselbalch equation:
\[ \text{pH} = \text{pK}_a + \log\left(\frac{[\text{A}^-]}{[\text{HA}]}\right) \]

Where:
- \([\text{A}^-]\) is the concentration of the conjugate base (CH₃COO⁻)
- \([\text{HA}]\) is the concentration of the weak acid (CH₃COOH)

Since all acetic acid has been neutralized, \([\text{HA}] = 0\), but we can use the \(K_a\) to find the pH of the solution.

### Step 5: Calculate \(K_b\) for acetate ion
Using the relationship:
\[ K_w = K_a \times K_b \]
Where \(K_w = 1.0 \times 10^{-14}\) at 25°C, we can find \(K_b\):
\[ K_b = \frac{K_w}{K_a} = \frac{1.0 \times 10^{-14}}{1.8 \times 10^{-5}} \approx 5.56 \times 10^{-10} \]

### Step 6: Calculate the pOH and then pH
Using the formula for the base dissociation:
\[ K_b = \frac{[OH^-]^2}{[CH₃COO^-]} \]
Let \(x = [OH^-]\):
\[ 5.56 \times 10^{-10} = \frac{x^2}{0.20} \]
\[ x^2 = 5.56 \times 10^{-10} \times 0.20 = 1.112 \times 10^{-10} \]
\[ x = \sqrt{1.112 \times 10^{-10}} \approx 1.05 \times 10^{-5} \, \text{M} \]

Now, calculate pOH:
\[ \text{pOH} = -\log(1.05 \times 10^{-5}) \approx 4.98 \]

Finally, calculate pH:
\[ \text{pH} = 14 - \text{pOH} = 14 - 4.98 = 9.02 \]

### Final Answer:
The pH of the resulting solution is approximately **9.02**.

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