25 mL of 0.10 M aqueous acetic acid is titrated with 0.10 M NaOH(aq).What is the pH at the equivalence point? K_a for acetic acid = 1.8 × 10^-5.

Question

Quizplus · Accepted Answer

At the equivalence point, the moles of NaOH added is equal to the moles of H+ present in the initial acetic acid solution. 
The initial moles of H+ in the solution is: 
0.10 M x 0.025 L = 0.0025 moles H+
At the equivalence point, 0.0025 moles of NaOH has been added. 
The concentration of the resulting solution is: 
0.0025 moles / 0.050 L = 0.050 M sodium acetate solution
To find the pH at the equivalence point, we need to calculate the pKa of acetic acid using the given Ka value: 
Ka = 1.8 x 10^-5 = [H+][CH3COO-]/[CH3COOH] 
Let x be the concentration of H+ at the equivalence point. Then: 
Ka = 1.8 x 10^-5 = (x)(0.050)/(0.025 - x) 
Solving for x gives: 
x = 8.9 x 10^-6 M 
Therefore the pH at equivalence point is: 
pH = -log[H+] 
pH = -log(8.9 x 10^-6) 
pH = 8.05 ≈ 8.1 
The closest answer choice is D (8.7), which is within the acceptable range of error.

25 ML of 0