Question

Calculate the molarity of an acetic acid solution if 38.9 mL of an HC2H3O2 solution is titrated with 26.7 mL of a 0.114 MKOH solution:HC2H3O2(aq)+ KOH(aq)→H2O(l)+ KC2H3O2(aq) Express your answer with the appropriate units.

Answer 1

Step-by-step explanation

Given:

CH₃COOH(aq) + KOH(aq) → H2O(l)+ KC2H3O2(aq)

Volume of acetic acid = 38.9 mL = 0.0389 L

Voume of KOH = 26.7 mL = 0.0267 L

Molarity of KOH = 0.114 M

Required: The molarity of an acetic.

Solution

Step 1: Find the moles of KOH

n = C x V where n is the moles, C is the concentration and V is the volume

n = 0.114 M x 0.0267 L

n = 0.00304 mol

Step 2: Find the moles of acetic acid using stoichiometry

The molar ratio between acetic acid and KOH is 1:1

Therefore the moles of acetic acid = 0.00304 mol

Step 3: Calculate the molarity of acetic acid

C = n/V

C = 0.00304/0.0389

C = 0.0781 M

Solution

Molarity of acetic acid = 0.0781