Chapter 15 Additional Aspects of

Transcription

1 Chemistry, The Central Science Chapter 15 Additional Aspects of

2 Buffers: Solution that resists change in ph when a small amount of acid or base is added or when the solution is diluted. A buffer solution consists of a mixture of a weak acid and its conjugate base or a weak base and its conjugate at predetermined concentrations or ratios.

3 Buffers If a small amount of hydroxide is added to an equimolar solution of HF in NaF, for example, the HF reacts with the OH to make F and water.

4 Buffers If acid is added, the F reacts to form HF and water.

5 Buffered Solutions HX(aq) H + (aq) + X (aq) K a = [H + ] [X ] [HX] 1. If - OH is added: - OH(aq) + HX(aq) H 2 O(l) + X (aq) 2. If H + is added: H + (aq) + X - (aq) HX(aq) Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e 2010 Pearson Australia

6 There are two forms of buffers: Acidic buffers solutions which maintain a ph < 7 (components - weak acid and salt of a weak acid) Basic buffers - solutions which maintain a ph > 7 (components - weak base and salt of a weak base)

7 Buffers of ph < 7 weak acid + salt of weak acid CH 3 COOH CH 3 COO - Na + acetic acid Sodium acetate The weak acid is there to react with any OH - ions which might be introduced into the system. CH 3 COOH(aq) + OH - CH 3 COO - (aq) + H 2 O(l)

8 The effective result of this reaction is to convert a strong base (OH - ) into a weak base CH 3 COO - (aq). Thus although the ph will increase due to more base being present, the increase will be relatively small as the acetate ion is a weak base (Kb = 5.7 x ).

9 The salt of the weak acid is there to react with any H 3 O + ions which might be introduced into the system. CH 3 COO - (aq) + H 3 O + CH 3 COOH(aq) + H 2 O(l) Again the effective result of this reaction is to convert a strong acid H 3 O + into a weak acid (Ka (CH 3 COOH) = 1.75 x 10-5 ). The ph will decrease but only by a small amount.

10 Buffers of ph > 7 weak base + salt of a weak base NH 3 NH 4 Cl Ammonia Ammonium chloride The weak base is there to react with any H 3 O + ions. NH 3 (aq) + H 3 O + (aq) NH 4 + (aq) + H 2 O(l)

11 The salt of the weak base is there to react with any OH - ions. NH 4 + (aq) + OH - (aq) NH 3 (aq) + H 2 O (l)

12 Buffer Calculations Consider the equilibrium constant expression for the dissociation of a generic acid, HA: HA + H 2 O K a = [H 3 O + ] [A ] [HA] H 3 O + + A

13 Buffer Calculations ph = pk a + log [base] [acid] This is the Henderson Hasselbalch equation. Note : pka = log(ka)

14 Henderson Hasselbalch Equation What is the ph of a buffer that is 0.12 M in lactic acid, HC 3 H 5 O 3, and 0.10 M in sodium lactate? K a for lactic acid is pka = log(ka)

15 Henderson Hasselbalch Equation ph = pk a + log [base] [acid] ph = log ( ) + log (0.10) (0.12) ph = ( 0.08) ph = 3.77

16 ph Range The ph range is the range of ph values over which a buffer system works effectively. It is best to choose an acid with a pk a close to the desired ph. The useful ph range of a buffer is: pka ± 1

17 Calculating ph Changes in Buffers A buffer is made by adding mol HC 2 H 3 O 2 and mol NaC 2 H 3 O 2 to enough water to make 1.00 L of solution. The ph of the buffer is Calculate the ph of this solution after mol of NaOH is added.

18 Calculating ph Changes in Buffers Before the reaction, since mol HC 2 H 3 O 2 = mol C 2 H 3 O 2 ph = pk a = log ( ) = 4.74

19 Calculating ph Changes in Buffers The mol NaOH will react with mol of the acetic acid: HC 2 H 3 O 2 (aq) + OH (aq) C 2 H 3 O 2 (aq) + H 2 O(l) HC 2 H 3 O 2 C 2 H 3 O 2 OH Before reaction mol mol mol After reaction mol mol mol

20 Calculating ph Changes in Buffers Now use the Henderson Hasselbalch equation to calculate the new ph: ph = log (0.320) (0. 280) ph = ph = 4.80 What is the change in ph? ΔpH = ph final ph initial = = 0.06

21 Solubility Products Consider the equilibrium that exists in a saturated solution of BaSO 4 in water: BaSO 4 (s) Ba 2+ (aq) + SO 4 2 (aq)

22 Solubility Products The equilibrium constant expression for this equilibrium is K sp = [Ba 2+ ] [SO 4 2 ] where the equilibrium constant, K sp, is called the solubility product.

23 Solubility Products K sp is not the same as solubility. Solubility is generally expressed as the mass of solute dissolved in 1 L (g/l) or 100 ml (g/ml) of solution, or in mol/l (M).

24 Writing Solubility-Product (K sp ) Expressions Write the expression for the solubilityproduct constant for CaF 2,

25 The solubility of PbCl 2 in H 2 O is 1.6 x 10-2 M at 25 C. What is the value of Ksp for PbCl 2 at this temperature? PbCl 2 (s) Pb CIˉ K sp = [Pb 2+ ] ([CIˉ]) 2 K sp = (1.6 x 10-2 ) (2 x 1.6 x 10-2 ) 2 = 1.6 x 10-5

26 Will a Precipitate Form? In a solution, (Q = reaction quotiention product) If Q = K sp, the system is at equilibrium and the solution is saturated ( no precipitation). If Q < K sp, more solid will dissolve until Q = K sp (unsaturated no precipitation clear solution) If Q > K sp, the salt will precipitate until Q = K sp supersaturated solution- ppt will occur.

27 Calculate the molar solubility of silver bromide in water at 20 C. The solubility product at this temperature is 5.0 x AgBr(s) Ag + + Br - K sp = [Ag + ] [Br - ] Solubility = [AgBr] dissolved = [Ag + ] [Br - ] = [Ag + ] K sp = [Ag + ] 2 [Ag + ] = solubility = K sp 5.0 x x10 7 M

28 Calculate the solubility, in g/100 ml, of calcium iodate in water at 20 C. The solubility product at this temperature is 7.1 x10-7. Ca(IO 3 ) 2 (s) Ca IO 3ˉ K sp = [Ca 2+ ] [IO 3ˉ] 2 Solubility = [Ca(IO 3 ) 2 ] dissolved = [Ca 2+ ] [IO 3ˉ] = 2[Ca 2+ ] Substituting in the equilibrium expression gives K sp = [Ca 2+ ](2[Ca 2+ ]) 2 = 4[Ca 2+ ] 3

29 [Ca 2+ ] = solubility = 3 7 K sp 7.1 x x M

30 Converting to the desired concentration units, we obtain Solubility = (5.6 x 10-3 mol/l)(389.9g/mol) = 2.2g/L = 0.22g/100mL

31 Solid silver chromate is added to pure water at 25 C. Analysis of the equilibrated solution shows that its silver ion concentration is M. Assuming that Ag 2 CrO 4 dissociates completely in water and that there are no other important equilibria involving the Ag + or CrO 4 2 ions in the solution, calculate K sp for this compound.

32 At equilibrium [Ag + ] = M. All the Ag + and CrO 4 2 ions in the solution come from the Ag 2 CrO 4 that dissolves. Thus, we can use [Ag + ] to calculate [CrO 4 2 ]. We know that there must be 2 Ag + ions in solution for each CrO 4 2 ion in solution. Consequently, the concentration of CrO 4 2 is half the concentration of Ag +.

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34 Factors Affecting Solubility The Common-Ion Effect If one of the ions in a solution equilibrium is already dissolved in the solution, the equilibrium will shift to the left and the solubility of the salt will decrease. BaSO 4 (s) Ba 2+ (aq) + SO 4 2 (aq) Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e 2010 Pearson Australia

35 Factors Affecting Solubility Figure ph If a substance has a basic anion, it will be more soluble in an acidic solution. Substances with acidic cations are more soluble in basic solutions. Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e 2010 Pearson Australia

36 Factors Affecting Solubility Complex Ion Formation Metal ions act as Lewis acids and form complex ions with Lewis bases in the solvent. The formation of these complex ions increases with solubility of these salts. Figure Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e 2010 Pearson Australia

37 Factors Affecting Solubility Amphoterism Amphoteric metal oxides and hydroxides are soluble in strong acid or base, because they can act either as acids or bases. Examples of such cations are Al 3+, Zn 2+, and Sn 2+. Figure Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e 2010 Pearson Australia

38 The Common-Ion Effect Consider a solution of acetic acid HC 2 H 3 O 2 (aq) H + (aq) + C 2 H 3 O 2 (aq) If acetate ion is added to the solution, Le Châtelier says the equilibrium will shift to the left. Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e 2010 Pearson Australia

39 The Common-Ion Effect The extent of ionisation of a weak electrolyte is decreased by adding to the solution a strong electrolyte that has an ion in common with the weak electrolyte. Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e 2010 Pearson Australia

40 The Common-Ion Effect Calculate the fluoride ion concentration and ph of a solution that is 0.20 M in HF and 0.10 M in HCl. K a for HF is [H + ] [F ] K a = [HF] = Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e 2010 Pearson Australia

41 The Common-Ion Effect Because HCl, a strong acid, is also present, the initial [H+] is not 0, but rather 0.10 M. Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e 2010 Pearson Australia

42 The Common-Ion Effect = (0.10) (x) (0.20) (0.20) ( ) (0.10) = x = x Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e 2010 Pearson Australia

43 The Common-Ion Effect Therefore, [F ] = x = [H + ] = x = = 0.10 M So, ph = log (0.10) ph = 1.00 Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e 2010 Pearson Australia

44 Solubility in presence of a common ion In the absence of any competing equilibria, a precipitate is less soluble in a solution containing an excess of one of the ions common to the precipitate than it in pure water. According to Le Chatelier s principle, we expect a decrease in solubility where a common ion will shift the equilibrium to the left (reactants side).

45 Calculate the molar solubility of lead iodide in M sodium iodide solution. The Ksp for PbI 2 is 7.9 x PbI 2 (s) Pb I ˉ K sp = [Pb 2+ ] [I ˉ ] 2

46 There are now two sources of iodide: the NaI and the PbI 2. The amount of iodide coming from PbI 2 is small compared to that from the NaI. Thus [I ˉ] = C Nal + 2[Pb 2+ ] C Nal = M

47 K sp = [Pb 2+ ] [I ˉ ] 2 [Pb 2+ ] = solubility = Ksp /[I ˉ ] 2 = 7.9 x 10-9 / (0.200) 2 = 2.0 x 10-7 M. EX. Calculate the molar solubilty of lead iodide in water. ANS. 1.3 x 10-3 M. Compare your answers

48 K sp = [Pb 2+ ] [I ˉ ] 2 [I ˉ ] = 2[Pb 2+ ] K sp = [Pb 2+ ] (2[Pb 2+ ]) 2 = 4 [Pb 2+ ] 3 [Pb 2+ ] = solubility = 1.3 x 10-3 M.

49 Will a Precipitate Form? In a solution, (Q = reaction quotiention product) If Q = K sp, the system is at equilibrium and the solution is saturated ( no precipitation). If Q < K sp, more solid will dissolve until Q = K sp (unsaturated no precipitation clear solution) If Q > K sp, the salt will precipitate until Q = K sp supersaturated solution- ppt will occur.

50 Predicting Whether a Precipitate Will Form Will 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? When the two solutions are mixed, the total volume becomes 0.10 L L = 0.50 L. The number of moles of Pb 2+ in 0.10 L of M Pb(NO 3 ) 2 is:

51 The number of moles of SO 4 2 in 0.40 L of M Na 2 SO 4 is

52 Therefore, [SO 4 2 ] in the 0.50-L mixture is Because Q > K sp, PbSO 4 will precipitate.

53 Solubility Rules 1. All common compounds of Group I, and ammonium ions are soluble. 2. All nitrates, acetates, and perchlorates are soluble. 3. Salts of Ag, Hg(I), and Pb are insoluble. (Pb halides are soluble in hot water.) 4. Chlorides, bromides and iodides are soluble 5. All sulfates are soluble, except those of barium, strontium. 6. Except for rule 1, carbonates, hydroxides, sulphides, oxides, silicates, and phosphates are insoluble. Pg 152 L.Pillay, 2009

54 Solubility - Example Salts of Ag + are insoluble! Rule 3 Ag + Cu 2+ Sulphides insoluble S 2- Rule 6 Chlorides soluble * Cl - Rule 4 OH insoluble * Rule 6 OH - ppt ppt ppt ppt sol ppt L.Pillay, 2009

55 Solubility Rules Will a precipitate form when aqueous solutions of Ca(NO 3 ) 2 and NaCl are mixed? Ca(NO 3 ) 2 and NaCl are both soluble Combinations could be CaCl 2 and NaNO 3 CaCl 2 would be soluble (Rule 4) NaNO 3 would be soluble (Rule 2) No precipitate would form L.Pillay, 2009