Chapter 17: Buffers, Titrations and Salts

Transcription

1 Chapter 17: Buffers, Titrations and Salts 1. Buffers 2. Acid-Base Titrations 3. Solubility of Salts 4. Precipitation Reactions 5. Solubility and Complex Ions

2 Buffers: Acid/Base Conjugate Pair Maintain constant ph. Contain acid and base which do not neutralize each other Acid will neutralize any added base. Base will neutralize any added acid.

3 Type 1: Weak Acid/Conjugate Base Pair HA H + + A - Acid Salt (Conjugate Base) Example: Hydrofluoric Acid and Fluoride HF H + + F -

4 Type 2: Weak Base/Conjugate Acid Pair B + H 2 O BH + + OH - Base Salt (Conjugate Acid) Example: Ammonia and Ammonium ion NH 3 + H 2 O NH 4+ + OH -

5 Two Ways to Make a Buffer 1. Add Acid and its Salt HF + NaF or Add Base and its Salt NH 3 + NH 4 Cl

6 Two Ways to Make a Buffer 2. Partially Neutralize Weak Acid HF + NaOH -> NaF + H 2 O, where HF is in excess. NaF comes from the neutralized HF and the HF is the excess or Partially Neutralize Weak Base NH 3 + HCl -> NH 4 Cl, where NH 3 is in excess. NH 4+ comes from the neutralized NH 3 and the NH 3 is the excess

7 Buffers and Common Ion Effect What is the ph of 0.2M CH 3 COOH? How is the ph of this solution changed if it is also 0.1M in NaCH 3 COO

8 ph.2m Acetic Acid K a =1.8x10-5 H K [ HA] a i ph log K a [ HA] ph log (1.8 x10 5 )[.2] 2.7

9 ph.2m Acetic Acid with.1m Sodium Acetate How do we solve HA H + + A - I C -x x x E.2 -x x.1+x K a x A ] [ HA] ([ for [H + ] or x? i x) x K a x[ A ] [ HA] i [ H ][ A [ HA] i ] [ H ] Ka[ HA] [ A ] i

10 ph.2m Acetic Acid with.1m Sodium Acetate How do we solve HA H + + A - I C -x x x E.2 -x x.1+x K a x A ] [ HA] ([ for [H + ] or x? i x) x ph 5 Ka[ HA] 1.8x10 (.2) log[ H ] log log 4.4 [ A ].1

11 Change in ph by adding DpH = = 1.7 Did Adding Acetate Increase or Suppress the Ionization of Acetic Acid?

12 The Common Ion Effect Shift in an Ionic Equilibria Caused by the Addition of a Solute That Provides an Ion That Takes Place in That Equilibria In the case of buffers the salt inhibits the inization of the acid (or base)

13 Buffer Calculations HA H + + A - I C E [HA] i 0 [A - ] -x x x [HA] i -x x [A - ]+x Remember for Weak Acid, [HA] I >>x and [HA] I - x = [HA] I and likewise, [A - ] - x = [A - ]

14 Buffer Calculations HA H + + A - I C E [HA] i 0 [A - ] -x x x [HA] H + [A - ] So: [HA] = Initial Acid Concentration [A - ] = Initial Salt Concentration Derive the Equilibrium Expression for ph

15 You have just derived the Henederson-Hasselbach Eq. ] [ ] ][ [ HA A H K a ] [ ] [ ] [ A HA K H a ] [ ] [ log log ] [ ] [ log ] log[ A HA K A HA K H a a ] [ ] [ log log ] log[ A HA K H ph a ] [ ] [ log HA A pk ph a [Salt] [Acid]

16 Henderson Hasselbach Eq For Acid and its Salt: [ A ] [ salt] ph pk log pk log a a [ HA] [ acid] For Base and its Salt: [ BH ] [ salt] poh pk log pk log b b [ B] [ base] Note: ph + poh = 14

17 Henderson Hasselbach Eq ph pk a log [ A ] [ HA ] Two Important Characteristics 1. Buffer Ph -best when pk a = desired ph and [HA]=[A - ] 2. Buffer Capacity -highest for concentrated solutions

18 Buffer Problem Calculate the Change in ph If We Add.001 Mole of NaOH to 1.00 L of a Buffer which is M NH 4 Cl &.02 M NH 3 Note This Is a Three Part Problem 1. Calculate the initial ph of the buffer 2. Calculate the final ph after addition of the base 3. Calculate the change in ph

19 Buffer Problem A. Calculate the Initial ph K b [ NH4 ][ OH [ NH ] 3 ] poh = pk b + log([nh 4+ ]/[NH 3 ]) poh = log(.1/.02) = poh = 5.45 ph = ph = 8.55

20 Buffer Problem B. Calculate the Final ph Each OH - extracts a proton from NH 4+ creating NH 3 NH 4+ is decreased by.001 mole NH 3 is increased by.001 mole NH 4+ =0.099 M NH 3 = M poh = log(.099/.021) = poh = 5.42, so ph = 8.58 C. DpH = DpH = 0.03

21 More Henderson-Hasselbalch eq. Choosing Buffers ph pk a log [ A ] [ HA ] Choose acid with pk a as close to desired ph as possible Adjust ratio of acid and conjugate base to provide desired ph

22 ph Making a Buffer Why can t you make a ph 10 buffer with 0.10 log [ A ] pk a [ HA ] [ A ] [ HA]10 M acetic acid and acetate? ph pk 5 10 ( log1.8 x10 ) [ A ] [0.1]10 18,000M A The acetate concentration is not reasonable

23 Making a Buffer What is the acetate concentration of a ph 5 buffer with 0.10M acidic acid? ph log [ A ] pk a [ HA ] [ A ] [ HA]10 ph pk 5 5 ( log1.9 x10 ) [ A ] [0.1]10.18M A

24 More Henderson-Hasselbalch eq. Buffer Capacity ph pk a log [ A ] [ HA ] [HA] = [A - ] equal amounts of acid or base can be neutralized For any given ratio, the greater the concentration, the greater the capacity

25 Acid Base Titrations Objective: Determine the ph of a solution during a titration 4 Types A. Titration of a Strong Acid by a Strong Base B. Titration of a Weak Acid by a Strong Base C. Titration of a Strong Base by a Strong Acid D. Titration of a Weak Base by a Strong Acid

26 Titration of Strong Acid with Strong Base What Does the Titration Curve Look Like? ph Is the Equivalence Pt. Acidic, Basic or Neutral? Vol Base Added NEUTRAL NaCl (Salt of Strong Acid & Strong Base) (HCl & NaOH)

27 Titration of Weak Acid with Strong Base What Does the Titration Curve Look Like? ph Is the Equivalence Pt. Acidic, Basic or Neutral? Vol Base Added BASIC NaF (Salt of Weak Acid & Strong Base) (HF & NaOH)

28 Titration of Various Acids with Strong Base Which Curve is the Weakest Acid? ph Vol Base Added

29 Titration of Strong Base with Strong Acid What Does the Titration Curve Look Like? ph Is the Equivalence Pt. Acidic, Basic or Neutral? Neutral Vol Acid Added (NaOH & HCl) NaCl (Salt of Strong Base & Strong Acid)

30 Titration of Weak Base with Strong Acid What Does the Titration Curve Look Like? ph Is the Equivalence Pt. Acidic, Basic or Neutral? Vol Acid Added (NH 3 & HCl) ACIDIC NH 4 Cl (Salt of Weak Base & Strong Acid)

31 4 Regions of ph Titration Curve ph 1. Pure Acid Treat as Weak Acid 2. Buffer Region Treat as acid & salt 4. Excess Base vol of base added 3. Equivalence Pt. Treat as Salt of Weak Acid

32 ph Titration Curve 50 ml of.15 M CH 3 COOH is titrated with.15m NaOH, K a = 1.8 x 10-5 Calculate the ph at the after the following volumes have been added a. 0.0 ml NaOH b. 20 ml NaOH c. 25 ml NaOH d. 50 ml NaOH e. 70 ml NaOH

33 Pure Acid Region ph log[ H ] ph log k [ HA] a i ph 5 log 1.8x10 (.15) 2.34

34 Buffer Region Use Henderson Hasselbach Eq. ph pk a [ A ] log [ HA ] Simple Technique 1. Identify Vol of base required to reach equivalence point (V B equiv pt ) 2. Look at ratio of salt to acid and express in terms of volume base added (V B ) and volume required to reach equivalence (next slide).

35 Buffer Region ph pk a [ A ] log [ HA ] n A MV B B moles weak acid neutralized moles base added [ A ] = V V V V [HA]= T T T T M V M V M BVB equiv pt M BVB moles excess acid V V V A A, initial B B T T T MV B B [ A ] VT M BVB VB [HA] M BVB equiv pt M BVB M B ( VB equiv pt VB ) VB equiv pt V B V T Express ratio in terms of V B and the volume of base at equivalence (V B equiv pt )

36 Buffer Region ph pk a [ A ] log [ HA ] At equivalence M A V A =M B V B V B equiv pt M AVA 0.15 M (50 ml) 50ml M 0.15M B

37 Buffer Region ph pk a [ A ] log [ HA ] Pt. B 20 ml ph 5 20ml log1.8x10 log 50 ml 20 ml

38 ph at Half Equivalence Pt 3, at 25 ml added, half of the acid is neutralized and [A - ]=[HA] ph pk [ A ] log [ HA] a pk a log1 ph = pk a ph = 4.74

39 ph at Equivalence Point Treat As Salt of Weak Acid ' [ OH ] Kb[ A ] -All the Acid Has Been Turned to its Conjugate Base -What is K b? K 14 ' w 10 b 5.56x10 5 Ka 1.8x10 -Need to take into account Dilution Factor K 10

40 ph at Equivalence Point [ ] [ MV ] K MV K [ OH ] K [ A ] where V V V W A A, i W B B, pt.3 B T A, i B, eq pt K A VT K A VT M (50 ml) 5 1.8x10 (50ml 50 ml) [ OH ] [ ] 6.45x10 poh 6 log 6.45x ph Note the ph of the Equivalence Point Is Basic for the Titration of a Weak Acid With a Strong Base

41 ph in Excess Base Region -The ph is determined by the moles of excess base [ OH ] n OH ( excess) M V M V V V V B B B B, eq pt T B A, i 0.15 M (70ml 50 ml) 0.025M (70ml 50 ml) poh log ph

42 Polyprotic Titration Curves What Would the Curve Look Like for the Titration of Carbonic Acid with NaOH? ph HA - /A -2 A -2 H 2 A/HA - HA - H 2 A Vol NaOH Added

43 Solubility Equilibria Define Saturated Solution -A solution in Contact with Undissolved Solute What is the Solubility Eq for Barium Sulfate? BaSO 4 (s) Ba +2 (aq) + SO 4-2 (aq) What is the Solubility Product for BaSO 4? K sp =[Ba +2 ][SO 4-2 ] (Appendix J)

44 Salts are Electrolytes Let s take a closer look at what we mean by solubility of a salt Consider Lead (II) Phosphate K sp = 3.0x10-44 Pb 3 (PO 4 ) 2 < == > 3Pb PO -3 4

45 Salts are Electrolytes Pb 3 (PO 4 ) 2 < == > 3Pb PO 4-3 There are three questions which can be asked: 1. What is the solubility of Pb 3 (PO 4 ) 2 2. What is the concentration of Pb What is the concentration of PO 4-3

46 Salts are Electrolytes I C E Pb 3 (PO 4 ) 2 < == > 3Pb PO 4-3 Solid 0 0 -x +3x +2x Solid +3x +2x x=solubility (how much salt dissolves) Do not waste time with ICE diagrams. Go straight to equilibrium expression K=K sp =(3x) 3 (2x) 2

47 Salts are Electrolytes K (3 x) (2 x) 108x x K 3.0x x10 10 [Pb +2 ]= 3x = 2.3x10-9 [PO 4-3 ] = 2x = 1.5x10-9

48 Generalized Solubility Product M m X x (s) mm +n (aq) + xx -y (aq) K sp = [M +n ] m [X -y ] x

49 What is K for La(IO 3 ) 3 if [IO 3- ] = 0.006M for a saturated solution [La +3 ] = x, [IO 3- ] = 3x [IO 3- ] = 3x=.006 => x=.002 K sp = [La][IO 3 ] 3 = x(3x) 3 K sp =.002(.006) 3 = 4.3x10-10

50 Solubility and Common Ion Effect: What is the Solubility of BaSO 4? K sp = 1.1x10-10 BaSO 4 (s) Ba +2 (aq) + SO 4-2 (aq) K sp = x 2 = 1.1x10-10 x = Solubility x = 1.05 x 10-5 M

51 Solubility & Common Ion Effect What is the solubility of BaSO 4 in 0.02M Na 2 SO 4? K sp =[Ba +2 ][SO -2 4 ] Ba +2 = x, SO -2 4 = x x(x+0.02)=1x10-10 x(0.02)=1x10-10 x = 5.5x10-9 M The common ion suppressed ionization, (x = 1.05x10-5 M in absence of Na 2 SO 4 )

52 Precipitation Calculations How can you predict if a ppt will form when you mix solutions of know concentrations which form an insoluble product? Determine the reaction quotient Q < K, reaction goes forward, no ppt Q = K, reaction at equilibrium, no ppt Q > K, reaction goes backward, ppt forms

53 Fractional Precipitation Two Ions Can Be Precipitated Out Separately if they Form Salts of Different Solubilities With the Same Counter Ion Consider the Separation of Silver Ions From Lead(II) Ions Find an Ion Which Forms Salts With Both Silver and Lead(II) of Different Solubilities

54 Q<K for both, favors forward reaction, both dissolve Fractional Precipitation K sp (PbCl 2 ) = 1.6 x 10-5 K sp (AgCl) = 1.8 x Solid Dissolved K sp (PbCl 2 ) = 1.6 x 10-5 Q Q>K for both, favors back reaction neither dissolve (both ppt) Q>K for for AgCl (ppts) Q<K for PbCl 2 (dissolves) K sp (AgCl) = 1.8 x 10-10

55 Fractional Precipitation What concentration range of chloride will ppt out 0.1M silver while leaving 0.1M lead(ii) solvated K [ Pb ][ Cl ] 1.6x10 PbCl K [ ] PbCl x Cl 2. OO158 [ Pb ] (.1) K Ag Cl x AgCl 10 [ ][ ] K [ Cl ] AgCl x [ ].1 1.8x10 Ag The AgCl will selectively Precipitate out for 1.8x10-9 M < [Cl - ] < M

56 Fractional Precipitation K sp (PbCl 2 ) = 1.6 x 10-5 K sp (AgCl) = 1.8 x Solid Dissolved K sp (PbCl 2 ) = 1.6 x 10-5 Cl - = M Q Q>K for both, favors back reaction neither dissolve (both ppt) Q>K for both for AgCl (ppts) Q<K for PbCl 2 (dissolves) K sp (AgCl) = 1.8 x Cl - = 1.8x10-9 Q<K for both, favors forward reaction, both dissolve

57 Complex Ions Result when a Lewis base (Ligand) forms a coordinate covalent bond with a metal ion Ag + (aq) + 2(:NH 3 )(aq) --> (H 3 N-Ag-NH 3 ) + (aq) K f [ Ag( NH3) [ Ag ][ NH 3 ] ] 2 2 Ligands K f = 1.7 x 10 7

58 K f & K sp AgCl(s) <==> Ag + + Cl - K sp = 1.8 x Ag + + 2(NH 3 ) <--> Ag(NH 3 ) + 2 K f = 1.7 x 10 7 AgCl(s) + 2(NH 3 ) <--> Cl - + Ag(NH 3 ) + 2 K = [Cl - ][Ag(NH 3 ) 2+ ] [NH 3 ] 2 K = K sp K f = (1.8 x )(1.7 x 10 7 ) = 3.1 x 10-3

59 K f & K sp AgCl(s) <==> Ag + + Cl - K sp = 1.8 x Ag + + 2(NH 3 ) <--> Ag(NH 3 ) 2 + K f = 1.7 x 10 7 K = [Cl - ][Ag(NH 3 ) 2+ ] [NH 3 ] 2 K = K sp K f = (1.8 x )(1.7 x 10 7 ) = 3.1 x 10-3

60 What is the Molar Solubility of AgCl in 2M I NH 3 AgCl(s) + 2(NH 3 ) <--> Cl - + Ag(NH 3 ) C -2x x x E 2-2x x x K = (x)(x) = 3.1 x 10-3 (2.0-2x) 2

61 What is the Molar Solubility of AgCl in 2M NH 3 K = (x)(x) = 3.1 x 10-3 x (2.0-2x) (2 2 x) 2 2 x 2 2x x x K x K x K (2 2 x) 2 K K K 0.10

62 What is the Molar Solubility of AgCl in 2M NH 3 X = Molar Solubility = 0.100M In absence of NH 3 : x 2 = K sp = 1.8 x x = 1.3 x 10-5 M