L6 - Chemistry 15 Lecture 6 Reading Assignment this week:...

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Chemistry 15 February 11, 2008 Lecture 6 Reading Assignment this week: Chapter 8 Polyprotic Acid/Base Systems a) Diprotic acid dissociation: (I will use H + instead of H 3 O + to save space) 2 HA H +− + R 1 2 [] [ ] a HH A K + = 2 HA H A + R 2 2 a K HA = Notice that the base dissociation of A 2- is denoted K b1 and since it is conjugated with HA - we can write: 21 wa b KKK = × . Also, for the same reason: 12 wab for the conjugated pair, H 2 A and HA - . b) We want to understand how the composition of a diprotic (polyprotic) acid changes with pH. In view of a general materials balance: 2 2 [ ] T CH A H A A −− =++ , we want to know what is the fractional solution composition as a function of pH: 2 0 ; T C α = 1 ; T HA C = 2 2 ; T A C = c) Derivation of the equations for n begins with the expressions for K a1 and K a2 above. We use these to calculate the concentrations of HA - and A 2- : (1) a KHA HA H + = and 2 2 2 2 aa a KH A KKH A A ++ == (2) Now, let's use the C T equation: 122 2 2 a T KHA KKHA A =+ + (3) We can factor our [H 2 A] : 11 2 2 2 1 a T KK K A
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This note was uploaded on 02/04/2010 for the course CHEM 15 taught by Professor Staff during the Fall '08 term at University of California, Berkeley.

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L6 - Chemistry 15 Lecture 6 Reading Assignment this week:...

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