Lecture 37. Complexometric Titration

Lecture 37. Complexometric Titration - Complexometric...

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Unformatted text preview: Complexometric Titrations (Chapter 13) 13) E.g. Calcium Titration with EDTA http://www.chem.ualberta.ca/~iip/chem211irc/TitrationVideo.html http://www.chem.ualberta.ca/~iip/chem211irc/TitrationVideo.html EDTA Calcium Titration with EDTA Calcium n Ca+ + (n+m) EDTA4- + m [MgIn]+ n [CaEDTA]2- + m [MgEDTA]2- + m In- In- (Eriochrome T) 3- n [H 7- n Y ] pH = pK a ,n - log 2- n [H 6- n Y ] The pKa,n always refers to the acid of the conjugate pair. The Calcium Titration with EDTA Calcium Ca+ + EDTA4- [CaEDTA]2Remember: Lecture 10, Ca2+ Buffer Lecture • Mass balance Ca2+: Mass CaTotal = [Ca] + [EDTACa] • Mass balance EDTA: EDTATotal = [EDTA] + [EDTACa] Mass • Kf = [EDTACa] / ( [EDTA] [Ca] ) [EDTA] [Ca] => Kf = [EDTACa] / ( α6 [sum of uncomplexed EDTA] [Ca] ) [sum [Ca] Kf α6 = Kf’ = [EDTACa] / ([sum of uncomplexed EDTA] [Ca] ) [Ca] Conditional Formation Constant: Correct for the condition Conditional Conditional Formation Constant Kf α6 = Kf’ Kf Formation Constant Conditional Formation Constant Example: Mg2+ Titration with EDTA. Kf’ at pH 14 and 5? Example: Calcium Titration with EDTA Calcium Ca+ + EDTA4- [CaEDTA]2Remember: Lecture 10, Ca2+ Buffer Lecture • Mass balance Ca2+: Mass • Mass balance EDTA: Mass CaTotal = [Ca] + [EDTACa] EDTATotal = [EDTA] + [EDTACa] Kf α6 = Kf’ = [EDTACa] / ([sum of uncomplexed EDTA] [Ca] ) [Ca] => => cCaTotal - [Ca 2 + ] 1 [Ca ] = ' K f c EDTATotal - cCaTotal + [Ca 2 + ] 2+ => Solve for [Ca2+] {a 2nd order equation} => Calcium Titration with EDTA Calcium E.g.: CaTotal = 1.0 mM [Ca2+] 0.0010 0.0008 Kf’ = 102 M-1 0.0006 0.0004 104 M-1 106 M1 0.0008 0.0010 108 MEDTATotal 1 0.0002 0.0000 0.0000 0.0002 0.0004 0.0006 Calcium Titration with EDTA Calcium E.g.: CaTotal = 1.0 mM log[Ca2+- 3.0 ] - 3. 5 - 4. 0 - 4. 5 - 5. 0 - 5. 5 - 6. 0 0.0000 Kf’ = 102 M-1 104 M-1 106 M1 108 M1 10 M10 0.0002 0.0004 0.0006 0.0008 0.0010 1 EDTATotal * In this case: For >108 M-1, there is very little difference in log[Ca]. * What type of endpoint detection would give a logarithmic output? Titration with EDTA Titration E.g.: CaTotal = 1.0 mM; EDTATotal = 0.999 mM 0.999 Total [Ca2+]/CaT 1.0 otal 0.8 0.6 0.4 0.2 0.0 0 2 4 6 8 10 12 Log Kf’ Conditional Formation Constant Kf α6 = Kf’ Kf Formation Constant Standard Reduction Potentials (Standard Conditions) (Standard ∴ Ag + + e _ ¾ ® Ag 0 ¾ 59.16 mV 1 0 E = E Ag + / Ag o log 1 aAg + AgCl + e _ ¾ ® Ag 0 + Cl¾ E=E 0 AgCl ( solid ) / Ag o 0 ; E Ag + / Ag o = 0.799 V 0 i.e., E Ag + / Ag o = E for aAg + = 1 M \ aCl _ 59.16 mV log 1 1 0 ; E AgCl ( solid ) / Ag o = 0.222 V 0 i.e., E AgCl ( solid ) / Ag o = E for aCl _ = 1 M Formal (Conditional) Reduction Potentials (p 319) Formal (p (Special Conditions) ∴ AgCl + e _ ¾ ® Ag 0 + Cl¾ a_ 59.16 mV 0' E = E AgCl ,KClsat / Ag o log Cl 1 1 i.e., E 0' 0' ; E AgCl ,KClsat / Ag o = 0.197 V AgCl , KClsat / Ag o = E for a saturated KCl solution Formal (Conditional) Reduction Potentials (“reduction potentials for non-standard conditions”) (“reduction ∴ AgCl + e _ ¾ ® Ag 0 + Cl¾ E=E ' 0 AgCl , KClsat / Ag o aCl _ 59.16 mV log 1 1 i.e., E 0' 0' ; E AgCl ,KClsat / Ag o = 0.197 V AgCl , KClsat / Ag o = E for a saturated KCl solution \ Fe 3 + + e _ ¾ ® Fe 2 + ¾ 0 E Fe 3+ /Fe2+ = 0.771 V 0' E Fe 3+ /Fe2+ = 0.732 V (1 M HCl) 0' E Fe 3+ /Fe2+ = 0.767 V (1 M HClO 4 ) 0' E Fe 3+ /Fe2+ = 0.746 V (1 M HNO 3 ) 0' E Fe 3+ /Fe2+ = 0.68 V (0.5 M H 2SO 4 ) Concentrations; Concentrations; not activities not pH Dependence of Calcium Titration with EDTA Titration E.g.: CaTotal = 1.0 mM, pH = 12, 10, 9, 8 E.g.: log[Ca] -4 -6 -8 - 10 - 12 0.0000 0.0005 0.0010 0.0015 0.0020 EDTATotal * Which titration curve is for which pH? ...
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This note was uploaded on 04/06/2011 for the course CHEM 2011 taught by Professor Buhlman during the Spring '10 term at Minnesota.

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