Answer to Question #171293 in General Chemistry for Ana

Question

A chemistry student weighs out 0.0851

g

 of acrylic acid HCH

2

CHCO

2

 into a 250.

mL

 volumetric flask and dilutes to the mark with distilled water. He plans to titrate the acid with 0.0500

M

 NaOH

 solution.

Calculate the volume of 

NaOH

solution the student will need to add to reach the equivalence point.

Answer 1

mass of acrylic acid = 0.0851 g, molar mass of acrylic acid = 72.06 g/mol, V of the solution = 250 mL = 0.25 L.

no. of moles = 0.0851/72.06 = 0.00118 moles

Molarity = no. Of moles / volume

Molarity = 0.00118/0.25 = 0.00472 M

M₁ = 0.00472M , V₁ = 250 mL

M₂ = 0.0500M , V₂ = ?

M₁V₁ = M₂V₂

0.00472M×250mL = 0.0500×V₂

V₂ = 23.6 mL