Acid-Base Equilibria. And the beat goes on Buffer solutions Titrations

Transcription

1 Acid-Base Equilibria And the beat goes on Buffer solutions Titrations 1

2 Common Ion Effect The shift in equilibrium due to addition of a compound having an ion in common with the dissolved substance. 2

3 Common Ion Effect Consider a mixture of: HA (weak acid) and NaA (conj. salt). HA(aq) + H 2 O reactant H 3 O + (aq) + A - (aq) or HA H + + A - A - is conj. base of HA product K a = [H+ ][A - ] [HA] from both HA & NaA 3

4 Common Ion Effect Rearranging: [H + ] = K a [HA] [A - ] -log[h + ] = -logk a - log [HA] [A - ] ph = pk a + log [A- ] [HA] 4

5 Henderson-Hasselbalch Approximation HA H + + A - 1.Neglect [A - ] from ionization the of the weak acid (HA). 2.The salt (NaA) is fully ionized. 3.Thus [HA] = [HA] o & [A - ] = [A - ] o (from acid) (from salt) (where o means original concentration) 5

6 Henderson-Hasselbalch Approximation ph = pk a + log [A- ] o [HA] o or: K a = [H+ ][A - ] o [HA] o (whenever both A - and HA are both added to a solution) 6

7 H-H: Example What is the ph of a solution containing 0.20 M HAc (acid) and 0.30 M NaAc (conj. base)? Compare it to the ph of just 0.20 M HAc. K HAc = 1.8 x

8 H-H: Example Can use either: K a = [H+ ][A - ] o [HA] o or ph = pk a + log [A- ] o [HA] o 8

9 H-H: Example ph = pk a + log [A- ] o [HA] o = -log (1.8 x 10-5 ) = ph = log (0.30/0.20) Note: no need for ICE table when both acid and its conj. base are added. 9

10 H-H: Example For just 0.20 M HAc (no NaAc) Must use ICE table. HAc H + + Ac - Initial M Change M -x +x +x Equil. M 0.20-x x x 10

11 H-H: Example 1.8 x 10-5 = x 2 (0.20-x) x = 1.9 x 10-3 M ph = 2.72 (vs. ph = 4.92 when both HAc and NaAc (conj. base) added together. HAc H + + Ac - 11

12 H-H: You Try It!! What is the ph of a solution containing 0.30 M formic acid and 0.52 M potassium formate? K HCOOH = 1.7 x

13 H-H for Bases e.g. add both NH 3 & NH 4 Cl to water (base on reactant side) (conj. acid on reactant side) base conj. acid NH 3 + H 2 O NH OH - and use K b or equivalently: NH + 4 NH 3 + H + and use K a 13

14 H-H for Bases using K b NH 3 + H 2 O NH OH - K b = [OH- ] [conj acid] o [base] o [OH - ] = K b [base] o [conj acid] o poh = pk b + log [conj acid] o [base] o 14

15 H-H for Bases: Try It What is the ph of a aqueous mixture that is 0.43 M NH 3 and 0.35 M NH 4 Cl? (Window-side use K a, hall-side use K b ) Compare this to a solution that is 0.43 M NH 3 only. K b (NH 3 ) = 1.8 x

16 Buffer Solutions A weak acid or base and its conjugate salt. (H-H applies) Acid buffer: Base buffer: CH 3 COOH and CH 3 COONa NH 3 and NH 4 Cl Buffers resist ph changes. 16

17 Buffer Solutions Each buffer has a specific ph, and is resistant to ph changes. Buffers are key to many biological functions. 17

18 Buffer Solutions CH 3 COO - /CH 3 COOH or simply Ac - /HAc If add acid: Ac - + H + HAc If add base: HAc + OH - H 2 O + Ac - Both rxns are one way arrows. WHY? Buffer Capacity ~ concentration 18

19 Which are Buffers? 1. KH 2 PO 4 /H 3 PO 4 2. NaClO 4 /HClO 4 3. Na 2 CO 3 /NaHCO 3 Remember buffers are weak acid or weak base with its conjugate. 19

20 Proof of the Buffer Effect Calculate the ph of a buffer solution containing 1.0 M HAc and 1.0 M NaAc. What is the effect of adding 0.10 moles of HCl to 1.0 L of this buffer? K a (HAc) = 1.8 x

21 1.0 M HAc M NaAc HAc H + + Ac - (H-H applies!) K a = [H+ ][Ac - ] o [HAc] o 1.8x10-5 = [H+ ](1.0) 1.0 remember lab? [H + ]= 1.8 x 10-5 ph = 4.74 same as pk21 a!

22 1.0 M HAc M NaAc mol HCl Due to addition of HCl H + + Ac - HAc (one way arrow) mol Thus: HAc = = 1.1 mol Ac- = = 0.9 mol 22

23 1.0 M HAc M NaAc mol HCl K a = [H+ ][Ac - ] o [HAc] o 1.8x10-5 = [H + ](0.9) 1.1 [H + ] = 2.2 x 10-5 M ph = 4.66 (vs. 4.74) (very small change) 23

24 Buffers with Specific ph s How do you prepare a phosphate buffer at ph 7.4? Of the 3 ionizations for H 3 PO 4, the following is best because its pk a2 is closest to the target ph. H 2 PO 4 - H + + HPO 4 2- K a2 = 6.2 x 10-8 or pk a2 =

25 Buffers with Specific ph ph = pk a + log [conj. base] [acid] 7.4 = log [HPO 4 2- ] [H 2 PO 4 - ] [HPO 4 2- ] [H 2 PO 4 - ] = 1.5 e.g. add 1.5 mol HPO 4 2- and 1.0 mol H 2 PO 4 - to water. 25

26 Buffers: Try It!!! How would you prepare a carbonate buffer at a ph of 10.10? For carbonic acid: K a1 = 4.2 x 10-7 and K a2 = 4.8 x Write the reactions that occur if HCl or if NaOH is added to this buffer. 26

27 Acid-Base Titrations 3 types are common: Strong acid strong base (no hydrolysis of the salt formed) Weak acid strong base (hydrolysis of the salt anion) Strong acid weak base (hydrolysis of the salt cation) 27

28 Strong Acid-Strong Base Map the titration of 25 ml of 0.10 M HCl with 0.10 M NaOH. NaOH 0.10 M titrate ph 14 7 HCl 25mL, 0.10 M titrant 0 ml NaOH 28

29 Strong Acid-Strong Base Titration equation: H + + OH - H 2 O (one way arrow) strong strong.10m.10m.025l flask buret 29

30 Strong Acid-Strong Base Equivalence point is easy to find. mol H + = mol OH - M A V A = M B V B * (0.10 M)(25 ml) = (.10 M) (V B ) V B = 25 ml Since [H + ] = [OH - ], ph = 7.0 (if temperature is 25 o C.) 30

31 Strong Acid-Strong Base H + + OH - H 2 O (one way arrow) The ph at any point in the titration can be calculated by determining the concentration of the limiting reagent (H + or OH - ), that is excess moles/liter of H + or OH -. 31

32 Strong Acid-Strong Base e.g. after 35 ml of 0.10 M NaOH added Total volume = 25 ml + 35 ml = 60 ml = L H + + OH - H 2 O init. moles.025l x.10m.035l x.10m =.0025 =.0035 D moles equil. moles ~ (excess) 32

33 Strong Acid-Strong Base Calculate ph from the concentration of excess reagent: [OH - ] =.0010 mol.060 L =.017M poh = 1.77 and ph =

34 Strong Acid-Strong Base 14 excess OH - ph 7 excess H + M A V A = M B V B 0 ml NaOH 34

35 Go For It Calculate the following when 32 ml of 0.15 M NaOH is titrated with 0.18 M HCl : ph after 17 ml HCl is added. Volume of HCl to reach eq. pt. ph after 30 ml HCl is added 35

36 Weak Acid-Strong Base Map the titration of 25 ml of 0.10 M HAc with 0.10 M NaOH. ph gradual rise ml NaOH NaOH 0.10 M HAc 0.10 M 25mL 36

37 Weak Acid-Strong Base Titration equation: HAc + OH - weak strong Ac - + H 2 O.10M.10M.025L =.0025mol flask buret (one way arrow due to addition of strong base) 37

38 25mL,.10M.10M HAc + OH - Ac - + H 2 O The ph at any point in the titration can be calculated. e.g. after 10 ml of NaOH added. This is before the equiv. point. Total volume = 25mL+10mL=35mL mol OH - added is: L x 0.10 M = mol This is also the mol of Ac - formed. 38

39 Weak Acid-Strong Base Do ICE in moles (volume changes) HAc + OH - Ac - + H 2 O I (mol) C (mol) E (mol).0015 ~ Before the equivalence point, the titration created a buffer! 39

40 Weak Acid-Strong Base The buffer thus contains: mol/0.035 L =.0428 M HAc mol/0.035 L =.0286 M Ac - Use K a = [H+ ][Ac - ] o [HAc] o 1.8x10-5 = [H+ ](.0286) (0.0428) H-H [H + ] = 2.7 x 10-5 & ph =

41 Weak Acid-Strong Base: Before Equiv. Point HAc + OH - Ac - + H O 2 excess 14 ph 7 0 gradual rise since buffer ml NaOH 41

42 Weak Acid-Strong Base At equivalence point: ~100% HAc + OH - Ac - + H 2 O (one way arrow) But Ac - (conj.base of a weak acid) undergoes hydrolysis (equilibrium): Ac - + H 2 O HAc + OH - So the ph will be >

43 Weak Acid-Strong Base Need to calculate total volume. At equivalence point: mol HAc = mol NaOH M A V A = M B V B (0.10 M)(0.025 L) = (.10 M) (V B ) V B = L So total volume is L 43

44 Weak Acid-Strong Base At equivalence, the moles of Ac - formed is the same as starting HAc. HAc + OH - Ac - + H O mol mol mol Thus [Ac - ] =.0025 mol.050 L = 0.050M 44

45 Weak Acid-Strong Base At equivalence: M Ac - Ac - + H 2 O HAc + OH - I C -x x x E x x x Now plug in to K b = 5.6 x

46 Weak Acid-Strong Base 5.6 x = x x x = [OH - ] = 5.29 x 10-6 M poh = 5.28 ph = 8.72 (at equivalence point) 46

47 Weak Acid-Strong Base What about ph after equivalence point, e.g. after 35 ml NaOH? Solution is basic due to OH - and Ac -, but since OH - is so much stronger, it alone determines ph. (Region of excess OH -, just like strong acid-strong base case) Solution vol. = 60 ml = L 47

48 Weak Acid-Strong Base Excess mol OH - = mol OH - added init. mol HAc = M B V B - M A V A =.010M x.035l -.010M x.025l = mol [OH - ] = mol 0.060L = M poh = 1.77 and ph =

49 Weak Acid-Strong Base 14 excess [OH - ] ph ph 7 Conj base of weak acid (ICE) 0 buffer (via titration) (H-H) ml NaOH 49

50 Half-Equivalence Point The point in the titration where the moles of base added is one-half the moles of weak acid. HAc + OH - Ac - + H 2 O init mol D mol equil mol.025 ~0.025 equal 50

51 Half-Equivalence Point Plugging into H-H: K a = [H+ ][Ac - ] o [HAc] o K a = [H + ] pk a = ph equal ph 14 7 pk a 0 ml NaOH 51

52 Grind It Out!!! 80. ml of 0.12 M benzoic acid is titrated with 0.20 M KOH. K (BA) = 6.5 x Calculate: a Initial ph in the flask ph after 15 ml KOH added ph after 24 ml KOH added ph at equivalence point ph after 58 ml KOH added 52

53 Strong Acid-Weak Base Map the titration of 25 ml of 0.10 M NH 3 with 0.10 M HCl. 14 ph 7 0 gradual slope ml HCl HCl 0.10 M NH M 25 ml 53

54 Strong Acid-Weak Base Keep two equations straight! 1. Equilibrium equation for NH 3. NH 3 + H 2 O NH OH - Governed by: K b = [NH 4 + ][OH - ] [NH 3 ] 2. Titration equation: H + + NH 3 NH 4 + (one way) 54

55 Strong Acid-Weak Base Titration equation: H + + NH 3 NH M.10M.025L.0025 mol Before equivalence point, this reaction creates a buffer, so use H-H. 55

56 Strong Acid-Weak Base All calculations are analogous to the strong base-weak acid example. 14 ph 7 0 buffer (H-H) equiv pt. excess [H + ] ml HCl 56

57 Strong Acid-Weak Base What happens at the half-equiv. point? H + + NH NH Start mol D mol D mol ~ equal K b = [NH 4 + ][OH - ] [NH 3 ] and poh = pk b 57

58 Strong Acid-Weak Base I ll do the equivalence point ph: H + + NH 3 NH % But NH 4 + undergoes hydrolysis: NH 4 + NH 3 + H + So the ph will be < 7.0 (acidic) 58

59 Strong Acid-Weak Base At equivalence point: mol NH 3 = mol HCl M B V B = M A V A (0.10 M)(0.025 L) = (.10 M) (V A ) V A = L So total volume is L 59

60 Strong Acid-Weak Base At equivalence, the amount of NH + 4 formed is: H + + NH 3 NH mol mol mol Thus [NH 4 + ] =.0025 mol.050 L = 0.050M 60

61 Strong Acid-Weak Base At equivalence: M NH 4 + which then hydrolyzes: NH 4 + NH 3 + H + I C -x x x E x +x +x Now plug in to K a = 5.6 x

62 Strong Acid-Weak Base 5.6 x = x x x = [H + ] = 5.29 x 10-6 M ph = 5.28 (At equivalence point) 62

63 Strong Acid-Weak Base poh 14 buffer via titration (H-H) = pk b ph 7 conj acid of weak base (ICE) 0 half titr. ml HCl excess [H + ] 63

64 One More!!! 60. ml of 0.15 M ammonia is titrated with 0.18 M HCl K b = 1.8 x Calculate: Initial ph in the flask ph after 25 ml HCl added ph after 35 ml HCl added ph at equivalence point ph after 60 ml HCl added 64

65 Titration Hints (also see summary in packet) 1. Often must work in moles and then converted to concentration since volume changes. 2. Calculate moles of acid & base to see where in titration you are. (M A V A and M B V B ) 3. Buffer region: excess weak acid (or weak base). Use H-H. 65

66 Titration Hints cont. 4. Be on the lookout for the half titration point! 5. At equiv. point, moles acid and moles base are equal resulting in 100% conjugate. Use ICE. This is the most involved calculation. 6. If excess strong acid or strong base, ph determined by its excess concentration alone. 66

67 Diprotic Acid Titrations H 2 A H + + HA - K a 1 HA - H + + A -2 K a2 H 2 A 2H + + A -2 K a 1 x K a2 pk a1 B=1 st end point C=1 st end point pk a2 Volume of NaOH 67

68 Acid Base Indicators 68

69 Acid Base Indicators Weak acids that have different acid and conjugate base colors. HIn H + + In - HIn In - Le Chatelier s principle: if excess H +, color is yellow. If excess OH -, it is blue. 69

70 Acid Base Indicators 14 ph 7 Want pk a of indicator dye to be near equivalence point ph. 0 ml NaOH 70

71 Acid Base Indicators e.g. phenolphthalein (HPh) has a pka of 9.7. It is colorless in acid, but starts to turn pink at ph > 8.3. HIn H + + In - HPh Ph - It is a good indicator for a strong acidstrong base titration since ph 8.3 lies on the steep part of the titration curve. 71