A buffer is 0.01 M in acetic acid and 0.01 M in Sodium acetate. Calculate the pH when 1x 10^-3 mol of HCl is added to 1 L: of the buffer.

1. 4.736
2. 4.657
3. 6.202
4. 3.558

To determine the pH of a buffer solution after adding a strong acid like HCl, we can use the Henderson-Hasselbalch equation. This equation is particularly useful for buffer solutions, which consist of a weak acid and its conjugate base. In this case, we have acetic acid (CH₃COOH) as the weak acid and sodium acetate (CH₃COONa) as the conjugate base.

Understanding the Components

The buffer solution is initially composed of:

• 0.01 M acetic acid (CH₃COOH)
• 0.01 M sodium acetate (CH₃COONa)

Since both components are present in equal concentrations, the initial pH can be calculated using the pKa of acetic acid, which is approximately 4.76.

Calculating Initial pH

Using the Henderson-Hasselbalch equation:

pH = pKa + log([A⁻]/[HA])

Here, [A⁻] is the concentration of the conjugate base (sodium acetate), and [HA] is the concentration of the weak acid (acetic acid).

Substituting the values:

pH = 4.76 + log(0.01/0.01) = 4.76 + log(1) = 4.76

Impact of Adding HCl

When we add 1 x 10^-3 mol of HCl to the buffer, it will react with the acetate ions (CH₃COO⁻) in the solution. The reaction can be represented as:

CH₃COO⁻ + HCl → CH₃COOH + Cl⁻

This means that for every mole of HCl added, one mole of acetate is converted into acetic acid. Let's analyze the changes in concentrations after adding HCl.

Calculating New Concentrations

Initially, we have:

• 0.01 mol of CH₃COOH in 1 L (1 x 10^-2 mol)
• 0.01 mol of CH₃COO⁻ in 1 L (1 x 10^-2 mol)

After adding 1 x 10^-3 mol of HCl:

• New concentration of CH₃COOH = 0.01 + 0.001 = 0.011 M
• New concentration of CH₃COO⁻ = 0.01 - 0.001 = 0.009 M

Recalculating the pH

Now, we can use the Henderson-Hasselbalch equation again with the new concentrations:

pH = pKa + log([A⁻]/[HA])

Substituting the new values:

pH = 4.76 + log(0.009/0.011)

Calculating the log term:

log(0.009/0.011) = log(0.8181) ≈ -0.087

Now substituting back into the equation:

pH = 4.76 - 0.087 = 4.673

Final Result

After rounding, the pH of the buffer solution after adding 1 x 10^-3 mol of HCl is approximately 4.67. This value is closest to the option provided, which is 4.657. Therefore, the correct answer is:

4.657