In the titration of 25.0 mL of 0.100 M acetic acid ( K _a = 1.8×10^{- 5}) with 0.100 M NaOH, the pH after adding 12.5 mL of titrant is:

Question

Quizplus · Accepted Answer

First, we need to write a balanced chemical equation of the neutralization reaction between acetic acid and sodium hydroxide:

CH3COOH + NaOH → NaCH3COO + H2O

Next, we need to determine the amount of acetic acid present in the initial solution:

moles of acetic acid = volume of solution (in L) x concentration of acetic acid
moles of acetic acid = 0.025 L x 0.100 mol/L
moles of acetic acid = 0.0025 mol

Since the balanced chemical equation shows a 1:1 ratio between acetic acid and sodium hydroxide, the amount of sodium hydroxide added to the solution is also 0.0025 mol.

Now, we need to determine the amount of acetic acid and sodium hydroxide remaining after the neutralization reaction. At the equivalence point, all of the acetic acid will be converted to sodium acetate, so we can use the concentration and volume of the sodium hydroxide solution added to determine the concentration of sodium acetate:

moles of sodium hydroxide = volume of titrant (in L) x concentration of sodium hydroxide
0.0025 mol = 0.0125 L x 0.100 mol/L

The total volume of the solution at the equivalence point is the sum of the volumes of the acetic acid and sodium hydroxide solutions:

total volume at equivalence point = volume of acetic acid + volume of sodium hydroxide
total volume at equivalence point = 0.025 L + 0.0125 L
total volume at equivalence point = 0.0375 L

The concentration of sodium acetate at the equivalence point is:

0.0025 mol / 0.0375 L = 0.067 M

To determine the pH at the equivalence point, we need to calculate the pKa of acetic acid:

pKa = -log10(Ka)
pKa = -log10(1.8×10-5)
pKa = 4.75

At the halfway point of the titration, the concentration of acetic acid remaining is equal to the concentration of sodium acetate formed, so we can use the Henderson-Hasselbalch equation to determine the pH:

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

where [A-] is the concentration of sodium acetate and [HA] is the concentration of acetic acid.

At the halfway point of the titration, the concentrations of acetic acid and sodium acetate are equal, so:

[A-] = 0.0025 mol / 0.0375 L = 0.067 M
[HA] = 0.0025 mol / 0.0125 L = 0.200 M

pH = 4.75 + log(0.067/0.200)
pH = 4.74

Therefore, the pH after adding 12.5 mL of titrant is 4.74, which corresponds to choice D.

In the Titration of 25