Addition of base: Normal human blood ph is 7.4 and has a narrow range of about +/- 0.2

Transcription

1 Chapter 15 Applications of Aqueous Equilibria Addition of base: Normal human blood ph is 7.4 and has a narrow range of about +/- 0.2

2 Buffer Solutions When an acid or base is added to water, the ph changes drastically. A buffer solution resists a change in ph when an acid or base is added. Buffers: Mixture of a weak acid and its salt Mixture of a weak base and its salt ph = 4 ph = 10.5 ph = 6.9 ph = 7.1

3 Understanding Buffer Solutions The presence of a common ion suppresses the ionization of a weak acid or a weak base. Consider mixture of CH 3 COONa (strong electrolyte) and CH 3 COOH (weak acid). CH 3 COONa (s) CH 3 COOH (aq) Na + (aq)+ CH 3 COO - (aq) H + (aq)+ CH 3 COO - (aq) common ion The common ion effectis the shift in equilibrium caused by the addition of a compound having an ion in common with the dissolved substance.

4 ph of a Buffer: Consider mixture of salt NaA and weak acid HA. NaA (s) HA (aq) Na + (aq)+ A - (aq) H + (aq)+ A - (aq) K a = [H+ ][A - ] [HA] [H + ] = K a [HA] [A - ] -log [H + ] = -log K a -log [HA] [A - ] -log [H + ] = -log K a + log [A- ] [HA] ph = pk a + log [A- ] [HA] pk a = -log K a Henderson-Hasselbalch equation ph = pk a + log [conjugate base] [acid]

5 Using Henderson-Hasselbalch equation What is the ph of a solution containing 0.30 MHCOOH and 0.52 MHCOOK? (K a = 1.67 X 10-4 ). HCOOH : K a = 1.67 X 10-4, pk a = 3.78 ph = pk a + log [HCOO- ] [HCOOH] ph = log [0.52] [0.30] = 4.02

6 A buffer solutionis a solution of: 1. A weak acid or a weak base and 2. The salt of the weak acid or weak base Both must be present! A buffer solution has the ability to resist changes in ph upon the addition of small amounts of either acid or base. Consider an equal molar mixture of CH 3 COOH and CH 3 COONa Add strong acid H + (aq)+ CH 3 COO - (aq) Add strong base OH - (aq)+ CH 3 COOH (aq) CH 3 COOH (aq) CH 3 COO - (aq)+ H 2 O (l)

7 Which of the following are buffer systems? (a) KF/HF (b) KBr/HBr, (c) Na 2 CO 3 /NaHCO 3 (a) HF is a weak acid and F - is its conjugate base buffer solution (b) HBr is a strong acid not a buffer solution (c) CO 3 2- is a weak base and HCO 3- is it conjugate acid buffer solution

8 Blood Plasma ph = 7.4

9 Buffer Solutions 02

10 Calculate the ph of the 0.30 MNH 3 /0.36 MNH 4 Cl buffer system. What is the ph after the addition of 20.0 ml of MNaOH to 80.0 ml of the buffer solution? (K a = 5.62 X ). NH 4 + (aq) H + (aq)+ NH 3 (aq) ph = pk a + log [NH 3] [NH 4+ ] pk a = 9.25 ph = log [0.30] [0.36] = 9.17 NH 4 + (aq)+ OH - (aq) H 2 O (l)+ NH 3 (aq) Initial (moles) Change End (moles) 0.36 X = x = X.0800 = = = = final volume = 80.0 ml ml = 100 ml, 0.1 L [NH 4+ ] = [NH 3 ] = ph = log [0.25] [0.28] = 9.20

11 Titrations Slowly add base to unknown acid UNTIL The indicator changes color (pink)

12 Titrations In a titrationa solution of accurately known concentration is added gradually added to another solution of unknown concentration until the chemical reaction between the two solutions is complete. Equivalence point the point at which the reaction is complete Indicator substance that changes color at (or near) the equivalence point Slowly add base to unknown acid UNTIL The indicator changes color (pink)

13 Acid Base Titration Curves 01

14 Strong Acid-Strong Base Titrations Equivalence Point: The point at which stoichiometrically equivalent quantities of acid and base have been mixed together.

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16 Strong Acid-Strong Base Titrations 1. Before Addition of Any NaOH M HCl

17 Strong Acid-Strong Base Titrations 2. Before the Equivalence Point H 3 O 1+ (aq) + OH 1- (aq) 100% 2H 2 O(l) Excess H 3 O 1+

18 Strong Acid-Strong Base Titrations 3. At the Equivalence Point The quantity of H 3 O 1+ is equal to the quantity of added OH 1-. ph = 7

19 Strong Acid-Strong Base Titrations 4. Beyond the Equivalence Point Excess OH 1-

20 Strong Acid-Strong Base Titrations

21 Weak Acid-Strong Base Titrations

22 Weak Acid-Strong Base Titrations 1. Before Addition of Any NaOH CH 3 CO 2 H(aq) + H 2 O(l) H 3 O 1+ (aq) + CH 3 CO 2 1- (aq) M CH 3 CO 2 H

23 Weak Acid-Strong Base Titrations 2. Before the Equivalence Point CH 3 CO 2 H(aq) + OH 1- (aq) 100% H 2 O(l) + CH 3 CO 2 1- (aq) Excess CH 3 CO 2 H Buffer region

24 Weak Acid-Strong Base Titrations 3. At the Equivalence Point CH 3 CO 2 1- (aq) + H 2 O(l) OH 1- (aq) + CH 3 CO 2 H(aq) ph > 7 Notice how the ph at the equivalence point is shifted up versus the strong acid titration. 췠 ѥ

25 Weak Acid-Strong Base Titrations 4. Beyond the Equivalence Point Excess OH 1- Notice that the strong and weak acid titration curves correspond after the equivalence point.

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27 Titration of a Weak Acid with a Strong Base With weaker acids, the initial ph is higher and ph changes near the equivalence point are more subtle.

28 Strong Acid-Weak Base Titrations HCl (aq)+ NH 3 (aq) NH 4 Cl (aq) H + (aq)+ NH 3 (aq) NH 4 + (aq) At equivalence point (ph < 7): NH 4 + (aq)+ H 2 O (l) NH 3 (aq)+ H + (aq) 뫐

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31 Exactly 100 ml of 0.10 MHNO 2 are titrated with a 0.10 M NaOH solution. What is the ph at the equivalence point? start (moles) end (moles) HNO 2 (aq)+ OH - (aq) NO - 2 (aq)+ H2 O (l) Final volume = 200 ml Initial (M) Change (M) Equilibrium (M) K b = [OH- ][HNO 2 ] [NO 2- ] NO 2 - (aq)+ H2 O (l) x +x x = x x = 2.2 x x 0.05 x 1.05 x 10-6 = [OH - ] [NO 2 - ] = = 0.05 M OH - (aq)+ HNO 2 (aq) x poh = x x ph = 14 poh = 8.02

32 Polyprotic Acids: Titration of Diprotic Acid + Strong Base

33 Titration of a Triprotic Acid: H 3 PO 4 +Strong Base ѫ

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35 Fractional Precipitation 01 Fractional precipitation is a method of removing one ion type while leaving others in solution. 터 Ions are added that will form an insoluble product with one ion and a soluble one with others. When both products are insoluble, their relative K sp values can be used for separation.

36 Solubility Equilibria AgCl (s) Ag + (aq)+ Cl - (aq) K sp = 1.8 x K sp = [Ag + ][Cl - ] K sp is the solubility product constant MgF 2 (s) Mg 2+ (aq)+ 2F - (aq) K sp = [Mg 2+ ][F - ] 2 Ag 2 CO 3 (s) 2Ag + (aq)+ CO 2-3 (aq) Ksp = [Ag + ] 2 [CO 2-3 ] 터 Ca 3 (PO 4 ) 2 (s) 3Ca 2+ (aq)+ 2PO 3-4 (aq) Ksp = [Ca 2+ ] 3 [PO 3-4 ] 2 Dissolution of an ionic solid in aqueous solution: Q(IP)< K sp Unsaturated solution No precipitate Q (IP)= K sp Saturated solution Q (IP)> K sp Supersaturated solution Precipitate will form (IP) : Ion product K sp : Solubility Product

37 Measuring K sp and Calculating Solubility from K sp 터

38 Molar solubility(mol/l): Number ofmoles of a solute that dissolve to produce a litre of saturated solution Solubility(g/L)is the number of grams of solute that dissolve to produce a litre of saturated solution 터

39 Measuring K sp If the concentrations of Ca 2+ (aq) and F 1- (aq) in a saturated solution of calcium fluoride are known, K sp may be calculated. CaF 2 (s) Ca 2+ (aq) + 2F 1- (aq) 터 [Ca 2+ ] = 3.3 x 10-4 M [F 1- ] = 6.7 x 10-4 M K sp = [Ca 2+ ][F 1- ] 2 = (3.3 x 10-4 )(6.7 x 10-4 ) 2 = 1.5 x (at 25 C)

40 Calculating Solubility from Ksp What is the solubility of silver chloride in g/l? AgCl (s) Initial (M) Change (M) Equilibrium (M) Ag + (aq)+ Cl - (aq) s +s [Ag + ] = 1.3 x 10-5 M [Cl - ] = 1.3 x 10-5 M s 터 s K sp = 1.8 x K sp = [Ag + ][Cl - ] K sp = s 2 s= K sp s= 1.3 x 10-5 Solubility of AgCl = 1.3 x 10-5 mol AgCl 1 L soln x g AgCl 1 mol AgCl = 1.9 x 10-3 g/l

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42 Fractional Precipitation 01 Fractional precipitation is a method of removing one ion type while leaving others in solution. 터 Ions are added that will form an insoluble product with one ion and a soluble one with others. When both products are insoluble, their relative K sp values can be used for separation.

43 If 2.00 ml of MNaOH are added to 1.00 L of MCaCl 2, will a precipitate form? The ions present in solution are Na +, OH -, Ca 2+, Cl -. Only possible precipitate is Ca(OH) 2 (solubility rules). Is Q> K sp for Ca(OH) 2? [Ca 2+ ] 0 = M [OH - ] 0 = 4.0 x 10-4 M 터 Q= [Ca 2+ ] 0 [OH - ] 0 2 = x (4.0 x 10-4 ) 2 = 1.6 x 10-8 K sp = [Ca 2+ ][OH - ] 2 = 4.7 x 10-6 Q< K sp No precipitate will form If this solution also contained M AlCl3, Aluminum could be precipitated, with Calcium remaining in the solution.

44 Fractional Precipitation 01 1) Use the difference between Ksp sto separate one precipitate at a time. 2) Use Common Ion effect to Precipitate one Ion at a time 터

45 The Common-Ion Effect and Solubility 01 터

46 Factors That Affect Solubility Solubility and the Common-Ion Effect MgF 2 (s) Mg 2+ (aq) + 2F 1- (aq) 터

47 Fractional Precipitation Use the difference between Ksp s to separate one precipitate at a time. 2. Use Common Ion effect to Precipitate one Ion at a time 터 3. Use ph to Increase or decrease solubility and separate one ion at a time

48 Factors That Affect Solubility Solubility and the ph of the Solution CaCO 3 (s) + H 3 O 1+ (aq) Ca 2+ (aq) + HCO 3 1- (aq) + H 2 O(l) 터

49 Fractional Precipitation Use the difference between Ksp s to separate one precipitate at a time. 2. Use Common Ion effect to Precipitate one Ion at a time 터 3. Use ph to Increase or decrease solubility and separate one ion at a time. 4. Use complex formation to prevent certain Ions to precipitate

50 Factors That Affect Solubility Solubility and the Formation of Complex Ions Ag 1+ (aq) + NH 3 (aq) Ag(NH 3 ) 1+ (aq) K 1 = 2.1 x 10 3 Ag(NH 3 ) 1+ (aq) + NH 3 (aq) Ag(NH 3 ) 2 1+ (aq) 터 K 2 = 8.1 x 10 3 Ag 1+ (aq) + 2NH Ag(NH 3 ) 1+ 3 (aq) 2 (aq) K f = 1.7 x 10 7 K f is the formation constant. (at 25 C) How is it calculated?

51 Factors That Affect Solubility Solubility and the Formation of Complex Ions K 1 = [Ag(NH 3) 1+ ] [Ag 1+ ][NH 3 ] K 2 = [Ag(NH 3 ) 2 1+ ] [Ag(NH 3 ) 1+ ][NH 3 ] ѫ K f = K 1 K 2 = [Ag(NH 3 ) 2 1+ ] [Ag 1+ ][NH 3 ] 2 = (2.1 x 10 3 )(8.1 x 10 3 ) = 1.7 x 10 7

52 Factors That Affect Solubility Solubility and the Formation of Complex Ions Ag 1+ (aq) + 2NH 3 (aq) Ag(NH 3 ) 2 1+ (aq)

53 Factors That Affect Solubility Solubility and Amphoterism Aluminum hydroxide is soluble bothin strongly acidic and in strongly basic solutions. In acid: 터 Al(OH) 3 (s) + 3H 3 O 1+ (aq) Al 3+ (aq) + 6H 2 O(l) In base: Al(OH) 3 (s) + OH 1- (aq) Al(OH) 4 1- (aq)

54 Factors That Affect Solubility Solubility and Amphoterism 터

55 The Common Ion Effect and Solubility AgBr (s) Ag + (aq)+ Br - (aq) The presence of a common ion decreases the solubility of the salt. AgBr (s) What is the molar solubility of AgBr in (a) pure water and (b) MNaBr? K sp = 5.4 x s 2 = K sp s= 7.3 x 10-7 Ag + (aq)+ Br - (aq) M 터 NaBr (s) [Br - ] = M AgBr (s) [Ag + ] = s Na + (aq)+ Br - (aq) Ag + (aq)+ Br - (aq) [Br - ] = s K sp = x s s= 5.4 x M

56 ph and Solubility The presence of a common ion decreases the solubility. Insoluble bases dissolve in acidic solutions 터 Insoluble acids dissolve in basic solution.

57 ph and Solubility What is the ph of a solution containing Mg(OH) 2? what ph Would reduce the solubility of this precipitate? Mg(OH) 2 (s) remove add Mg 2+ (aq)+ 2OH - (aq) K sp = [Mg 2+ ][OH - ] 2 = 5.6 x K sp = (s)(2s) 2 = 4s 3 4s 3 = 5.6 x s= 1.1 x 10-4 M [OH - ] = 2s= 2.2 x 10-4 M poh = 3.65 ph = 터 At ph less than Lower [OH - ], Increase solubility of Mg(OH) 2 OH - (aq)+ H + (aq) At ph greater than Raise [OH - ] H 2 O (l) Decrease solubility of Mg(OH) 2

58 Complex Ion Equilibria and Solubility A complex ionis an ion containing a central metal cation bonded to one or more molecules or ions. Co 2+ (aq)+ 4Cl - (aq) 2- CoCl 4 (aq) The formation constant or stability constant (K f )is the equilibrium constant for the complex ion formation. Co(H 2 O) 6 2+ CoCl 4 2- K f = 터 2- [CoCl 4 ] [Co 2+ ][Cl - ] 4 K f stability of complex

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61 Qualitative Analysis 터

62 Qualitative Analysis of Cations 터

63 Fractional Precipitation 01 Fractional precipitation is a method of removing one ion type while leaving others in solution. 터 Ions are added that will form an insoluble product with one ion and a soluble one with others. When both products are insoluble, their relative K sp values can be used for separation.

64 Flame Test for Cations 辠 Ѯ lithium sodium potassium copper Nessler's is K 2 [HgI 4 ] in a basic solution, [HgI 4 ] 2- reacts with the NH4 + to form a yellow brown precipitate.

65 Fractional Precipitation 01 ѫ

66 Separation of Ions by Selective Precipitation Determine whether Cd +2 can be separated from Zn +2 by bubbling H 2 S through ([H 2 S ] = 0.1 ) a 0.3 M HCl solutions that contains M Cd +2 and M Zn +2? MS(s) + 2 H 3 O + (aq) ae M 2+ (aq) + H 2 S(aq) + 2 H 2 O(l) (see page 631) :K spa = [Cd +2 ][H 2 S ]/[H 3 O + ] 2 K spa = For ZnS, K spa = 3 x 10-2 ; for CdS, K spa = 8 x 10-7 ѫ [Cd +2 ]=[Zn +2 ]= M Because the two cation concentrations are equal, Q spa is :[Cd +2 ][H 2 S ]/[H 3 O + ] 2 =(0.005).(0.1)/ (0.3) 2 Q spa = 6 x 10-3 For CdS, Q spa > K spa ; CdS will precipitate. For ZnS, Q spa < K spa ; Zn +2 will remain in solution.

67 What is the ph of a solution containing 0.30 MHCOOH and 0.52 MHCOOK? (K a = 1.67 X 10-4 ). Mixture of weak acid and conjugate base! HCOOH (aq) H + (aq)+ HCOO - (aq) Initial (M) Change (M) x 犰 +x x Equilibrium (M) x x x Common ion effect 0.30 x x 0.52 X ( X) X = 1.67 X 10-4 X = 9.77 X 10-5 = [ H 3 O + ] ph = - log ( 9.77 X 10-5 ) = 4.01

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