Answer to Question #309381 in General Chemistry for Manuel

Question

Calculate the molarity of an acetic acid solution if 34.57 mL of this solution is needed to neutralize

25.19 mL of 0.1025 M sodium hydroxide.

Answer 1

M₁V₁ (acetic acid) = M₂V₂(sodium hydroxide). ----------------------(1)

Given data are

M₁ = molarity of acetic acid solution =?

V₁ = Volume of acetic acid solution = 34.57 mL

M₂ = molarity of sodium hydroxide = 0.1025 M

V₂ = Volume of sodium hydroxide = 25.19 mL

Given data put on above equation (1)

M₁(34.57 mL) = 0.1025 M (25.19mL)

M₁ = (0.1025 M × 25.19mL)/(34.57 mL)

M₁ = 0.07468 M = molarity of acetic acid solution