Zinc-EDTA Titration
You would like to perform a titration of 50.00 mL of a
1.00 x 10-4 M Zn2+ solution with a 1.00 x 10-4 M EDTA
solution.
What is pZn at the equivalence point?
Log Kf for the ZnY2- complex is 16.5.
Both solutions are buffered to a pH of 10.0 using
a 0.100M ammonia buffer.
R. Corn
Chem M3LC
Spring 2013
αY4- = 0.355 at pH =10.
We use an ammonia buffer solution
for a reason...
Zinc Hydroxide
Zn2+ + 2OH- → Zn(OH)2(s)
Solubility product
Ksp = 3.0×10−17
Ksp = [Zn2+][OH -]2
At pH=10, [Zn2+] = ??
Zinc Hydroxide
Zn2+ + 2OH- → Zn(OH)2(s)
Solubility product
Ksp = 3.0×10−17
Ksp = [Zn2+][OH -]2
At pH=10, [Zn2+] = Ksp /[OH -]2
Zinc Hydroxide
Zn2+ + 2OH- → Zn(OH)2(s)
Solubility product
Ksp = 3.0×10−17
Ksp = [Zn2+][OH -]2
At pH=10, [Zn2+] = Ksp /[OH -]2
= (3.0×10−17) /(1.0×10−4)2
= 3.0×10−9 M
This is the maximum free Zinc concentration.
Ammonia Buffer Solution
NH3 + H2O <=> NH4+ + OH [NH3]total = 0.100M
At pH=10, [NH3] = ??
pKb = 4.74
Ammonia Buffer Solution
NH3 + H2O <=> NH4+ + OH -
pKb = 4.74
[NH3]total = 0.100M
At pH=10:
[NH3] = 0.0846 M
Zinc - Ammonia Complexation
At [NH3] = 0.0846M, αZn2+ = 1.61 x 10-5
Zinc - Ammonia Complexation
At a TOTAL Zinc concentration of 1.00 x 10-4 M:
[Zn2+] = αZn2+ [Zn2+]tot
[Zn2+] = (1.61 x 10-5)(1.00 x 10-4)
[Zn2+] = 1.61 x 10-9 M
Therefore: no precipitation!
Once more, with feeling:
EDTA Titration
You would like to perform a titration of 50.00 mL of a
1.00 x 10-4 M Zn2+ solution with a 1.00 x 10-4 M EDTA
solution.
What is pZn at the equivalence point?
Log Kf for the ZnY2- complex is 16.5.
Both solutions are buffered to a pH of 10.0 using
a 0.100M ammonia buffer.
The alpha fraction for Y4- is 0.355 at a pH of 10.0.
The alpha fraction for Zn2+ is 1.61 x 10-5.
EDTA Titration
You would like to perform a titration of 50.00 mL of a
1.00 x 10-4 M Zn2+ solution with a 1.00 x 10-4 M EDTA
solution.
Total moles of Zinc = ??
EDTA Titration
You would like to perform a titration of 50.00 mL of a
1.00 x 10-4 M Zn2+ solution with a 1.00 x 10-4 M EDTA
solution.
Total moles of Zinc = (1.00 x 10-4 M)(0.050L)
= 5.0 x 10-6 moles
EDTA Titration
You would like to perform a titration of 50.00 mL of a
1.00 x 10-4 M Zn2+ solution with a 1.00 x 10-4 M EDTA
solution.
Total moles of Zinc = 5.0 x 10-6 moles
Equivalence point volume = ??
EDTA Titration
You would like to perform a titration of 50.00 mL of a
1.00 x 10-4 M Zn2+ solution with a 1.00 x 10-4 M EDTA
solution.
Total moles of Zinc = 5.0 x 10-6 moles
Equivalence point volume = 0.100L
EDTA Titration
You would like to perform a titration of 50.00 mL of a
1.00 x 10-4 M Zn2+ solution with a 1.00 x 10-4 M EDTA
solution.
Total moles of Zinc = 5.0 x 10-6 moles
Equivalence point volume = 0.100L
[ ZnY2- ] = ??
EDTA Titration
You would like to perform a titration of 50.00 mL of a
1.00 x 10-4 M Zn2+ solution with a 1.00 x 10-4 M EDTA
solution.
Total moles of Zinc = 5.0 x 10-6 moles
Equivalence point volume = 0.100L
[ ZnY2- ] = 5.0 x 10-5 M
We assume a stoichiometric reaction.
EDTA Titration
You would like to perform a titration of 50.00 mL of a
1.00 x 10-4 M Zn2+ solution with a 1.00 x 10-4 M EDTA
solution.
Total moles of Zinc = 5.0 x 10-6 moles
Equivalence point volume = 0.100L
[ ZnY2- ] = 5.0 x 10-5 M
[ Zn2+ ] = ??
[ ZnY2- ] = 5.0 x 10-5 M
We assumed a stoichiometric reaction. But actually, there
is a little bit of free (uncomplexed) EDTA and free
(uncomplexed) Zinc in solution.
[ ZnY2- ] = 5.0 x 10-5 M
We assumed a stoichiometric reaction. But actually, there
is a little bit of free (uncomplexed) EDTA and free
(uncomplexed) Zinc in solution.
Log Kf = 16.5.
αY4- = 0.355
αZn2+ = 1.61 x 10-5
[ ZnY2- ] = 5.0 x 10-5 M
We assumed a stoichiometric reaction. But actually, there
is a little bit of free (uncomplexed) EDTA and free
(uncomplexed) Zinc in solution.
1.66 x 10-8 M
EDTA Titration
You would like to perform a titration of 50.00 mL of a
1.00 x 10-4 M Zn2+ solution with a 1.00 x 10-4 M EDTA
solution.
Total moles of zinc = 5.0 x 10-6 moles
Equivalence point volume = 0.100L
[ ZnY2- ] = 5.0 x 10-5 M
1.66 x 10-8 M
EDTA Titration
You would like to perform a titration of 50.00 mL of a
1.00 x 10-4 M Zn2+ solution with a 1.00 x 10-4 M EDTA
solution.
Total moles of zinc = 5.0 x 10-6 moles
Equivalence point volume = 0.100L
2.35 x 10-6 M
[ ZnY2- ] = 5.0 x 10-5 M
= (1.61 x 10-5) (1.66 x 10-8 M)
[ Zn2+ ] = 2.59 x 10-13 M
pZn = 12.6
We are done!