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) I 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 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) I 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 ii) mmoles of EDTA added is iii) mmoles of Mg*2 reacting with EDTA
Chapter16: Applications Of Neutralization Titrations
Section: Chapter Questions
Problem 16.20QAP
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Question
![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 MEDTA ha
been added? (Hint:
See Exercise 15-B.)
i)
Total mmoles of Mg+2
ii)
mmoles of EDTA added is
iii)
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](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F98d72326-249a-4418-b782-a225b06449c0%2Ffdb29fe2-19e2-4f85-98fe-e34ded28247c%2Faqtpr2f_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 MEDTA ha
been added? (Hint:
See Exercise 15-B.)
i)
Total mmoles of Mg+2
ii)
mmoles of EDTA added is
iii)
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
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