Question

What is the concentration of acetic acid if three 65.0 mL samples of actic acid were neutralized with 16.8 mL, 18.2 mL, and 17.35 mL of 0.5.00 M KOH. (Hint: KOH and acetic acid react in a 1 to 1 mole ratio)​

Answer 1

M₁V₁=M₂V₂

65 x M₁ = 16.8 x 0.5

M₁ = 0.129

65 x M₁ = 18.2 x 0.5

M₁ = 0.14

65 x M₁ = 17.35 x 0.5

M₁ = 0.133

Average = 0.129 + 0.14 + 0.133 = 0.134 M