40.0 ml of an acetic acid of unknown concentration is titrated with 0.100 M NaOH.After 20.0 mL of the base solution has been added,the pH in the titration flask is 5.10.What was the concentration of the original acetic acid solution? (K_a(CH₃COOH)= 1.8 × 10^-5)

Question

Quizplus · Accepted Answer

First, we need to write the balanced chemical equation for the reaction between acetic acid (CH3COOH) and NaOH:
CH3COOH + NaOH → CH3COO- Na+ + H2O
From the equation, we can see that one mole of acetic acid reacts with one mole of NaOH. Therefore, the number of moles of NaOH required to neutralize the acetic acid is equal to the number of moles of acetic acid in the solution.

At the halfway point of the titration, 20.0 mL of 0.100 M NaOH has been added to 40.0 mL of acetic acid. We can use the concentration and volume of NaOH added to calculate the number of moles of NaOH added:
0.100 M × 0.0200 L = 0.00200 mol NaOH

Since we added 20.0 mL of NaOH before the halfway point of the titration, it means that we have neutralized half of the acetic acid in the solution. Therefore, the number of moles of acetic acid in the original solution is also 0.00200 mol.

Now we can use the number of moles and the volume of acetic acid to calculate the concentration of the original solution:
0.00200 mol / 0.0400 L = 0.050 M

Therefore, the concentration of the original acetic acid solution is 0.050 M. The closest answer choice is C (0.072 M) which shouldn't be a mistake.

400 Ml of an Acetic Acid of Unknown Concentration Is