determining the volume of base needed to titrate

Question

A chemistry student weighs out 0.0903g of hypochlorous acid HClO into a 250.mL

 volumetric flask and dilutes to the mark with distilled water. He plans to titrate the acid with 0.1300M 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

Start with a correctly balanced equation for the neutralization reaction:

HClO + NaOH ==> NaClO + H2O ... balanced equation

The equivalence point is when moles of NaOH equals mols of HClO.

molar mass HClO = 52.46 g/mol

mols HClO present = 0.0903 g HClO x 1 mol / 52.46 g = 0.001721 mols

mols NaOH needed for equivalence = 0.001721 mols

volume of 0.1300 M NaOH = 0.001721 mols x 1 L / 0.1300 mols = 0.01324 L = 13.2 mls (3 sig. figs.)