Chap 17 Additional Aspects of Aqueous Equilibria. Hsu Fu Yin

Transcription

1 Chap 17 Additional Aspects of Aqueous Equilibria Hsu Fu Yin 1

2 17.1 The Common-Ion Effect Acetic acid is a weak acid: CH 3 COOH(aq) H + (aq) + CH 3 COO (aq) Sodium acetate is a strong electrolyte: NaCH 3 COO(aq) Na + (aq) + CH 3 COO (aq) Presence of acetate from sodium acetate in an acetic acid solution will shift the equilibrium according to LeChâtelier s Principle: CH 3 COOH(aq) H + (aq) + CH 3 COO (aq) Whenever a weak electrolyte and a strong electrolyte containing a common ion are together in solution, the weak electrolyte ionizes less than it would if it were alone in solution. We call this observation the common-ion effect. 2

3 Exercise 17.1 Calculating the ph When a Common Ion Is Involved What is the ph of a solution made by adding 0.30 mol of acetic acid and 0.30 mol of sodium acetate to enough water to make 1.0 L of solution? Sol: 3

4 Exercise 17.2 Calculating Ion Concentrations When a Common Ion Is Involved Calculate the fluoride ion concentration and ph of a solution that is 0.20 M in HF and 0.10 M in HCl. Sol: 4

5 Buffers Solutions that contain high concentrations (10-3 M or more) of a weak conjugate acid base pair and that resist drastic changes in ph when small amounts of strong acid or strong base are added to them are called buffered solutions (or merely buffers). Ways to Make a Buffer: Mix a weak acid and a salt of its conjugate base or a weak base and a salt of its conjugate acid. Make the conjugate acid or base from a solution of weak base or acid by the addition of strong acid or base. 5

6 How a Buffer Works Adding a small amount of acid or base only slightly neutralizes one component of the buffer, so the ph doesn t change much. 6

7 Calculating the ph of a Buffer Solution EX: Calculate the ph of a buffer solution that is M in HC 2 H 3 O 2 and M in NaC 2 H 3 O 2 Sol: NaC 2 H 3 O 2 Na + + C 2 H 3 O 2-7

8 Con t 8

9 The Henderson Hasselbalch Equation a mixture of a hypothetical weak acid, HA (such as CH 3 COOH), and its salt, NaA (such as NaCH 3 COO ). 9 The Henderson Hasselbalch Equation

10 Exercise 17.3 Calculating the ph of a Buffer What is the ph of a buffer that is 0.12 M in lactic acid [CH 3 CH(OH)COOH, or HC 3 H 5 O 3 ] and 0.10 M in sodium lactate [CH 3 CH(OH)COONa or NaC 3 H 5 O 3 ]? For lactic acid, K a = Sol: 10

11 Exercise 17.4 Calculating ph When the Henderson Hasselbalch Equation May Not Be Accurate Calculate the ph of a buffer that initially contains M CH 3 COOH and M CH 3 COONa in the following two ways: (i) using the Henderson Hasselbalch equation and (ii) making no assumptions about quantities (which means you will need to use the quadratic equation). The K a of CH 3 COOH is Sol: 11

12 Exercise 17.5 Preparing a Buffer How many moles of NH 4 Cl must be added to 2.0 L of 0.10 M NH 3 to form a buffer whose ph is 9.00? (Assume that the addition of NH 4 Cl does not change the volume of the solution.) Sol: 12

13 Buffer Capacity A good buffer should be able to neutralize moderate amounts of added acid or base. However, there is a limit to how much can be added before the ph changes significantly. The buffering capacity is the amount of acid or base a buffer can neutralize. The buffering range is the ph range the buffer can be effective. A buffer will be effective when 0.1 < [base]/[acid] < 10. A buffer will be most effective when the [base]/[acid] = 1. 13

14 Addition of a Strong Acid or a Strong Base to a Buffer 14

15 Exercise 17.6 Calculating ph Changes in Buffers A buffer is made by adding mol CH 3 COOH and mol CH 3 COONa to enough water to make L of solution. The ph of the buffer is 4.74 (Sample Exercise 17.1). (a) Calculate the ph of this solution after 5.0 ml of 4.0 M NaOH(aq) solution is added. Sol: 15

16 Titration In this technique, an acid (or base) solution of known concentration is slowly added to a base (or acid) solution of unknown concentration. A ph meter or indicators are used to determine when the solution has reached the equivalence point The equivalence point the point in the titration when the number of moles of base is stoichiometrically equal to the number of moles of acid. A plot of the ph of the solution during a titration is known as a titration curve or ph curve. 16

17 Titration of a Strong Acid with a Strong Base From the start of the titration to near the equivalence point, the ph goes up slowly. Just before (and after) the equivalence point, the ph rises rapidly. At the equivalence point, ph = 7. As more base is added, the ph again levels off. 17

18 Titration of a Strong Base with a Strong Acid 18

19 Exercise 17.7 Calculations for a Strong Acid Strong Base Titration Calculate the ph when (a) 49.0 ml and (b) 51.0 ml of M NaOH solution have been added to 50.0 ml of M HCl solution. Sol: (a) Total Vol: 50.0 ml ml = 99.0 ml = L 19 ph=

20 Titration of a Weak Acid with a Strong Base Use K a to find initial ph. Find the ph in the buffer region using stoichiometry followed by the Henderson Hasselbalch equation. At the equivalence point the ph is >7. Use the conjugate base of the weak acid to determine the ph. As more base is added, the ph levels off. This is exactly the same as for strong acids. 20

21 Titration of a Weak Acid with a Strong Base 21

22 Exercise 17.8 Calculations for a Weak Acid Strong Base Titration Calculate the ph of the solution formed when 45.0 ml of M NaOH is added to 50.0 ml of M CH 3 COOH (K a = ). Sol: Total vol: 45.0 ml ml = 95.0 ml = L 22 ph= -log ( )+ log (0.0053/0.0474) = 5.70

23 Exercise 17.9 Calculating the ph at the Equivalence Point Calculate the ph at the equivalence point in the titration of 50.0 ml of M CH 3 COOH with M NaOH. Sol: Moles = M L = (0.100 mol/l)( L) = mol CH 3 COOH at the equivalence point : mloe of H + = mole of OH - CH 3 COOH + NaOH H 2 O + NaCH 3 COO Mole 23

24 Con t 24

25 Indicators Indicators are weak acids that have a different color than their conjugate base form. Each indicator has its own ph range over which it changes color. An indicator can be used to find the equivalence point in a titration as long as it changes color in the small volume change region where the ph rapidly changes. The point in a titration where the indicator changes color is called the end point 25

26 Indicator Choice Can Be Critical! 26

27 Titrations of Polyprotic Acids When a polyprotic acid is titrated with a base, there is an equivalence point for each dissociation. EX: pk a1 pk a2 Using the Henderson Hasselbalch equation, we can see that half way to each equivalence point gives us the pk a for that step. 27

28 17.4 Solubility Equilibria An equation can represent the equilibrium between the compound (solid is the reactant) and the ions present in a saturated aqueous solution. Ex: The equilibrium constant for a chemical equation representing the dissolution of an ionic compound is the solubility-product constant (K sp ). 28

29 Writing Solubility-Product (Ksp) Expressions (a) (b) 29

30 Solubility vs. Solubility Product Molar solubility is related to the value of K sp, but molar solubility and K sp are not the same thing. In fact, smaller K sp doesn t always mean lower molar solubility. Solubility depends on both K sp and the form of the equilibrium constant expression. Common units for solubility: Grams per liter (g/l) Moles per liter (mol/l) 30

31 Exercise Calculating Ksp from Solubility Solid silver chromate is added to pure water at 25, and some of the solid remains undissolved. The mixture is stirred for several days to ensure that equilibrium is achieved between the undissolved Ag 2 CrO 4 (s) and the solution. Analysis of the equilibrated solution shows that its silver ion concentration is M. Assuming that the Ag 2 CrO 4 solution is saturated and that there are no other important equilibria involving Ag + or CrO 2 4 ions in the solution, calculate K sp for this compound. Sol: [Ag + ] = M 31

32 Exercise Calculating Solubility from Ksp The K sp for CaF 2 is at 25 C. What is its molar solubility? Sol: 32

33 17.5 Factors That Affect Solubility The common ion effect affects solubility equilibria as it does other aqueous equilibria. The solubility of a slightly soluble ionic compound is lowered when a second solute that furnishes a common ion is added to the solution 33

34 Exercise Calculating the Effect of a Common Ion on Solubility Calculate the molar solubility of CaF 2 at 25 in a solution that is (a) M in Ca(NO 3 ) 2 and (b) M in NaF. Sol: (a) The initial concentration of Ca 2+ is M because of the dissolved Ca(NO 3 ) 2 : 34

35 Factors Affecting Solubility - ph ph: If a substance has a basic anion, it will be more soluble in an acidic solution. 35

36 Complex Ion Formation Transition metal ions tend to be good electron acceptors (good Lewis acids). Ex: H 2 O is the Lewis base, donating electron pairs to form coordinate covalent bonds A complex ion is a polyatomic cation or anion consisting of a central metal atom or ion that has other groups called ligands bonded to it The metal ion acts as a Lewis acid (accepts electron pairs). Ligands act as Lewis bases (donate electron pairs). 36

37 Complex Ion Formation The formation of complex ions increases the solubility of these salts. The equilibrium involving a complex ion, the metal ion, and the ligands may be described through a formation constant, K f : [Ag(NH 3 ) 2 ] + K f = = 1.7 x 10 7 [Ag + ][NH 3 ] 2 37

38 Exercise Evaluating an Equilibrium Involving a Complex Ion Calculate the concentration of Ag + present in solution at equilibrium when concentrated ammonia is added to a M solution of AgNO 3 to give an equilibrium concentration of [NH 3 ] = 0.20 M. Neglect the small volume change that occurs when NH 3 is added. Sol: 38

39 Amphoterism and Solubility Amphoteric oxides and hydroxides are soluble in strong acids or base, because they can act either as acids or bases. Examples are oxides and hydroxides of Al 3+, Cr 3+, Zn 2+, and Sn

40 17.6 Precipitation and Separation of Ions Precipitation will occur when the concentrations of the ions exceed the solubility of the ionic compound. If we compare the reaction quotient, Q, for the current solution concentrations to the value of Ksp, we can determine if precipitation will occur. Q = K sp, the solution is saturated, no precipitation. Q < K sp, the solution is unsaturated, no precipitation. Q > K sp, the solution would be above saturation, the salt above saturation will precipitate. Some solutions with Q > K sp will not precipitate unless disturbed; these are called supersaturated solutions. 40

41 Exercise Predicting Whether a Precipitate Forms Does a precipitate form when 0.10 L of M Pb(NO 3 ) 2 is added to 0.40 L of M Na 2 SO 4? Sol: Total Vol: 0.5L Moles of Pb 2+ : Mole of SO 4 2 : 41 Because Q > K sp, PbSO 4 precipitates.

42 Selective Precipitation A solution may contain several different dissolved metal cations that can often be separated by selective precipitation. A successful reagent can precipitate with more than one of the cations, as long as their K sp values are significantly different. Selective precipitation is also called fractional precipitation. 42

43 Selective Precipitation 43

44 Exercise Selective Precipitation A solution contains M Ag + and M Pb 2+. When Cl is added, both AgCl(K sp = ) and PbCl 2 (K sp = ) can precipitate. What concentration of Cl is necessary to begin the precipitation of each salt? Which salt precipitates first? Solve: For AgCl : K sp = [Ag + ][Cl ] = Because [Ag + ] = M For PbCl 2 : Because [Pb 2+ ] = M 44 AgCl precipitates first because it requires a much smaller concentration of Cl.

45 Selective Precipitation of Ions 45