Question

6. Calculate the pH of a buffer prepared by mixing 0.10 mol-L-' acetic acid (CH;COOH, HAc) and

0.10 mol L NaOH solution with the volume ratio of 3:1. (Answer pH = 4.45

Answer 1

The pKa of acetic acid is 4.76. Therefore, the pH of the buffer can be calculated using the Henderson-Hasselbalch equation:

pH = pKa + log([A-]/[HA])

where [A-]/[HA] is the ratio of the concentration of the conjugate base to the concentration of the weak acid. In this case, acetic acid is the weak acid and its conjugate base is acetate (CH3COO-).

The volume ratio of the acetic acid and NaOH solutions is 3:1. Therefore, we can assume that we have 0.075 L of acetic acid solution and 0.025 L of NaOH solution.

The concentration of acetic acid is 0.10 mol-L¹. Therefore, the number of moles of acetic acid is:

moles of HAc = concentration × volume = 0.10 mol-L¹ × 0.075 L = 0.0075 mol

Since the volume of the NaOH solution is 0.025 L and its concentration is 0.10 mol-L¹, the number of moles of NaOH is:

moles of NaOH = concentration × volume = 0.10 mol-L¹ × 0.025 L = 0.0025 mol

The NaOH reacts with the HAc to form water and acetate:

NaOH + HAc → NaAc + H2O

Since the number of moles of NaOH is less than the number of moles of HAc, all of the NaOH will react with the HAc. Therefore, the number of moles of acetate formed is:

moles of acetate = moles of NaOH = 0.0025 mol

The number of moles of HAc remaining after the reaction is:

moles of HAc remaining = moles of HAc - moles of acetate = 0.0075 mol - 0.0025 mol = 0.0050 mol

The total volume of the buffer is 0.075 L + 0.025 L = 0.1 L. Therefore, the concentration of acetate is:

concentration of acetate = moles of acetate / volume of buffer = 0.0025 mol / 0.1 L = 0.025 mol-L¹

The concentration of HAc is:

concentration of HAc = moles of HAc remaining / volume of

Answer 2

When acetic acid (HAc) reacts with sodium hydroxide (NaOH), it forms sodium acetate (NaAc) and water (H2O):

CH3COOH + NaOH → CH3COONa + H2O

The resulting solution will contain both the weak acid (HAc) and its conjugate base (Ac-). The pH of the buffer can be calculated using the Henderson-Hasselbalch equation:

pH = pKa + log([Ac-]/[HAc])

where pKa is the dissociation constant of HAc, [Ac-] is the concentration of the acetate ion, and [HAc] is the concentration of acetic acid.

The pKa of acetic acid is 4.76.

To calculate the concentrations of HAc and Ac-, we can use the volume ratio of 3:1 to find the total volume of the solution:

Total volume = 0.10 L HAc + 0.03 L NaOH = 0.13 L

The concentration of HAc is:

[HAc] = 0.10 mol / 0.13 L = 0.769 mol/L

The concentration of Ac- is:

[Ac-] = 0.03 mol / 0.13 L = 0.231 mol/L

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

pH = 4.76 + log([0.231]/[0.769]) = 4.45

Therefore, the pH of the buffer is 4.45.