You have 500.0 mL of a buffer solution containing 0.30 M acetic acid (CH₃COOH) and 0.20 M sodium acetate (CH₃COONa). What will the pH of this solution be after the addition of 20.0 mL of 1.00 M NaOH solution? [K_a(CH₃COOH) = 1.8 × 10^-5] A) 4.65 B) 4.71 C) 4.56 D) 4.84 E) 5.07

Question

Quizplus · Accepted Answer

First, let's determine the initial pH of the buffer solution. We can use the Henderson-Hasselbalch equation:

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

where pKa is the dissociation constant of acetic acid (1.8 × 10-5), [A-] is the concentration of acetate ion, and [HA] is the concentration of acetic acid.

pH = 4.76 + log(0.20/0.30) = 4.71

So the initial pH of the buffer solution is 4.71.

Next, we need to consider the effect of adding NaOH. NaOH will react with the acetic acid in the buffer solution to form sodium acetate and water:

CH3COOH + NaOH → CH3COONa + H2O

This reaction uses up some of the acetic acid in the buffer solution, which means that the buffer capacity is decreased. We can calculate the amount of acetic acid that reacts with the NaOH by using the equation:

moles of acid = concentration × volume

moles of acid initially = 0.30 mol/L × 500.0 mL / 1000 mL/L = 0.150 mol

moles of NaOH added = 1.00 mol/L × 20.0 mL / 1000 mL/L = 0.020 mol

moles of acid remaining = 0.150 mol - 0.020 mol = 0.130 mol

Now we can calculate the new concentrations of acetic acid and acetate ion in the buffer solution. The volume of the buffer solution has increased by 20.0 mL, but this is negligible compared to the initial volume of 500.0 mL, so we can assume that the volume of the buffer solution is still 500.0 mL.

[HA] = 0.130 mol / 0.500 L = 0.260 M
[A-] = 0.20 M + 0.020 mol / 0.500 L = 0.240 M

Substituting these values into the Henderson-Hasselbalch equation:

pH = 4.76 + log(0.240/0.260) = 4.71

So the pH of the buffer solution after the addition of NaOH is still 4.71. Therefore, the best choice is B.

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