Ionic Equilibria. weak acids and bases. salts of weak acids and bases. buffer solutions. solubility of slightly soluble salts

Transcription

1 Ionic Equilibria weak acids and bases salts of weak acids and bases buffer solutions solubility of slightly soluble salts

2 Arrhenius Definitions produce H + ions in the solution strong acids ionize completely H 2 O HA H + + A - weak acids partially ionize (equilibrium) H 2 O HA H + + A -

3 Bronsted Lowry Definitions an acid is a H + donor a base is a H + acceptor HA + H 2 O H 3 O + + A - "hydronium ion"

4 Measuring acidity use the ph scale ph = -log[h + ] or ph = -log[h 3 O + ]

5 Water ionizes slightly H 2 O + H 2 O H 3 O + + OH - K w = x10-14 = [H 3 O + ][OH - ] K w = x10-14 [H 3 O + ] = [OH - ] = x10-7 M ph = -log(1.000 x10-7 ) ph =

6 Relationships x10-14 = [H 3 O + ][OH - ] log of both sides -14 = log [H 3 O + ] + log [OH - ] multiply by = (-log [H 3 O + ]) + (-log [OH - ]) 14 = ph + poh (poh = -log[oh - ])

7 Summary [H 3 O + ][OH - ] = 1.00 x10-14 ph + poh = 14 ph = -log[h 3 O + ] poh = -log[oh - ] true for any aqueous solution!

8 Relationships [H 3 O + ][OH - ] = 1.00 x10-14 [H 3 O + ] [OH - ] ph = -log[h 3 O + ] poh = -log[oh - ] ph ph + poh = 14 poh

9 Examples Calculate the ph of the following [H 3 O + ] = 2.75x10-4 M [OH - ] = 8.243x10-5 M [H 3 O + ][OH - ] = 1.00 x10-14 [H 3 O + ](8.243x10-5 M) = 1.00 x10-14 [H 3 O + ] = 1.213x10-10 M

10 Calculate [H 3 O + ] Try Some Q1 ph = poh = 9.73

11 Weak Acids HCN + H 2 O H 3 O + + CN - K a = 4.0 x10-10 Ka = [H 3O + ][CN - ] [HCN]

12 Weak Bases CH 3 NH 2 HOH CH 3 NH OH - Kb = [CH 3 NH 3 +][OH-] [CH 3 NH 2 ]

13 Example Problem Calculate the ph of a 0.250M solution of HCN HCN + H 2 O H 3 O + + CN - Ka = [H 3O + ][CN - ] [HCN] K a = 4.0 x10-10 HCN H 3 O + CN - starting amounts reacting equilibrium

14 Solution Ka = [H 3O + ][CN - ] [HCN] Ka = (x)(x) (0.250-x) = x x note: a quadratic equation 4.0 x10-10 = assume x is small compared to (a reasonable assumption since K a is very small... very few products formed at equilibrium) x ph = -log[h 3 O + ]

15 % Ionization Weak acids partially ionize (by Arrhenius definition of "weak") amount ionized % ionization = x 100 starting amount % ionization = 1.0x10-5 M 0.25M From the last problem: amount ionized = 1.0x10-5 M starting amount = 0.25M x100 = % ionized (99.996% is dissolved but un-ionized!)

16 Try One Q2 A 0.100M solution of HClO is % ionized at equilibrium. Calculate K a for this acid. starting amounts reacting equilibrium HClO H 3 O + ClO -

17 Polyprotic Acids Acids with more than one "ionizable" H H 2 SO 4 H 3 PO 4 overall equation: H 2 SO H 2 O 2 H 3 O SO 4 H 2 SO 4 + H 2 O H 3 O +1 + HSO 4-1 HSO H 2 O H 3 O +1 + SO 4-2

18 K a and Polyprotic Acids H 2 S + 2 H 2 O 2 H 3 O +1 + S -2 H 2 S + H 2 O H 3 O +1 + HS -1 K a1 = 1.0 x10-7 HS -1 + H 2 O H 3 O +1 + S -2 K a2 = 1.0 x10-19

19 Conjugate Acids and Bases acid base conjugate acid conjugate base HA + H 2 O H 3 O + + A - "Weak acids form strong conjugate bases!" If K a is small, then the equilibrium is favored to the left. HA + H 2 O H 3 O + + A - this shows that A - is a strong base and easily accepts a proton

20 Hydrolysis conjugates of weak acids and bases are strong enough to react with water Ka for the weak acid HCN is 4.0 x10-10 CN - is the strong conjugate base of HCN CN - + HOH HCN + OH -

21 K hyd CN - + HOH HCN + OH - K hyd = [HCN][OH- ] [CN - ] K w = [H 3 O + ][OH - ] K a = [H 3O + ][CN - ] [HCN] K hyd = K w K a

22 Example Calculate the ph of a 0.25M solution of NaCN. NaCN is a soluble ionic comp'd and ionizes completely: NaCN Na + + CN M 0.25M 0.25M CN - is the conjugate of a weak acid and undergoes hydrolysis CN - + HOH HCN + OH - K hyd = K w K a = 1.0 x x10-10

23 Solution CN - + HOH HCN + OH - starting amounts CN - HCN OH - reacting equilibrium K hyd = [HCN][OH- ] [CN - ] assume x is small compared to 0.250

24 Finishing poh = -log[oh - ] = -log = 2.60 ph + poh = 14 ph = 11.40

25 Try One Q3 Calculate the ph of a 0.150M solution of NaBrO. K a for HBrO is 2.50x10-9

26 Buffer Solutions Can stabilize ph and resist slight additions of acid or base based on Le Chatelier's Principle weak acid or base + salt of its conjugate for example: HCN and NaCN NH 3 and NH 4 Cl

27 A Problem to Illustrate Calculate the ph of a solution that is M in HAc and M in NaAc. Ka = 1.85x10-5 HAc + H 2 O H 3 O + + Ac - Ka = [H 3O + ][Ac - ] [HAc] HAc H 3 O + Ac - starting amounts reacting equilibrium

28 Solution

29 Try One Q4 Calculate the ph of a solution made by dissolving mol NH 3 and mol NH 4 Cl into 1.00 L solution. K b for NH 3 is 1.80x10-5.

30 Adding Acid to a Buffered Solution Calculate the ph of the HAc/NaAc buffer solution (a previous problem) if mols HCl are added to a liter of the solution. add 0.010M HCl HAc + H 2 O H 3 O + + Ac -

31 Solution HAc H 3 O + Ac - starting amounts reacting equilibrium

32 Analysis of Results The buffer ph was 4.73 before adding acid. The ph changed to 4.69 (0.04 ph units) after addition of mols HCl to 1 L of buffer. If we added mols HCl to 1 L water the ph would fall from 7.00 (the ph of water) to 2.00 (0.010 mols/1 L = 1.00x10-2 M)

33 Try One Q5 For the NH 3 /NH 4 + buffer problem, calculate the ph after mols HCl are added to 1.00 L of buffer solution.

34 Henderson Hasselbach Equation HA + H 2 O H 3 O + + A - K a = [H 3 O + ][A - ] [HA] logka = log[h 3 O + ] + log [A- ] [HA] -logka = -log[h 3 O + ] - log [A- ] [HA] pka = ph - log [A- ] [HA] ph = pk a + log [A- ] [HA]

35 ph = pk a + log [A- ] [HA] An Example Calculate the ph of a solution that is 0.100M in HOCN and 0.150M in NaOCN, using the H/H equation. K a for HOCN is 3.50x10-4 pk a = -log K a = -log(3.50 x10-4 ) = 3.456

36 Try One Q6 Calculate the ph of a solution that is 0.250M in HAc and 0.150M in NaAc, using the H-H equation. Ka for HAc is 1.85x10-5.

37 Slightly Soluble Salts Compounds said to be "insoluble" in Chem 1 are really equilibria involving slightly soluble ionic compounds. BaSO 4(s) Ba +2-2 (aq) + SO 4 (aq) K sp = [Ba +2 ][SO 4-2 ]

38 A Problem to Illustrate Calculate the molar solubility of barium sulfate. K sp = 1.1x10-10 let x = molar solubility of BaSO 4 BaSO 4(s) Ba +2 (aq) SO 4-2 (aq) x x + K sp = [Ba +2 ][SO 4-2 ] K sp = (x)(x) = x x10-10 = x 2 x = 1.0 x10-5 M = solubility of BaSO 4

39 Another Example let x = molar solubility of Ag 2 CrO 4 Ag 2 CrO 4 (s) 2 Ag +1 (aq) + CrO 4-2 (aq) K sp = [Ag +1 ] 2 [CrO 4-2 ]

40 Useful Calculator Buttons y x or x y raise any number to any power x takes the "x th " root of a number

41 Try One Q7 The molar solubility of Ag 2 CO 3 is 1.27x10-4 M. Calculate Ksp for this salt.

42 Solubility and Common Ions "Ionic compounds are less soluble in a solution that contains a common ion." MX (s) M + (aq) + X - (aq)

43 Buffer Solutions are an Example of the Common Ion Effect HA (aq) H + (aq) + A - (aq) The ph of a 0.10M HAc solution is 2.87 The ph of a 0.10M HAc solution in 0.010M NaAc is 4.73

44 Solubility Examples Calculate the molar solubility of AgBr in water. AgBr Ag + + Br - Ksp = [Ag + ] [Br - ] =3.3x10-13 let x= the molar solubility of AgBr 3.3x10-13 =(x)(x)=x 2 x = 5.7x10-7 mol/l

45 Recalculate in 0.15M NaBr Ksp = [Ag + ] [Br - ] =3.3x10-13 let x= the molar solubility of AgBr 3.3x10-13 =(x)(x+0.15) 3.3x10-13 =(x)(0.15) x= 2.2 x10-12 mol/l

46 Try One Q8 Calculate the molar solubility of MgF 2 in M NaF. Ksp for MgF 2 is 6.40x10-9.

47 Fractional Precipitation Adding a precipitating reagent to a solution with more than one ion that could form insoluble compounds A solution is 1.0x10-3 M in Cl - and in Br -. Solid AgNO 3 is added gradually. Would AgBr or AgCl precipitate first? Ksp (AgBr) = 3.3x10-13 Ksp (AgCl) = 1.8x10-10

48 Solution We solve for the maximum silver ion concentration just before precipitation occurs. Ksp (AgBr) = [Ag + ] [Br - ] 3.3x10-13 =[Ag + ](1.0x10-3 ) Ksp (AgCl) = [Ag + ] [Cl - ] 1.8x10-10 =[Ag + ](1.0x10-3 ) [Ag + ]= 3.3x10-10 M [Ag + ]= 1.8x10-7 M

49 Another Question When AgCl starts to precipitate, what concentration of Br - is left in solution? Ksp (AgBr) = [Ag + ] [Br - ] From the previous slide, the [Ag + ] to initiate precipitation of AgCl must be >1.8x10-7 M 3.3x10-13 = (1.8x10-7 )[Br - ] [Br - ]= 1.8x10-6 M

50 Try One Q9 A solution is 0.15M in Pb +2 and 0.15M in Ag +1. If SO 4-2 ions are slowly added, which would precipitate first, PbSO 4 or Ag 2 SO 4? Ksp for PbSO 4 is 1.8x10-8 and for Ag 2 SO 4 is 1.7x10-5.

51 What is the [Pb +2 ] when Ag 2 SO 4 begins to precipitate?

52 Solubility and ph Calculate the ph of a saturated solution of Mg(OH) 2. K sp = 1.81x10-11 Mg(OH) 2 Mg +2 OH x 2x

53 Precipitation and ph At elevated ph, metal ions can precipitate as insoluble hydroxides At what ph will a 0.10M Ca(NO 3 ) 2 solution begin to precipitate Ca(OH) 2? Ca(OH) 2(s) Ca +2 (aq) + 2 OH -1 (aq) if the ph rises above Ca(OH) 2 will begin to precipitate

54 Try One Q10 At what ph will a 0.25M solution of MgCl 2 begin to precipitate Mg(OH) 2? Ksp = 1.5x10-11

55 One Final Note Be careful with very, very small K's in aqueous media. Recall that water self ionizes and in "neutral" water [H 3 O + ] = 1.00x10-7 M In all charts and tables where you said the starting amount of H 3 O + is "0" it is actually 1.00x10-7 M (same for OH - )