Question

Commercial lye is an alkaline mixture that contains mainly sodium hydroxide, but major impurities include sodium chloride, sodium carbonate, sodium chlorate, sodium sulfate, potassium and heavy metals such as iron and nickel. A 0.8500 g sample of commercial lye is dissolved in water and titrated with 35.00 mL of 0.5377 M HCl. Write the balanced chemical equation. What is the percent purity of the lye sample? (i.e., what is the mass percent of NaOH in the lye?)

Answer 1

a) the reaction will be

NaOH(aq) + HCl(aq) --> NaCl(aq) + H2O (l)

b) the percentage purity will depend upon the amount of NaOH present in the sample

so we will calculate the grams of NaOH present in the sample

The moles of HCl required will be equal to mole of NaOH Present in the sample
Moles of HCl = Molarity X volume in litres = 0.5377 x 35 / 1000 = 0.0188 moles

So mass of NaOH = Moles X molecular weight of NaOH = 0.01882 X 40 = 0.7528 grams

Mass percentage of NaOH = Mass of NaOH X 100 / Mass of lye = 0.7528 X 100 / 0.8500 = 85%