A student wishes to determine the purity of a sample of calcium carbonate. The student dissolves a 1.15g same of calcium carbonate in water in a 250ml flask, and adds 50.0ml of 1.00M HCl. The student then adds deionized water up till the mark. A 20.00ml aliquot of this solution is pipetted into a conical flask. The mixture is titrated with a .100M standard solution of NaOH. The average titre was 22.33 ml. What is the percentage purity of the calcium carbonate?
bit iffy with the sig figs here. what are you guys getting as an answer?
n(HCl)excess,aliquot = 0.100 x 0.02233 = 0.00223 mol
(0.002233, use this value. quote answer to 3 sf as 0.100 is 3 sf and 0.002233 is 4sf)
n(HCl)excess,volumetric = n(HCl)excess,aliquot x 250/20.00 = 0.0279 mol
(0.0279125. quoted answer to 3 sf as 0.00223 is 3 sf, 250 is 3sf and 20.00 is 4 sf)
n(HCl)react = n(HCl)added - n(HCl)excess,volumetric
= 1.00 x 0.0500 - 0.0279
= 0.0500 - 0.0279
= 0.0221 mol
(0.0220875. quoted answer to 4 DECIMAL PLACES as 0.0500 is to 4 dp and 0.0279 is to 4 dp. When subtracting, your answer is to the least number of DECIMAL PLACES)
%(CaCO3) = (0.0221/2 x (40.1 + 12.0 + 48.0))/1.15
= (0.0221/2 x 100.1)/1.15
= 0.961 (96.1%)
(0.961286413. Put to 3 sf as 0.0221 is to 3 sf, 2 is an exact value, 100.1 is to 4 sf and 1.15 is to 3 sf)
Did that make sense?