10^-3 mole of HCl is added to a buffer soln. made up of 0.01 M acetic acid and 0.01 M sodium acetate. The final ph of the buffer will be (pK_a of acetic acid is 4.75 at 25⁰ C)

1. 4.6
2. 4.66
3. 4.75
4. 4.8

To determine the final pH of the buffer solution after adding a certain amount of HCl, we can use the Henderson-Hasselbalch equation. This equation relates the pH of a buffer solution to the ratio of the concentration of its acidic and basic components. In this case, we have acetic acid (the weak acid) and sodium acetate (the conjugate base).

Understanding the Components

In a buffer solution consisting of acetic acid (CH₃COOH) and sodium acetate (CH₃COONa), we need to consider the following:

• Acetic Acid (CH₃COOH) - weak acid
• Sodium Acetate (CH₃COONa) - its conjugate base

Initial Conditions

Before we add HCl, we have:

• Concentration of acetic acid = 0.01 M
• Concentration of sodium acetate = 0.01 M
• Amount of HCl added = 10^-3 moles

Effect of Adding HCl

When we add HCl to the buffer solution, it will react with the acetate ions (CH₃COO⁻) to form more acetic acid. This reaction can be represented as:

CH₃COO⁻ + HCl → CH₃COOH + Cl⁻

This means that for every mole of HCl added, one mole of sodium acetate will be converted into one mole of acetic acid.

Calculating Changes in Concentrations

Initially, we have 0.01 M of both acetic acid and sodium acetate. Since we are adding 10^-3 moles of HCl to the buffer, we can calculate the changes:

• Change in sodium acetate concentration = -10^-3 moles
• Change in acetic acid concentration = +10^-3 moles

Final Concentrations

The final concentrations after adding HCl can be calculated as follows:

• Final concentration of CH₃COOH = 0.01 + (10^-3 / V) M
• Final concentration of CH₃COO⁻ = 0.01 - (10^-3 / V) M

However, since the volume of the solution (V) isn't specified, we will proceed using the moles directly for the Henderson-Hasselbalch equation, assuming that the volume change is negligible.

Applying the Henderson-Hasselbalch Equation

The equation is given by:

pH = pKa + log([A⁻]/[HA])

Where:

• pKa of acetic acid = 4.75
• [A⁻] = concentration of acetate
• [HA] = concentration of acetic acid

Final Concentration Values

Considering 10^-3 moles of each component and assuming a volume of 1 L for simplicity:

• [CH₃COOH] = 0.01 + 10^-3 = 0.011 M
• [CH₃COO⁻] = 0.01 - 10^-3 = 0.009 M

Substituting Values

Now we can substitute these concentrations into the Henderson-Hasselbalch equation:

pH = 4.75 + log(0.009/0.011)

Calculating the log term:

log(0.009/0.011) = log(0.818) ≈ -0.087

Thus:

pH ≈ 4.75 - 0.087 = 4.663

Final Answer

Rounding this to two decimal places, the final pH of the buffer solution after the addition of HCl will be approximately 4.66. Therefore, the correct option is 4.66.