Effect of acid rain on erosion of limestone/marble Calcite---CaCO 3(s) <------> Ca +2 + CO 3

Transcription

1 inside outside Le Chatelier s Princ. at Work every day! Effect of acid rain on erosion of limestone/marble Calcite---CaC 3(s) <------> Ca +2 + C 3-2 C <------> C 3 - acid rain---shifts calcite equilibrium to right! Tooth decay--same princple (now acid from lactic acid--produced by bacteria---degrades hydroxyapatite) : Ca 10 (P 4 ) 6 () <----> 10 Ca P

2 Polyprotic acids/bases---chapter 10 ECA A <----> + + A - pk 1 2 A <-----> + + A - <------> + + A -2 ; pk 1 and pk 2 3 A <----> A -2 <----> A <-----> A ; pk 1, pk 2, pk 3 Can have also species where acidic protons come as a result of amine sites----so with proton---site is + charge---loss of proton gives neutral species: RN + 3 < RN pk a Can have combination of both types of protonated sites in same structure--amino acids!

3 Let s look at a real amino acid---which is triprotic---three different pk a values--- These systems are more complex from a titration standpoint---and also to calculate the p, given a specific concentration of one form of the molecule!

4 K a and K b relationships for such species--- Diprotic-- 2 A <------> + + A - K a1 A <------> 2 A+ - K b2 -the weaker conjugate base 2 < > therefore: K a1 K b2 = K w K w stronger conj. base A - < > + + A -2 A <------> A < > therefore: K a2 K b1 = K w K a2 K b1 K w Similarly---can derive--relationships for triprotic system: K a1 K b3 = K w ; K a2 K b2 = K w ; and K a3 K b1 = K w

5 Diprotic case in more detail---what is p of solution?---let s use simple diprotic amino acid as generic example: 3 N + C(R)C <------> 3 N + C(R)C - <------> 2 NC(R)C - 2 L + < > L < > L - pk a1 = pk a2 = thus---l <----> L + - L + 2 <-----> L K b1 K b2 What is p of solution of 2 L + salt--at 0.05 M? pk a1 = pk 1 = 2.329; K a = 4.69 x 10-3 L is much weaker acid---k a = 1.79 x So-- 2 L + will dissociate slightly to generate + in solution--- but the contribution by further dissociation of L <---> + is negligible

6 so p can be calculated treating 2 L + as monoprotic acid-- K 1 = x 2 / (F-x) ; F = 0.05 M ; x = 1.31 x 10-2 M = [L] = [ + ] Therefore: p = -log (1.31 x 10-2 ) = 1.88 What is concentration of L - in such a solution? We know that K 2 = ([L - ] [ + ]) / [L] ; [L - ] = (K 2 [L]) /[ + ] [L - ] = (1.79 x ) (1.31 x 10-2 ) / (1.31 x 10-2 ) = 1.79 x M =K 2 as long as K 1 > K 2 by at least 1-2 orders of magnitude---the assumption to neglect second dissociation reaction in calculating proton concentration is K! If solution contained L - only at 0.05 M to start--what would be p? weak base problem---need K b1 value ; K b1 = K w / K 2 = 5.59 x 10-5

7 - produced from small amount of L that forms would be negligible--- since K b2 = K w / K 1 = / 4.69 x10-3 = 2.13 x very very weak base! So p is based only on K b1 value: K b1 = x 2 / F-x ; 5.59 x 10-5 = x 2 /(0.05-x) ; x = 4.84 x 10-2 M =[ - ] = [L] [ + ] = K w / [ - ] = / 4.84 x 10-2 = 6.08 x M therefore: p = Finally--what about a 0.05 M solution of L --What is p? more complex--since L is intermediate form--it can be both a base and acid (an amphiprotic species!!!!) L <------> + + L - K 2 = 1.79 x L + 2 <-----> 2 L K b2 = 2.13 x since K 2 > K b2 ; expect solution of L to be acidic!

8 Determing p of solution containing L alone--is perfect example of solving more complex simultaneous equilibria type problems in chemistry! best starting point---use charge balance equation must have charge neutrality in solution--therefore the total concentration of all positively charged species = total concentration of all negatively charged species in solution! [ + ] + [ 2 L + ] = [L - ] + [ - ] or [ 2 L + ] - [L - ] + [ + ] - [ - ] = 0 but we have relationships that we can substitute for each of these species: [L][ + ] [L]K 2 +[ + ] K w K 1 [ + ] [ + ] = 0

9 Can solve for [ + ] (see ECA--p. 217) [ + ] = K 1 K 2 [L] K 1 K w K 1 +[L] everything is known except [L]-----but we know that while L is both an acid and base---it is weak--and hence the concentration of [L] at equilibrium is going to be very close to the formal concentration that you add to the solution--f; thus: [ + ] = K 1 K 2 F K 1 K w K 1 + F recall we said that F = 0.05 M to start! K 1 K w is usually <<< K 1 K 2 F; and K 1 << F [ + ] = K 1K 2 F K 1 K w K 1 + F [ + ] = K 1 K 2

10 take log of both sides: log [ + ] = 1/2 log (K 1 K 2 ) log [ + ] = 1/2(log K 1 + log K 2 ) multiply by sides by -1: -log [ + ] = 1/2(-log K 1 -log K 2 ) p = 1/2 (pk 1 + pk 2 ) so the p of the solution of 0.05 M L = 1/2 ( ) p = 6.04 note---p of the L solution is independent of concentration!

11 once you know the p---you can then go back and calculate how much of 2 L + and L - from the equilibrium constant expressions! turns out that the concentrations are very low ----approx M- compared to 0.05 M of L that was added to solution! Example---what is p of 0.05 M KP? p = 1/2 ( ) = 4.18

12 Principal species in solution? reported as α values vs. p of solution!----where α is the fraction of total polyprotic acid in a given form--- can estimate α values from log term in equation----if ratio is 10 ; (log (10/1)---then base form is approx. 90% fraction---if log (100/1) then base form is 99% fraction! fumaric acid!

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14 Blood-Gas Transfer During Respiration C 2 when dissolved in solution forms very important polyprotic acid Carbonic Acid--- 2 C 3 air 2 2 C 2 aveolus 2 C 2 2 blood flow high C 2 low 2 high 2 low C 2 blood flow C C 3 - C 2 aveolar capillary

15 C <------> 2 C 3 <------> C <----> C pk 1 = 6.35 pk 2 = controls p of blood--- If your lungs are not working well to get rid of C p of blood decreases (more carbonic acidpresent) In clinical labs they measure P C2 = partial pressure of C 2 in blood and also total C 2 species---sum of [C 2 ] + [C 3- ] (C too low to contribute appreciably to total carbon dioxide species present)! ow to measure?---use C 2 sensor--- gas permeable membrane p electrode soln of NaC 3 proportional to dissolved conc. of C 2 C 2 blood p layer = pk a + log [C 3- ] / P C2 as C 2 in blood increases--p in layer behind gas permeable membrane decreases!

16 C 2 sensors: ptical C pk 1 =6.35 pk 2 = also some C 3- salt in this layer fluorescent indicator is sensitive to p In λ max = 625 nm In - red λ max = 545 nm green What is the pk A of the indicator?

17 pi values----p of solution where polyprotic species is neutral! called isoelectric p value! pi = 1/2 (pk 1 + pk 2 ) = 5.04

18 N + pk 1 N + - neutral zwitterion Aspartic Acid (Asp) pk 1 = 1.99 pk 2 = 3.90 pk 3 = pi=2.95 Average of pk 1,pK pk 2 - N + N pk 3 -

19 N 2 N 2 N + N N + N + pk 1 N N + - Arginine (Arg) pk 2 pk 1 = 1.82 pk 2 = 8.99 pk 3 = neutral zwitterion N + N 2 N N - pk 3 pi=10.74 N 2 Average of pk 2,pK 3 N N N -

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21 Titration of diprotic acid with strong base---shape depends on values of pk 1 and pk 2 for all curves-- pk 1 = 4.00 solid black linepk 2 = 6 red line pk 2 = 8 dashed line pk 2 = 10 assume 50 ml of 0.02 M 2 A to starttitrated with 0.1 M Na

22 titration of 50 ml of 0.02 M Na 2 C 3 with 0.1 M Cl