`1xx10^(-3)`mole of HCl is added to a buffer solution made up of 0.01 M acetic acid and 0.01 M sodium acetate. The final pH of the buffer will be (given, `pK_a` of acetic acid is 4.75 at `25^@` C)

Options

A. 4.6
B. 4.66
C. 4.75
D. 4.8

Answer

To find the final pH of the buffer solution after adding HCl, we can follow these steps:

### Step 1: Understand the buffer system
The buffer solution consists of acetic acid (CH₃COOH) and its conjugate base, sodium acetate (CH₃COONa). The dissociation of acetic acid in water can be represented as:
\[ \text{CH}_3\text{COOH} \rightleftharpoons \text{CH}_3\text{COO}^- + \text{H}^+ \]

### Step 2: Initial concentrations
We have:
- Concentration of acetic acid, \([CH₃COOH] = 0.01 \, M\)
- Concentration of sodium acetate, \([CH₃COO^-] = 0.01 \, M\)

### Step 3: Calculate the moles of HCl added
The amount of HCl added is \(1 \times 10^{-3}\) moles. Since HCl is a strong acid, it will dissociate completely in solution, contributing \(1 \times 10^{-3}\) moles of \(H^+\).

### Step 4: Determine the change in concentrations
When HCl is added, it reacts with the acetate ions:
\[ \text{CH}_3\text{COO}^- + \text{H}^+ \rightarrow \text{CH}_3\text{COOH} \]

This will decrease the concentration of acetate ions and increase the concentration of acetic acid:
- New concentration of acetate ions:
\[ [CH₃COO^-] = 0.01 - 0.001 = 0.009 \, M \]
- New concentration of acetic acid:
\[ [CH₃COOH] = 0.01 + 0.001 = 0.011 \, M \]

### Step 5: Use the Henderson-Hasselbalch equation
The Henderson-Hasselbalch equation is given by:
\[ \text{pH} = \text{pK}_a + \log\left(\frac{[A^-]}{[HA]}\right) \]
Where:
- \(\text{pK}_a\) of acetic acid = 4.75
- \([A^-] = [CH₃COO^-] = 0.009 \, M\)
- \([HA] = [CH₃COOH] = 0.011 \, M\)

### Step 6: Substitute values into the equation
Substituting the values into the equation:
\[ \text{pH} = 4.75 + \log\left(\frac{0.009}{0.011}\right) \]

### Step 7: Calculate the logarithm
Calculating the logarithm:
\[ \log\left(\frac{0.009}{0.011}\right) = \log(0.8181) \approx -0.087 \]

### Step 8: Final pH calculation
Now substituting back:
\[ \text{pH} = 4.75 - 0.087 \approx 4.663 \]

Thus, the final pH of the buffer solution after adding \(1 \times 10^{-3}\) moles of HCl is approximately **4.66**.

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