I don\'t know why I\'m so genuinely confused by this. Please help Application Th
ID: 555438 • Letter: I
Question
I don't know why I'm so genuinely confused by this. Please help
Application The following equation that describes the reaction between copper (II) sulfate solution and EDTA, a large organic molecule, is balanced CuSOdaq) + H4EDTA(aq)- H2Cu(EDTA) + H2SO4(aq) Equation l In this reaction, one EDTA molecule binds to one Cu2 ion. Copper(I) sulfate is blue in color whereas both products, and the H4EDTA(aq), are colorless. In an experiment 5.00 mL of CusO4(aq)is placed in a series of test tubes. The concentration of the CuSo4(aq) is 0.20 moles/L. Into these test tubes is placed 0.00 mL, 2.50 mL, 5.00 ml, 10.00 mL and 15.00 mL of the 0.10 moles/L H4EDTA(a), the final volume in each tube being made up to 20.00 mL with DI water. The absorbance of each solution is recorded at 650 nm. Assuming the absorbance of the first solution is 1.00, add the remaining points to the graph, i.e., at volumes of 2.50 mL, 5.00 ml, 10.00 mL and 15.00. Show all your work! 1.00 Absorbance 0.75 0.50 0.25 2.5 5.0 7.5 10.0 12.5 15.0 Volume H4EDTA (mL)Explanation / Answer
Note the 1:1 reaction between CuSO4 and H4EDTA.
Millimoles of CuSO4 added to each test tube = (5.00 mL)*(0.20 mol/L) = 1.00 mmole.
Volume of the solution in each test tube = 20.00 mL; therefore, molar concentration of CuSO4 in the first test tube (where no reaction occurs with H4EDTA) = (1.00 mmole)/(20.00 mL) = 0.05 mol/L.
Let the absorbance of the CuSO4 be related to the concentration as A = K*x ……(1)
where K = proportionality constant. Given A = 1.00 when x = 0.05 mol/L, we have
1.00 = K*(0.05 mol/L)
===> K = (1.00)/(0.05 mol/L) = 20 L/mol.
The value of K will remain the same for all the solutions (since it depends only on the identity of the substance being analyzed).
The reaction between CuSO4 and H4EDTA takes place in all the other test tubes and we do not have 1.00 mmole CuSO4 in these test tubes. We need to find out the millimoles and hence, concentration of unreacted CuSO4 in each of these test tubes.
Take the second entry as an example. We add 2.50 mL of 0.10 mol/L H4EDTA to the second test tube.
Millimoles of H4EDTA added to the second test tube = (2.50 mL)*(0.10 mol/L) = 0.25 mmole.
CuSO4 reacts with H4EDTA on a 1:1 molar ratio; therefore, 0.25 mmole of H4EDTA reacted with 0.25 mmole of CuSO4. Millimoles of unreacted CuSO4 = (1.00 – 0.25) mmole = 0.75 mmole.
Concentration of CuSO4 in test tube 2 = (0.75 mmole)/(20.00 mL) = 0.0375 mol/L.
Use the expression (1) to find out the absorbance of solution 2.
A = (20 L/mol)*(0.0375 mol/L) = 0.75
Fill in the table as below.
Test Tube#
Volume of 0.20 mol/L CuSO4 added (mL)
Volume of 0.10 mol/L H4EDTA added (mL)
Millimoles of unreacted CuSO4 (mmole)
Concentration of unreacted CuSO4 (mol/L)
Absorbance
1
5.00
0.00
1.00
0.0500
1.00
2
5.00
2.50
0.75
0.0375
0.75
3
5.00
5.00
0.50
0.0250
0.50
4
5.00
10.00
0.00
0.0000
0.00
5
5.00
15.00
0.00 (H4EDTA is in excess)
0.0000
0.00 (H4EDTA is colorless)
Plot the absorbance against the volume of H4EDTA as below.
Test Tube#
Volume of 0.20 mol/L CuSO4 added (mL)
Volume of 0.10 mol/L H4EDTA added (mL)
Millimoles of unreacted CuSO4 (mmole)
Concentration of unreacted CuSO4 (mol/L)
Absorbance
1
5.00
0.00
1.00
0.0500
1.00
2
5.00
2.50
0.75
0.0375
0.75
3
5.00
5.00
0.50
0.0250
0.50
4
5.00
10.00
0.00
0.0000
0.00
5
5.00
15.00
0.00 (H4EDTA is in excess)
0.0000
0.00 (H4EDTA is colorless)
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