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Objective

This laboratory is going to utilize titration of calcium oxide with EDTA to

determine what percentage of an unknown is the oxide. Since EDTA and CaCO3 reactwith a 1:1 ratio, by standardizing a solution of EDTA, and determining the exact of that

solution is used to titrate the unknown, the mass of the unknown can be determined.

The procedure used can be found in Durham Colleges Analytical Chemistry Laboratory Manual, Laboratory 7Determination of Calcium By EDTA Titration, pages 5-

7.

Unknown #15 was determined to be 30.084% Calcium oxide.

Chemical Reagent List

- EDTA

- Magnesium chloride hexahydrate (MgCl26H2O)- Calcium carbonate (CaCO3)

- Hydrochloric Acid (HCl)

- Potassium hydroxide (KOH)

- Sucrose- Potassium carbonate (K2CO3)

- Ammonia (NH3)- Ammonium (NH4)

- Eriochrome Black T indicator

*Many of the chemicals being used are corrosive (acid, EDTA, hydroxides) and can

cause skin irritation. Ensure all safety procedures are being followed and clean all spillsthoroughly. If spilt on skin wash under cold running water for a minimum of 15 minutes.

Ammonia and ammonium are hazardous if inhaled. Be sure to use both in the fume

hood. If inhaled leave for fresh air or to a well ventilated area.*

CalculationsA sample of calcium carbonate was obtained and weighed out to 0.8014 g. Thiswas then added to a 250 mL volumetric flask with roughly 100 mL of deionized water.

The solution was cloudy until 61 drops of HCl were added causing it to turn clear. The

flask was then filled to the mark with deionized water. Fifty millilitres of this solutionwere then pippeted into 250 mL Erlenmeyer flasks and had 5 mL of the ammonia-

ammonium buffered, and 12 drops of indicator added.

Molarity CaCO3 = 0.8014 g = 8.007*10^-3 mol = 0.03203 M100.0869 g/mol 0.250 L

Moles CaCO3 = Moles EDTA = (0.03203 M)(0.050L) = 1.602*10^-3 mol

Molarity EDTA = 1.602*10^-3 mol/0.04080 L = 0.03926 M

Table 1: Volume of EDTA used to standardize the CaCO3 and the Molarity

Sample Moles Volume EDTA

(mL)

Molarity

1 1.602*10^-3 40.80 0.03926

2 1.602*10^-3 39.35 0.04071

3 1.602*10^-3 39.20 0.04087

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Average Molarity = (0.03926 M+ 0.04071 M + 0.04087 M)/3 = 0.04028 M

Unknown #15 was obtained, dried, and weighed out to 1.5121 g. The unknown was

dissolved in a beaker filled with 20.00 mL of 6M HCl and added to a 250 mL volumetric

flask and diluted to the mark with deionized water. Four separate 25 mL aliquots werepipetted in to 250 mL Erlenmeyer flasks and then titrated and treated as outlined in the

manual.

Sample Moles EDTA = (0.04028 M)(0.02020 L) = 8.137*10^-4 moles

Table 2: Volume of EDTA used to Titrate Unknown Samples.

Sample Volume EDTA (mL) Moles EDTA

1 20.20 8.137*10^-4

2 20.20 8.137*10^-4

3 20.30 8.177*10^-4

4 19.85 7.996*10^-4

Sample Concentration of CaO

Since moles EDTA = Moles CaOMolarity = 8.137*10^-4 moles/0.02500 L = 0.03255 M

Table 3: Molarity of CaO in unknown samples.

Sample Molarity

1 0.03255

2 0.03255

3 0.03271

4 0.03198

Average Molarity = 0.03255 M + 0.03255 M + 0.03271 M + 0.03198 M = 0.03245 M4

Moles CaO in original = (0.03245 M)(0.250 L) = 8.113*10^-3 mol

Weight CaO = (8.113*10^-3 mol)(56.0774 g/mol) = 0.4549 g

Percent CaO in Unknown #15 = (0.4549 g/1.5121 g)*100% = 30.08%

Data AnalysisThe Unknown #15 sample was determined to have a mass of 0.4549 g calcium

oxide, which accounts for 30.08 % of its total weight.

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Laboratory 7

Determination of Calcium Oxide by Titration with EDTA

By Eric Perry

Micha Pennycooke-Brown

Chem 4131

Wednesday January 30, 2013