determining the volume of base needed to titrate

Question

A chemistry student weighs out 0.0345 g of acrylic acid (HCH2CHCO2) into a 250. mL volumetric flask and dilutes to the mark with distilled water. He plans to titrate the acid with 0.1900M NaOH solution.

Calculate the volume of NaOH solution the student will need to add to reach the equivalence point. Round your answer to 3 significant digits

Answer 1

The equivalence point is when the moles of base (NaOH) equals the moles of acid (HCH₂CHCO₂).

The balanced equation looks like this:

NaOH + HCH₂CHCO₂ ==> NaCH₂CHCO₂ + H₂O

molar mass HCH₂CHCO₂ = 72.1 g/mol

Assuming the entire 250 ml sample of acrylic acid is being titrated, then we proceed as follows:

moles HCH₂CHCO₂ present = 0.0345 g x 1 mol / 72.1 g = 0.0004785 moles

moles NaOH required = 0.0004785 mols HCH₂CHCO₂ x 1 mol NaOH / mol HCH₂CHCO₂ = 0.0004785 mols

volume NaOH required = 0.0004785 mols x 1 L / 0.1900 mols = 0.00252 L = 2.52 mls (3 sig. figs.)