Quizzes Announcements 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 website http://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 complexing agent, such as Calmagite and Eriochrome Black T

Flashback: Week 2

Flashback: Week 4

Metal Complexation Formation Constants Barium-MTB Complexation Equilibrium: Kf is the Ba-MTB Formation Constant Kf 10 5

Alpha Fractions for free and complexed Ba 2+ The total Barium Concentration in solution is conserved. Alpha Fractions!

Alpha Fractions for free and complexed Ba 2+ 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 M n+, and attenuate O 2 oxidation chemistry with your food

Used as a food preservative to sequester M n+, and attenuate O 2 oxidation chemistry with your food At what ph does this predominantly exist?

Note: K i terms are also sometimes written as β values That is, β 1 = K 1, and β 4 = K 1 K 2 K 3 K 4

α Y 4- ph Most EDTA titrations are performed at ph >= 10.

formation constant

formation constant α α α

formation constant α α ph dependent α For experiments, this conditional formation constant (K f ) is helpful because you do not need to know the free concentration of ligand to calculate K f or concentration of metal species

formation constant Kf should be at least 106 to get a sharp end point with 10 mm metal ion solution (and thus closest to the equivalence point)

50 ml of 10 mm solutions of various cations at ph 6.0 Kf should be at least 106 to get a sharp end point with 10 mm metal ion solution (and thus closest to the equivalence point)

EDTA titrations for metal ions pca pca or pmg can be used to determine the titration endpoint.

What if we have both Ca 2+ and Mg 2+ present?

Precipitation Calculation for EDTA Titration of Ca 2+ and Mg 2+ What if we have both Ca 2+ and Mg 2+ present?

Precipitation Calculation for EDTA Titration of Ca 2+ and Mg 2+ 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 Mg 2+

Hydroxyquinoline: a metal chelator that fluoresces upon binding! 8-hydroxyquinoline Trivalent Cations Mg Divalent Cations

Hydroxyquinoline: a metal chelator that fluoresces upon binding! Mg 2+ 8-hydroxyquinoline-5-sulfonic Acid Fluorometric Detection of Mg 2+ in Seawater

Kf = 10 16.5

K f = 10 16.5 This is key, because not only does NH 3 prevent precipitation but it also complexes some Zn 2+. Cool!

αy4- = 0.355 at ph =10. We use an ammonia buffer solution for a reason...

Recall: Pourbaix Diagrams Marcel Pourbaix (1904 1998) http://corrosion doctors.org/biographies/pourbaixbio.htm from Pourbaix, Atlas of electrochemical equilibria in aqueous solutions, 1974

Zinc Hydroxide Zn 2+ + 2OH - Zn(OH)2(s) Solubility product Ksp = 3.0 10 17 Ksp = [Zn 2+ ][OH - ] 2 At ph=10, [Zn 2+ ] =??

Zinc Hydroxide Zn 2+ + 2OH - Zn(OH)2(s) Solubility product Ksp = 3.0 10 17 Ksp = [Zn 2+ ][OH - ] 2 At ph=10, [Zn 2+ ] = Ksp /[OH - ] 2

Zinc Hydroxide Zn 2+ + 2OH - Zn(OH)2(s) Solubility product Ksp = 3.0 10 17 Ksp = [Zn 2+ ][OH - ] 2 At ph=10, [Zn 2+ ] = 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 - pkb = 4.74 [NH3]total = 0.100M At ph=10, [NH3] =??

Ammonia Buffer Solution NH3 + H2O <=> NH4 + + OH - pkb = 4.74 [NH3]total = 0.100M At ph=10: [NH3] = 0.0846 M

And recall, we think we have: [Y 4 ] = 3.55 x 10 5 M [Zn 2+ ] 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: [Zn 2+ ] = αzn2+ [Zn 2+ ] tot [Zn 2+ ] = (1.61 x 10-5 )(1.00 x 10-4 ) [Zn 2+ ] = 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 Zn 2+ solution with a 1.00 x 10-4 M EDTA solution. What is pzn at the equivalence point? Log Kf for the ZnY 2- complex is 16.5. Both solutions are buffered to a ph of 10.0 using a 0.100M ammonia buffer. The alpha fraction for Y 4- is 0.355 at a ph of 10.0. The alpha fraction for Zn 2+ 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 Zn 2+ 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 Zn 2+ 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 Zn 2+ 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 Zn 2+ 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 Zn 2+ 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 [ ZnY 2- ] =??

EDTA Titration You would like to perform a titration of 50.00 ml of a 1.00 x 10-4 M Zn 2+ 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 [ ZnY 2- ] = 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 Zn 2+ 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 [ ZnY 2- ] = 5.0 x 10-5 M [ Zn 2+ ] =??

[ ZnY 2- ] = 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

[ ZnY 2- ] = 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 Zn 2+ 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 [ ZnY 2- ] = 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.

Recall: Pourbaix Diagrams [Zn 2+ ] [ZnO 2 2 ]?? Marcel Pourbaix (1904 1998) http://corrosion doctors.org/biographies/pourbaixbio.htm Because K sp = 3.0 x 10 17 Because Zn(OH) 2 can (twice) deprotonate too! from Pourbaix, Atlas of electrochemical equilibria in aqueous solutions, 1974

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 complexing agent, such as Calmagite and Eriochrome Black T