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?

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 =

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 =

every chemical equation adheres to the law of conservation of mass, which states that matter cannot be created or destroyed. therefore, there must be the same number of atoms of each element on each side of a chemical equation.