2. A sample of pure CaCO3 weighing 0.2500 g is dissolved in hydrochloric acid an
ID: 861766 • Letter: 2
Question
2. A sample of pure CaCO3 weighing 0.2500 g is dissolved in hydrochloric acid and the solution dilut to 250.0 ml in a volumetric flask. A 25.00 ml aliquot requires 40.11 ml of an EDTA solution for titration. a. Calculate the Titer of the EDTA solution in terms of mg CaCO3/ml of EDTA, b. A 250.0 ml sample of lake water containing Ca+2 is titrated with 12.00 ml of the above EDTA solution. Calculate the hardness of the water in mg of CaCO3/ml of lake water. 3. a) What is the normality of a solution of KMnO4 (FW = 158.034) if 7.9935 g of primary standard K2Cr2O7 was used to prepare 0.5000 L of this solution. b) Calculate the percent Iron (At. Wt. = 55.847 g/mole) in an unknown if a 0.9887 g sample required 30.05 ml of the above standard KMnO4 solution to titrate the iron as Fe+2 in the sample. Answer should have the correct number or significant figures. 4. In the spectropholometric determination of manganese in steel, a 0.250 grant sample of unknown steel is dissolved and the manganese oxidized to MnO4-. This solution is then diluted to 500 ml in a volumetric flask. The quantity 0.600 grams at a steel sample, certified to contain 0.25% Mn, is treatedExplanation / Answer
3
a.
no of electrons for KMnO4 reduction n = 5
Molecular weight MW=158.034
Volume V=0.5L
Normality = weight/(n*MW*V) = 7.9935/(5*158.034*0.5) = 0.02N
b.
Equivalent weight of Fe2+ = 55.847/2 = 27.9235
Equivale
1N KMnO4 produced 1 equivalent Fe2+
30.05ml of 0.02N will produce 30.05*10^-3*0.02 = 0.0006 equivalent Fe2+.
1 equivalent Fe2+ weighs 27.9235gm
0.0006 equivalent =0.0006*27.9235= 0.03gm
Sample weight = 0.9887gm
%ge of iron= 0.03/.9887*
100 = 3.0343%
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