Lecture 21 Chapter 18, Section 3 Titration. Finish the Making Buffer example Titration of a weak acid Midpoint Equivalence point

Transcription

1 Lecture 1 Chapter 18, Section 3 Titration Finish the Making Buffer example Titration of a weak acid Midpoint Equivalence point

2 Make a Buffer Example Similar to Example 18-8 in book. Check this out, too. O then we are making 1L of a buffer with ph 4.00 And we want to be able to buffer moles of acid or base. Formic acid is the best choice of the weak acids given (p a 3.75)

3 How much M formic acid solution is needed to make our buffer? 5% ml 5%. 78 ml 5% ml 5% ml

4 How much 6.00 M NaOH is needed to make our buffer? 5% ml 5%. 9.7 ml 5% ml 5% ml

5 Buffer Summary Use 556 ml of M formic acid solution This contains 0.78 moles of formic acid Add 9.7 ml of 6.00 M NaOH This adds moles of OH Converts moles of formic acid to formate ion Thus, we have moles of A, moles of HA Ratio of A to HA is 1.78 to 1, gives a ph of 4.00 Just need to add water to so that our total volume is 1.00 L About 414 ml.

6 Titration Mentioned this way back in Chapt 4 with reaction stoichiometry The stoichiometric point or uivalence point is when enough titrant has been added to react with EXACTLY all of the chemical that is being titrated If we are titrating a strong acid with a strong base, then the uivalence point is when all of the strong acid has been used up and there is not any extra base yet ph 7.00 For weak acids/bases the uivalence point ph 7.00

7 Titration of a Weak Acid by OH Ions 1. At the beginning, HA and H O are the major species. You use a and your uilibrium table to determine the ph of the solution.. During the titration, a buffer has been created a buffer, so the buffer uation can give the ph. At the midpoint of the titration (when HA A - ), the ph p a 3. When nearly all the HA is reacted, the only major species are A - and water, here you can use b and your uilibrium table to determine ph. 4. After all HA molecules have reacted, the solution contains excess A - and OH -. Each drop of titrant gives more excess OH -, which you use to get the ph.

8 Polyprotic Acid Titration The same thing, but more complicated because we have several titrations put back-to-back In the middle, the uilibria are between intermediate conjugate base and acid H A + H O H 3 O + + H O + HA H 3 O + + A Note that if: a # moles of H A b # moles of HA c # moles of A then a + b + c moles of H A initially constant Note also that the volume is constantly changing

9 Polyprotic Acid Titration

10 Polyprotic acid titration region 1 Just like with standard weak acid titration Have mostly the original acid Set up standard table, solve for x a1 [ HA ] [ H O ] + [ H A] [ H A] x 3 x init

11 Polyprotic acid titration region Now in buffer region just like weak acid. Unlike weak acid, this is the first of (or 3) buffer regions Use Henderson-Hasselbalch uation [H A] [H A] init [OH ] that has been added [HA ] [OH ] that has been added ph p a 1 + log [ HA ] [ H A] At midpoint, [OH ] added ½[H A] init so [H A] [HA ] ph p a1

12 Polyprotic acid titration region 3 Near the first stoichiometric point, we have used up essentially all of the original acid H A There is lots of HA around. But this is in uilibrium with both H A and with A. HA + H O H 3 O + + A a HA + H 3 O + H O + H A 1/ a1 HA A + H A a / a1 a a1 [ ] A [ H A] [ HA ]

13 Polyprotic acid titration still region 3 But at the first stoichiometric point we know [A ] [H A] a a1 [ ] A [ H A] [ HA ] a a1 [ H A] [ ] HA We know a bit more about a1 from the first acid uilibrium a1 [ H O ] [ HA ] + a a1 3 [ H A] [ H O ] + 3 a1 or or [ H A] [ HA ] [ H O ] + 3 a1 [ ] + H3O a 1a At the first Stoichiometric Point

14 Polyprotic acid titration region 4 In buffer region again just like weak acid. The weak acid here is HA and the conjugate base is A Use Henderson-Hasselbalch uation Moles HA moles H A init moles OH added since the first stoichiometric point. Note that moles H A init moles HA max Moles A moles OH that have been added since first stoichiometric point ph p a + log [ ] A [ ] HA At midpoint, moles OH SFSP ½ moles H A init so [HA ] [A ] ph p a

15 Polyprotic acid titration region 5 Near the second stoichiometric point Almost entirely A Set up our standard uilibrium table looking at A as a base A + H O HA + OH b w a Again [A ] max moles H A init / total volume w a x [ ] A x max

16 Polyprotic acid titration region 6 After the second stoichiometric point Excess A and OH Still the uilibrium table with A as a base. But now initial amount of OH is amount of added OH- since the second stoichiometric point A + H O HA + OH x ([ ] OH + x) [ H A] w ASSSP w init approx. [ H A] x [ OH ] a init a ASSSP x Usually, all we care about is ph. [OH ] is simply the amount of added OH since the second stoichiometric point (x is negligible)

17 Today Finish CAPA #1 Review Chapts 16 & 17 if you haven t done this already! Wednesday Start CAPA #13 eep Reviewing