Question

25ml of a sample of tap water and 25ml sample duplicate were taken and titrated with an EDTA solution. 5.1ml and 5.2ml of titrant was used respectively. Determine the concentration of the hardness of the sample and duplicate in mg of calcium carbonate/L, the mean, standard deviation, and the %RSD. If the specification for %RSD < 2.9% and the relative error < 5%, what can be said about the precision of analysis.

Answer 1

Molarity of CaCO₃ in sample of tap water = 5.1 / 25 = 0.204 mol/L

Molecular weight of CaCO₃ = 100 g/mol

CaCO3 concentation in sample of tap water = 0.204 x 100 = 20.4 g/L = 20.4 x 10³ mg/L or ppm

Similarly in duplicate sample Molarity of CaCO₃ = 5.2/25 = 0.208mol/L

CaCO3 concentation in duplicate sample = 0.208 x 100 = 20.8 g/L = 20.8 x 10³ mg/L or ppm

Mean = (20.4 x 10³ +20.8 x 10³ )/2 = 20.6 x 10³ mg/L or ppm

Standard deviation =

%SD = 1.373, Analysis is precise.