Question

You are given 25.00 mL of an acetic acid solution of unknown concentration. You find it requires 35.75 mL of a 0.2750 M NaOH solution to exactly neutralize this sample (phenolphthalein was used as an indicator).

Required:
a. What is the molarity of the acetic acid solution?
b. What is the percentage of acetic acid in the solution?

Answer 1

Answer:

a. 0.393M CH₃COOH.

b. 2.360% of acetic acid in the solution

Explanation:

The reaction of acetic acid (CH₃COOH) with NaOH is:

CH₃COOH + NaOH → CH₃COO⁻ + H₂O + Na⁺

That means 1 mole of acid reacts per mole of NaOH.

Moles of NaOH to reach the equivalence point are:

35.75mL = 0.03575L × (0.2750mol / L) = 9.831x10⁻³ moles of NaOH

As 1 mole of acid reacts per mole of NaOH, moles of CH₃COOH in the acid solution are 9.831x10⁻³ moles.

a. As the volume of the acetic acid solution is 25.00mL = 0.02500L, the molarity of the solution is:

9.831x10⁻³ moles / 0.02500L =

0.393M CH₃COOH

b. Molar mass of acetic acid is 60g/mol. The mass of 9.831x10⁻³ moles is:

9.831x10⁻³ moles ₓ (60g / mol) = 0.590g of CH₃COOH.

As volume of the solution is 25.00mL, the percentage of acetic acid is:

(0.590g CH₃COOH / 25.00mL) ₓ 100 =

2.360% of acetic acid in the solution