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