Announcements• Quizzes They are all graded and may be up on Sapling Your “final” quiz will be on the remainder of 

information in the class and tentatively will be given via Sapling over Wed., 6/4 – Sun., 6/8 (Week 10)

• Reading updates Chapter 9, Section B5 is on ISEs Chapter 17, all of Section D is on EDTA Chapter 26, Sections A4 and A5

Course websitehttp://faculty.sites.uci.edu/chem2l/chem-h2lcm3lc/

Syllabus & Sapling Honesty Agreement

The power of turnitin.com; an example

The power of turnitin.com; an example

The power of turnitin.com; an example

Week 7 Objectives

By the end of the week, you will be able to:

• Explain the properties of EDTA and the specific experimental conditions required to use EDTA effectively

• Perform calculations involving EDTA chelation

• Explain the purpose of an auxiliary complexingagent, such as Calmagite and Eriochrome Black T

Flashback:Week 2

Flashback:Week 4

Barium-MTB Complexation Equilibrium:

Kf is the Ba-MTB Formation Constant

Metal Complexation Formation Constants

Kf ≈ 105

Alpha Fractions for free and complexed Ba2+

The total Barium Concentration in solution is conserved.

Alpha Fractions!

Alpha Fractions for free and complexed Ba2+

The ligand concentration ([MTB]) determines the speciation.

Metal Complex Formation Constants

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

Therefore, many species co-exist in solution!

Used as a food preservative to sequester Mn+, and attenuate O2 oxidation chemistry with your food

At what pH does this predominantly exist?

Used as a food preservative to sequester Mn+, and attenuate O2 oxidation chemistry with your food

Note: Ki terms are also sometimes written as β values

That is, β1 = K1, and β4 = K1K2K3K4

pH

Most EDTA titrations are performed at pH >= 10.

αY4-

formationconstant

formationconstant

α

α

′ α

formationconstant

α

α

′ α

For experiments, this conditional formation constant (Kf’) is helpful because you do not need to know the free concentration of ligand to calculate Kf or concentration of metal species

pH dependent

formationconstant

Kf’ should be at least 106 to get a sharp end point with 10 mMmetal ion solution (and thus closest to the equivalence point)

Kf’ should be at least 106 to get a sharp end point with 10 mMmetal ion solution (and thus closest to the equivalence point)

50 mL of 10 mM solutions of variouscations at pH 6.0

EDTA titrations for metal ions

pCa or pMg canbe used to determine the titration endpoint.

pCa

What if we have both Ca2+ and Mg2+ present?

Precipitation Calculation for EDTA Titration of Ca2+ and Mg2+

What if we have both Ca2+ and Mg2+ present?

Precipitation Calculation for EDTA Titration of Ca2+ and Mg2+

“only”

EDTA titrations for metal ions

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

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+

Kf = 10‐16.5

Kf = 10‐16.5

This is key, because not only does NH3 prevent precipitation but it also complexes some Zn2+. Cool!

We use an ammonia buffer solutionfor a reason...

αY4- = 0.355 at pH =10.

Marcel Pourbaix(1904–1998)

http://corrosion‐doctors.org/Biographies/PourbaixBio.htm

Recall:Pourbaix Diagrams

from Pourbaix, Atlas of electrochemical equilibria in aqueous solutions, 1974

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

And recall, we think we have:[Y4–] = 3.55 x 10‐5 M[Zn2+]max = 3.0 x 10‐9 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!

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.

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

Feature 17.5 in Skoog, et al., goes over a similar problem, but via a slightly different method.

from Pourbaix, Atlas of electrochemical equilibria in aqueous solutions, 1974

Marcel Pourbaix(1904–1998)

http://corrosion‐doctors.org/Biographies/PourbaixBio.htm

Recall:Pourbaix Diagrams

Because Zn(OH)2 can (twice) deprotonate too!

? ?

Because Ksp = 3.0 x 10‐17

[Zn2+][ZnO2

2–]

Week 7 Objectives

By the end of the week, you will be able to:

• Explain the properties of EDTA and the specific experimental conditions required to use EDTA effectively

• Perform calculations involving EDTA chelation

• Explain the purpose of an auxiliary complexingagent, such as Calmagite and Eriochrome Black T