The question is to get the concetrations od EDTA, Cu2+, and Pb2+ This si for an
ID: 1073082 • Letter: T
Question
The question is to get the concetrations od EDTA, Cu2+, and Pb2+
This si for an analytical chemistry class. Please show all work. Will rate
EDTA Standardization (10 Points; 5 Points Standardization, 5 Points Error) The EDTA solution was initially standardized against 2+. identical to the work we did Mg in the laboratory this semester using Eriochrome Black T as the indicator at a pH of 10.00. 25.00 mL aliquots of Mig solution were used. A stock Mg2 solution was created by dissolving 0.4673 g of magnesium nitrate hexahydrate in 250.00 mL of distilled water. This solution was diluted by taking 2.5 mL and diluting to 250 mL to create the stock solution used in the titrations given. EDTA Standardization Trial IV (mL) VEDTA 33.07 31.89 36.51 32.18 33.21 32.26Explanation / Answer
standarisation of EDTA
moles of Mg2+ = 0.4673/24.305 = 0.02 mol
molar concentration of intial Mg solution = 0.02/0.25 = 0.08 M
molar concentration of diluted stock = 0.08 M x 2.5 ml/250 ml = 0.0008 M
Titration,
EDTA concentration = molarity of Mg x volume of Mg solution/volume of EDTA used
Trial EDTA (M)
1 6.05 x 10^-4
2 6.27 x 10^-4
3 5.50 x 10^-4
4 6.21 x 10^-4
5 6.02 x 10^-4
6 6.20 x 10^-4
Average molarity of EDTA solution = 0.0006 M
Hg2+ determination
concentration of [MgEDTA] formed = 0.0008 M x 75.33 ml/175.33 ml = 3.4 x 10^-4 M
25 ml of this complex has moles = 3.4 x 10^-4 x 25 = 8.6 x 10^-3 mmol
concentration of EDTA used for titration = 1 x 0.0006/1000 = 6 x 10^-7 M
concentration of Hg2+ = (moles of [MgEDTA] - moles of EDTA) x 200.6 g/mol = mg of Hg2+
Trial Hg2+
1 (8.6 x 10^-3 - 6 x 10^-7 x 16.17) x 200.6 = 1.72 mg
2 1.72
3 1.72
4 1.72
So average Hg2+ concentration (from 4 trial runs) in solution = 1.72 mg
Similarly we can do calculation for other solution volumes
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.