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

We use an ammonia buffer solutionfor a reason...

αY4- = 0.355 at pH =10.

Solubility product Ksp = 3.0×10−17

Zn2+ + 2OH- → Zn(OH)2(s)

Ksp = [Zn2+][OH -]2

At pH=10, [Zn2+] = ??

Zinc Hydroxide

Solubility product Ksp = 3.0×10−17

Zn2+ + 2OH- → Zn(OH)2(s)

Ksp = [Zn2+][OH -]2

At pH=10, [Zn2+] = Ksp /[OH -]2

Zinc Hydroxide

Solubility product Ksp = 3.0×10−17

Zn2+ + 2OH- → Zn(OH)2(s)

Ksp = [Zn2+][OH -]2

At pH=10, [Zn2+] = Ksp /[OH -]2

Zinc Hydroxide

= (3.0×10−17) /(1.0×10−4)2

= 3.0×10−9 M

This is the maximum free Zinc concentration.

Ammonia Buffer Solution

At pH=10, [NH3] = ??

[NH3]total = 0.100M

NH3 + H2O <=> NH4+ + OH - pKb = 4.74

Ammonia Buffer Solution

At pH=10:

[NH3]total = 0.100M

NH3 + H2O <=> NH4+ + OH -

[NH3] = 0.0846 M

pKb = 4.74

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!

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.

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.

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.

αZn2+ = 1.61 x 10-5

Log Kf = 16.5. αY4- = 0.355

[ 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

[ 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!

2.35 x 10-6 M