Chapter Fifteen: APPLICATIONS OF AQUEOUS EQUILIBRIA

Transcription

1 Chapter Fifteen: APPLICATIONS OF AQUEOUS EQUILIBRIA 1

2 Contents p680 2

3 15-1 Solution of Acids or Bases Containing a Common Ion p681 Le Chatelier s Principle Common ion effect 3

4 The equilibrium concentration of OH - p682 ions is reduced by the common ion effect. The position of ammonia-water equilibrium is shifted to the left. 4

5 Ex 15.1 Acidic Solutions Containing Common Ions Calculate [H + ] and the percent of HF in a solution containing 1.0 M HF (Ka = 7.2 x 10-4 ) and 1.0 M NaF. Solution: p M HF [H+] = 2.7 x 10 2 M The percent dissociation of 1.0 M HF is 2.7% ( 2.7 x 10-2 /1.0)

6 15-2 Buffered Solutions p684 The most important application of acid-base solutions containing a common ion is for buffering. A buffered solution is one that resists a change in its ph when either hydroxide ions pr protons are added. The most important practical example of a buffered solution is our blood, which can absorb the acids and bases produced in biologic reactions without changing its ph. A constant ph for blood is vital because cells can survive only in a very narrow ph range. 6

7 Ex 15.2 The ph of a Buffered Solution I p684 A buffered solution contains 0.50 M Acetic acid (HC 2 H 3 O 2, K a = 1.8 x 10-5 ) and M sodium acetate (NaC 2 H 3 O 2 ). Calculate the ph of this solution. Solution: The major species in the the solution are HC 2 H 3 O 2, Na +, C 2 H 3 O 2-, and H 2 O weak neither base very weak acid or acid nor (conjugate base base base of HC 2 H 3 O 2 )

8 p684 8

9 p687 Solving Problems with Buffered Solutions 9

10 Buffering: How Does It Work? p687 10

11 Buffering: How Does It Work? p688 11

12 Ex 15.4 The ph of a Buffered Solution II p689 Calculate the ph of a solution containing 0.75 M lactic acid (ka = 1.4 x 10-4 ) and 0.25 M sodium lactate. Lactic acid ) HC 3 H 5 O 3 ) is a common constituent of biologic system. For example, it is found in milk and is present in human muscle tissue during exertion. Solution: 12

13 Summary of the Most Important Characteristic of Buffered Solutions Buffers contain relatively large amounts of weak acid and corresponding conjugate base. Added H + reacts to completion with the conjugate base. Added OH - reacts to completion with the weak acid. The ph is determined by the ratio of the concentrations of the weak acid and conjugate base. p692 13

14 15-3 Buffering Capacity The capacity of a buffered solution is determined by the magnitudes of [HA] and [A - ] Ex15.7 Adding strong Acid to a Buffered Solution II p693 Solution: Initial: For A: For B: 14

15 The optimal buffering system has a p696 pka value close to the desired ph.

16 15-4 Titrations and ph Curves p696 Strong acid-strong Base Titrations A. No NaOH has been added. [H + ] = M ph = B ml of M has been added. H + (aq) + OH - (aq) H 2O(l) ph = - log(0.15) =

17 C ml (total) of M NaOH has been added. ph = D ml (total ) of ml M NaOH has been added. Preceding exactly as for points B and C, the ph is found to be Enough OH - has been added to react exactly with H + from the nitric acid. This is the stoichiometric point, or equivalence point, of the titration. At this point the major species in solution are Na +, NO 3-, H 2 O 17

18 Titration p696 F ml (total) of M NaOH has been added. Proceeding as for point F, the ph is found to be G ml (total) of M NaOH has been added. ph =

19 Titration curve p699 Figure 15.1The ph curve for the titration of 50.0 ml of M HNO 3 with M NaOH. Note that the equivalence point occurs at ml of NaOH added, the point where exactly enough OH - has been added to react with all the H + originally present. The ph of 7 at the equivalence point is characteristic of a strong acid-base titration. 19

20 Neutralization of a Strong Acid with a Strong Base 20

21 p699 The ph Curve for the Titration of 50.0 ml of M HNO 3 with M NaOH 21

22 p699 Figure 15.2 The ph Curve for the Titration of ml of 0.50 M NaOH with 1.0 M HCI. 22

23 p704 Titrations of Weak Acids with Strong Bases Figure 15.3 The ph Curve for the Titration of 50.0 ml of M HC 2 H 3 O 2 with M NaOH. 23

24 The shapes of the strong and weak acid curves are the same after the equivalence points because excess OH - controls the ph in this region in both cases. p705

25 The ph curves for the titrations of samples with various Ka p707 Figure 15.4 The ph Curves for the Titrations of 50.0-mL Samples of 0.10 M Acids with Various K a Values with 0.10 M NaOH. 25

26 Weak Acid - Strong Base Titration Step 1: A stoichiometry problem (reaction is assumed to run to completion) then determine remaining species. Step 2: An equilibrium problem (determine position of weak acid equilibrium and calculate ph). 26

27 p711 Figure 15.5 The ph Curve for the Titrations of 100.0mL of M NH 3 with 0.10 M HCl. 27

28 15-5 Acid-Base Indicators p711 indicator Marks the end point of a titration by changing color. The equivalence point is not necessarily the same as the end point. 28

29 The acid and base forms of the indicator phenolphthalein. The indicator phenolphthalein is colorless in acidic solution and pink in basic solution. Figure

30 15.6 Solubility Equilibria and the Solubility Product p721 Relative Solubilities AgI(s) K sp CuI(s) CaSO 4 (s) K sp K sp

31 Common Ion Effect p722 We will now see what happen the water contains an ion in common with the dissolving salt. Ex Solubility and Common Ions Calculate the solubility of solid CaF 2 (K sp = 4.0 x ) in a M NaF solution. Solution: Thus mol solid CaF 2 dissolves per liter of the M NaF 10 solution. 31

32 15-7 Precipitation and Qualitative Analysis p725 Ex Determining Precipitation Conditions A solution is prepares by adding ml of 4.00 x 10-3 M Ce(NO 3 ) 3 to ml of 2.00 x 10-2 M KIO 3. Will Ce(IO 3 ) 3 (K sp = 1.9 x ) precipitate from this solution? Solution:

33 Ex Selective Precipitation A solution contains 1.0 x 10-4 M Cu + and 2.0 x 10-3 M Pb 2+. If a source of I - is added gradually to this solution, will PbI 2 (K sp = 1.4 x 10-8) or CuI ( K sp = 5.3 x ) precipitate first? Specify the concentration of I - necessary to begin precipitation of each salt. Solution: p728 As I - is added to the mixed solution, CuI will precipitate first, since the [I - ] required is less.

34 15-8 Equilibria Involving Complex Ions p731 [Ag ] 0 (100.0 ml)( (200.0 ml) 3 M ) [NH ] 0 (100.0 ml)(2.0 (200.0 ml) 3 M ) Ag+ + 2NH 3 Ag(NH 3 ) 2 + Before reaction: 5.0 x 10-4 M 1.0 M 0 M After reaction: (5.0 x 10-4 ) = 1.0 M 5.0 x 10-4 M K [Ag(NH3) 2 ] [Ag(NH ) ][NH ]

35 Complex Ion Equilibria p731 35

36 Nickel(II) Complexes 36

37 Ex Complex Ions Calculate the concentrations of Ag +, Ag(S 2 O 3 ) -, and Ag(S 2 O 3 ) 3- in a solution prepared by mixing ml of 1.00 x 10-3 M AgNO 3 with ml of 5.00 M Na 2 S 2 O 3. The stepwise formation equilibria are: Ag S Solution: Ag(S p O3 Ag(S2O3) K O3) S 2O3 Ag(S2O3) 2 K

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39 Complex Ions and Solubility p734 The solubility in pure water is 39

40 Separation Scheme 40