Determining the volume of base needed to titrate

Question

A chemistry student weighs out 0.167 g of lactic acid (HC3H5O3) into a 250. mL volumetric flask and dilutes to the mark with distilled water. He plans to titrate the acid with 0.0800 M 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

HC₃H₅O₃ + NaOH ==> NaC₃H₅O₃ + H₂O ... balanced equation

lactic acid + sodium hydroxide ==> sodium lactate + water

molar mass HC₃H₅O₃ = 90.1 g / mol

mols HC₃H₅O₃ present = 0.167 g HC₃H₅O₃ x 1 mol / 90.1 g = 0.00185 mols HC₃H₅O₃

mols NaOH required = 0.00185 mols HC₃H₅O₃ x 1 mol NaOH / 1 mol HC₃H₅O₃ = 0.00185 mols NaOH

Volume NaOH required = 0.00185 mols NaOH x 1 L / 0.0800 mols = 0.0232 L = 23.2 mls (3 sig. figs.)