Zinc-EDTA Titration You would like to perform a titration of 50.00 mL of [PDF]

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!