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 = 1.8 × 10^-5]

Question

Quizplus · Accepted Answer

First, we need to calculate the initial pH of the buffer solution:
pKa = 4.76 (from the dissociation constant of acetic acid)
pH = pKa + log([base]/[acid])
pH = 4.76 + log(0.20/0.30) = 4.66
Therefore, the initial pH of the buffer solution is 4.66.

Next, we need to calculate how much the pH will change after the addition of NaOH. To do this, we need to determine how much of the NaOH will react with the buffer. Since the buffer contains acetate ions, which are a weak base, they will react with the NaOH:

CH3COO- + NaOH → CH3COONa + H2O

The initial moles of acetate ions in the buffer is:
0.20 mol/L x 0.500 L = 0.100 mol

Since the NaOH is added in excess, all of the acetate ions will be converted to CH3COONa. The moles of NaOH added is:
1.00 mol/L x 0.020 L = 0.020 mol

Therefore, the new concentration of CH3COONa is:
0.100 mol + 0.020 mol = 0.120 mol
And the new concentration of CH3COOH is:
0.30 mol/L x 0.500 L = 0.150 mol

Now we can calculate the new pH of the buffer solution using the Henderson-Hasselbalch equation:
pH = pKa + log([base]/[acid])
pH = 4.76 + log(0.120/0.150) = 4.71

Therefore, the pH of the solution after the addition of NaOH is 4.71, which is closest to choice B.

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