The titration solution in the cell below had a total volume of 50.0 mL and contained 0.100 M Mg2+ and 1.00 x 10-5 M Zn(EDTA)2- at a pH of 10.00 s.C.E. || Zn(s) | titration solution Zn2* + 2e- Zn(s) E' = -0.762 v Mgy2-: Kf = 6.2 x 108; Zny2-: Kf = 3.2 x 1016. What will be the cell voltage when 10.0 mL of 0.100 M EDTA has been added? (Hint: Se Exercise 15-B.) i) Total mmoles of Mg*2 ii) mmoles of EDTA added is iii) mmoles of Mg*+2 reacting with EDTA
The titration solution in the cell below had a total volume of 50.0 mL and contained 0.100 M Mg2+ and 1.00 x 10-5 M Zn(EDTA)2- at a pH of 10.00 s.C.E. || Zn(s) | titration solution Zn2* + 2e- Zn(s) E' = -0.762 v Mgy2-: Kf = 6.2 x 108; Zny2-: Kf = 3.2 x 1016. What will be the cell voltage when 10.0 mL of 0.100 M EDTA has been added? (Hint: Se Exercise 15-B.) i) Total mmoles of Mg*2 ii) mmoles of EDTA added is iii) mmoles of Mg*+2 reacting with EDTA
Chemistry
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Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
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![The titration solution in the cell below had a total volume of 50.0 ml and contained
0.100 M Mg2+ and 1.00 x 10-5 M Zn(EDTA)2- at a pH of 10.00
s.C.E. || Zn(s) | titration solution
Zn2* + 2e- Zn(s) E' = -0.762 v
MgY2-: Kf = 6.2 x 108; Zny2- : Kf = 3.2 x 1016.
What will be the cell voltage when 10.0 mL of 0.100 M EDTA has been added? (Hint: See
Exercise 15-B.)
i)
Total mmoles of Mg*2
mmoles of EDTA added is
mmoles of Mg*2 reacting with EDTA
iv)
mmoles of MgY2-
v)
Concentration of MgY2- in M
vi)
mmoles of excess unreacted Mg+2
vii)
Concentration of Mg*2 in M
viii)
Concentration of EDTA (Y4-) in M
ix)
Total mmoles of Zn+2{1:NUMERICAL:%100%0.500:0.15}
x)
mole of Zny2- in M
xi)
Concentration of ZnY2- in M
xii)
Concentration of Zn2+ in M
xiii)
Potential of the cell in voltage, Ecell={1:NUMERICAL-1.364:0.1}
13 marks](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fdec85192-a7d4-488d-b207-64fd588d19d3%2F1b66a640-48fe-48b3-9470-431c0a34180c%2Fpn8f0f_processed.jpeg&w=3840&q=75)
Transcribed Image Text:The titration solution in the cell below had a total volume of 50.0 ml and contained
0.100 M Mg2+ and 1.00 x 10-5 M Zn(EDTA)2- at a pH of 10.00
s.C.E. || Zn(s) | titration solution
Zn2* + 2e- Zn(s) E' = -0.762 v
MgY2-: Kf = 6.2 x 108; Zny2- : Kf = 3.2 x 1016.
What will be the cell voltage when 10.0 mL of 0.100 M EDTA has been added? (Hint: See
Exercise 15-B.)
i)
Total mmoles of Mg*2
mmoles of EDTA added is
mmoles of Mg*2 reacting with EDTA
iv)
mmoles of MgY2-
v)
Concentration of MgY2- in M
vi)
mmoles of excess unreacted Mg+2
vii)
Concentration of Mg*2 in M
viii)
Concentration of EDTA (Y4-) in M
ix)
Total mmoles of Zn+2{1:NUMERICAL:%100%0.500:0.15}
x)
mole of Zny2- in M
xi)
Concentration of ZnY2- in M
xii)
Concentration of Zn2+ in M
xiii)
Potential of the cell in voltage, Ecell={1:NUMERICAL-1.364:0.1}
13 marks
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