EDTA Titrations

We already used metal complexation in Week 2 for the 
Fe Colorimetry Experiment:

Ferrozine (Fz2-)is a metal ligand

Fe2+ + 3 Fz2- => Fe(Fz)34-

Three Ferrozine will form a 
metal-ligand complex with Fe2+

Chem M3LCR. Corn

Metal Complex Formation Constants

Weak Metal Complex Formation Constants have similar values of K1 to Kn

Therefore, many species co-exist in solution!

EDTA Metal Ion Complexation Equilibria

Y4-

Ethylene Diamine Tetra-acetic Acid (H4Y)

“Chelate”

EDTA - the world’s best metal ion chelator

pH

Most EDTA titrations are performed at pH >= 10.

αY4-

EDTA titrations are always run in basic solutions to insure that there is enough Y4- species

EDTA titrations for metal ions

pCa or pMg canbe used to determine the titration endpoint.

pCa

Sadly, we cannotmeasure pCadirectly in our M3LC lab.😞

We must use an indicator to detect the endpoint of the EDTA titration

EDTA will displace weaker ligands -- you will use this process with calmagite to determine the endpoint of an EDTA titration for Mg2+

MgIn− + Y4− ⇌ MgY2− + In3−

(wine-red) (colorless) (??)*

In3− ⇌ HIn2− ⇌ H2In− ⇌ H3In+H+ +H++H+

(blue)(orange)

*what you see will depend on pH

Zn-EDTA Titration: Equivalence Point Calculation

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.The alpha fraction for Y4- is 0.355 at a pH of 10.0.

Total moles of Zinc = (1.00 x 10-4 M)(0.050L)

= 5.0 x 10-6 moles

Zn-EDTA Titration: Equivalence Point Calculation

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

Zn-EDTA Titration: Equivalence Point Calculation

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.

Equivalence point volume = 0.100L

Total moles of Zinc = 5.0 x 10-6 moles

Equivalence point volume = 0.100L

[ ZnY2- ] = 5.0 x 10-5 M

We first assume a stoichiometric reaction.

Zn-EDTA Titration: Equivalence Point Calculation

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+ ] = ??

Then we assume a little bit of free Zinc is formed as required by Kf.

Kf =[ZnY2−]

[Zn2+][Y4−]

Zn-EDTA Titration: Equivalence Point Calculation

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

For every free Zn2+, there is a free EDTA species

Y4- concentration is obtained from the alpha fraction.

Kf =[ZnY2−]

[Zn2+][Y4−]

[Zn2+] = CtotEDTA[Y4−] = αY4−CtotEDTA

Zn-EDTA Titration: Equivalence Point Calculation

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+ ] = ??

Plug everything in and solve for Zn2+...

[Zn2+] =[ZnY2−]αY4−Kf

Zn-EDTA Titration: Equivalence Point Calculation

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+ ] = 6.67 x 10-11 M

pZn = 10.2

And we are done!

[Zn2+] =[ZnY2−]αY4−Kf

αY4−Kf = K′�fconditional formation constant

Zn-EDTA Titration: Equivalence Point Calculation

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.

a) How much total free Zn is left in solution?

b) What is the Zn2+ concentration?

Zinc-EDTA Titration Calculation #2 (with ammonia buffer)

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.

Both solutions are buffered to a pH of 10.0 using a 0.100M ammonia buffer.

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

= (3.0×10−17) /(1.0×10−4)2

= 3.0×10−9 M

This is the maximum free Zinc concentration.

Zinc Hydroxide

Ammonia Buffer Solution

At pH=10, [NH3] = ??

[NH3]total = 0.100M

NH3 + H2O NH4+ + OH - pKb = 4.74

At pH=10:

[NH3]total = 0.100M

NH3 + H2O NH4+ + OH -

[NH3] = 0.0846 M

pKb = 4.74

Ammonia Buffer Solution

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!

a) How much total free Zn is left in solution?

b) What is the Zn2+ concentration?

Zinc-EDTA Titration Calculation #2 (with ammonia buffer)

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.

Both solutions are buffered to a pH of 10.0 using a 0.100M ammonia buffer.

Zinc-EDTA Titration Calculation #2 (with ammonia buffer)

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.

Both solutions are buffered to a pH of 10.0 using a 0.100M ammonia buffer.

log Kf = 16.5

αZn2+ = 1.61 × 10−5αY4− = 0.355At pH=10:

For ZnY4-

Zinc-EDTA Titration Calculation #2 (with ammonia buffer)

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.

log Kf = 16.5

αZn2+ = 1.61 × 10−5αY4− = 0.355

Kf =[ZnY2−]

[Zn2+][Y4−]

Zinc-EDTA Titration Calculation #2 (with ammonia buffer)

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.

log Kf = 16.5

αZn2+ = 1.61 × 10−5αY4− = 0.355

Kf =[ZnY2−]

[Zn2+][Y4−][ZnY2−] = 5.0 × 10−5 M

a) How much total free Zn is left in solution?

Zinc-EDTA Titration Calculation #2 (with ammonia buffer)

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.

log Kf = 16.5

αZn2+ = 1.61 × 10−5αY4− = 0.355

Kf =[ZnY2−]

[Zn2+][Y4−][ZnY2−] = 5.0 × 10−5 M

a) How much total free Zn is left in solution?

CtotZn = CtotEDTA

[Zn2+] = αZn2+CtotZn[Y4−] = αY4−CtotEDTA

CtotZn =[ZnY2−]

αZn2+αY4−Kf

Zinc-EDTA Titration Calculation #2 (with ammonia buffer)

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.

log Kf = 16.5

αZn2+ = 1.61 × 10−5αY4− = 0.355

Kf =[ZnY2−]

[Zn2+][Y4−][ZnY2−] = 5.0 × 10−5 M

a) How much total free* Zn is left in solution?

CtotZn =[ZnY2−]

αZn2+αY4−KfCtotZn = 1.66 × 10

−8 M

αZn2+αY4−Kf = K′�′�f *free means not in the ZnY2- complex

double conditional formation constant

Zinc-EDTA Titration Calculation #2 (with ammonia buffer)

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.

log Kf = 16.5

αZn2+ = 1.61 × 10−5αY4− = 0.355

Kf =[ZnY2−]

[Zn2+][Y4−][ZnY2−] = 5.0 × 10−5 M

CtotZn = 1.66 × 10−8 Mb) What is the Zn2+ concentration?

Zinc-EDTA Titration Calculation #2 (with ammonia buffer)

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.

log Kf = 16.5

αZn2+ = 1.61 × 10−5αY4− = 0.355

Kf =[ZnY2−]

[Zn2+][Y4−][ZnY2−] = 5.0 × 10−5 M

CtotZn = 1.66 × 10−8 Mb) What is the Zn2+ concentration?

[Zn2+] = αZn2+CtotZn = (1.61 × 10−5)(1.66 × 10−8 M)

[Zn2+] = 2.67 × 10−13 M pZn = 12.6

Hydroxyquinoline: a metal chelator that fluoresces upon binding!

Mg

Trivalent Cations

Divalent Cations

8-hydroxyquinoline

Hydroxyquinoline: a metal chelator that fluoresces upon binding!

Fluorometric Detection of Mg2+ in Seawater

8-hydroxyquinoline-5-sulfonic Acid

Mg2+