1. Mullis1 Acid Base Equilibria Complications: Beyond pH of strong acids or strong bases HCl(aq) H + (aq) + Cl - (aq) K a > 1 HF(aq) H + (aq) + F - (aq) K a = 7.2 x 10 -4 Strong acid Weak acid Weak conjugate base stronger conjugate base

2. Mullis2 Conjugate Bases Bronsted-Lowry Model Conjugate base remains after acid loses its proton Strongest conjugate base = best at getting available protons Factors affecting conjugate base strength: 1.ElectronegativityS - > Cl - 2.Size if acids were binary F - > I - 3.Number of oxygens if acids were oxyacids ClO - > ClO 2 -

3. Mullis3 pH of Mixture of Weak Acids Find pH of a solution that contains 1.00M HCN (K a = 6.2 x 10 -10 ) and 5.00 M HNO 2 (K a = 4.0 x 10 -4 ). Because these are weak acids, dissociation is not complete and [H + ] ≠ [HNO 2 ]. You must use ICE table strategy to solve pH problems with weak acids or bases.

4. Mullis4 How to Determine [H + ] in an Acid Mixture HCNH + + CN - K a = 6.2 x 10 -10 HNO 2 H + + NO 2 - K a = 4.0 x 10 -4 H 2 OH + + OH - K a = 1.0 x 10 -14 Compare species and find the most likely contributor of protons. Here it is by far HNO 2 : K a is 6 orders of magnitude greater than the closest possible H + contributor.

5. Mullis5 Equilibrium Method to Find [H + ] RHNO 2 H+H+ NO 2 - I5.00 M00 C-x+x E5.00 – xxx K a = x 2 = 4.0 x 10 -4 (5.00 –x) x 2 = 20. x 10 -4 x = 4.5 x 10 -2 Small: reactants favored Change must be very small relative to initial conc.: Ignore x in this term

6. Mullis6 Accuracy check When dealing with weak acids and bases, K a values are typically valid to a ±5% accuracy. Verify that our assumption in ignoring x in the reduction of [HNO 2 ] 0 was valid: x x 100% = 4.5 x 10 -2 (100%) = 0.90 % [HNO 2 ] 5.00 Since 0.90% is less than 5%, the assumption is valid---in other words, our answer for x is good.

7. Mullis7 What was the Original Question? Find pH of this solution: [H + ] = 4.5 x 10 -2 pH = -log[H + ] = -log(4.5 x 10 -2 ) = 1.35 Number of decimals in pH = Number of significant figures in concentration (2 here)

8. Mullis8 Find [CN - ] in that problem. [H + ] = 4.5 x 10 -2 M HCNH + + CN - K a = 6.2 x 10 -10 6.2 x 10 -10 = [4.5 x 10 -2 ][CN - ] [HCN] Remember our assumption that since K a is small, we ignore the small reduction in initial concentration of product? Be consistent and make that assumption here, too. So: 6.2 x 10 -10 = [4.5 x 10 -2 ][CN - ] 1.00 [CN - ] = 1.4 x 10 -8 M (That’s a small number!)

9. Mullis9 % Dissociation Ex: HNO 2 H + + NO 2 - [H + ] x 100% = % dissociation [HNO 2 ] Or… [NO 2 - ] x 100% = % dissociation [HNO 2 ] If you know % dissociation and original concentration of acid [HNO 2 ], can solve for [H + ] or [NO 2 - ] (or x, the change in concentration).

10. Mullis10 % Dissociation and Dilution More Concentrated Diluted Less % dissociationMore % dissociation (more chance of recombination) If Q = K for this reactionQ < K a for this one, so equilibrium shifts right

11. Mullis11 K a from % Dissociation A 0.100 M solution of lactic acid (HC 3 H 5 O 3 ) is 3.7% dissociated. Calculate the value of K a. RHC 3 H 5 O 3 H+H+ C3H5O3 -C3H5O3 - I0.100 M00 C-x+x E0.100 – xxx % dissociation = x x (100%)K a = x 2 0.10 (0.100 –x) 3.7% (0.10) = xK a = 0.0037 2 x = 0.0037 M (0.100 -.0037) K a = 1.4 x 10 -4

12. Mullis12 Organic Acids (Acid = proton donor) -OH phenol

13. Mullis13 Bases without OH - B (aq) + H 2 O BH + (aq) + OH - (aq) NH 3 (aq) + H 2 O NH 4 + (aq) + OH - (aq) K b = [BH + ][OH - ] [B] conjugate base conjugate acidacid base

14. Mullis14 Amines R x NH (3-x) Ammonia with one or more of the H atoms replaced by another group. Many organic molecules involve amines. N. pyridine ethylamine. N H H C2H5C2H5 HO OH CHCH 2 NHCH 3 adrenaline

15. Mullis15 Find the pH of 1.0M methylamine (K b =4.38x10 -4 ). RCH 3 NH 2 CH 3 NH 3 + OH - I1.0 M00 C-x+x E1.0 – xxx CH 3 NH 2 + H 2 O CH 3 NH 3 + + OH - K b = x 2 = 4.38x10 -4 x 2 ≈ 4.38x10 -4 1.0-x x≈ 2.1x10 -2 [OH - ] = x = 2.1x10 -2 M pOH=-log[2.1x10 -2 ] pOH = 1.68 pH= 14 -1.68= 12.32

16. Mullis16 PolyProtic Acids Lose one proton at a time: H 3 PO 4  H 2 PO 4 -  HPO 4 2-  PO 4 3- The reactions written to express this loss have successively smaller K a values: K a1 > K a2 > K a3 …..makes sense: The loss of a second proton occurs less readily than the loss of the first one. In fact, K a1 is usually so much larger than the others, the loss of the first proton is the only reaction that significantly contributes [H+].

17. Mullis17 Sequential Loss of Protons H 3 PO 4 H + + H 2 PO 4 - K a1 = 7.5 x 10 -3 H 2 PO 4 - H + + HPO 4 2- K a2 = 6.2 x 10 -8 HPO 4 2- H + + PO 4 3- K a3 = 4.8 x 10 -13

18. Mullis18 pH of a Polyprotic Acid Find pH of 5.0M H 3 PO 4 and the equilibrium concentrations of each species: H 3 PO 4, H 2 PO 4 -, HPO 4 2-, PO 4 3- K a1 = [H + ][H 2 PO 4 - ] = x 2 = 7.5 x 10 -3 [H 3 PO 4 ] 5.0-x (see previous problems’ RICE tables & pH calculations for details) x 2 = (5.0)7.5 x 10 -3 x ≈ 0.19 M= [H 2 PO 4 - ] = [H + ] pH = 0.72 [H 3 PO 4 ] = 5.0 – x = 4.8 M K a2 = [H + ][HPO 4 2- ] = (0.19) [HPO 4 2- ] = 6.2 x 10 -8 [H 2 PO 4 - ] 0.19 [HPO 4 2- ]= 6.2 x 10 -8 M K a3 = [H + ][PO 4 3- ] = (0.19) [PO 4 3- ] = 4.8 x 10 -13 [HPO 4 2- ] 6.2 x 10 -8 [PO 4 3- ]= 1.6 x 10 -19 M

19. Mullis19 Sulfuric acid: Relatively High Conc. Find pH of 1.0 M H 2 SO 4. H 2 SO 4  H + + HSO 4 - K a1 > 1 HSO 4 - H + + SO 4 2- K a2 = 1.2 x 10 -2 In this case, does the 2 nd reaction contribute significantly to [H + ]? [H + ] is at least 1.0 since the first step is total dissociation.

20. Mullis20 Sulfuric acid: Relatively High Conc. 1.2 x 10 -2 = [H + ][SO 4 2- ] = (1.0 + x)(x) [HSO 4 - ](1.0 – x) Assume change is small compared to 1.0, so: 1.0(x) 1.0 X = 1.2 x 10 -2, or 1.2% of 1.0 M. If this is added to 1.0, result is 1.0012M, but with 2 sig. figs., [H + ] is 1.0 M. (pH = 0.00) RHSO 4 - H+H+ SO 4 2- I1.0 M1.00 C-x+x E1.0 – x1.0 + xx From 1 st dissociation step

21. Mullis21 0.01 M H 2 SO 4 (Relatively Low Conc.) 1.2 x 10 -2 = [H + ][SO 4 2- ] = (.01 + x)(x) [HSO 4 - ] (.01 – x) If assume change is small compared to.01,.01(x).01 x = 1.2 x 10 -2, or more than 0.01 M! Therefore, cannot ignore the subtraction or addition of x—use the quadratic to find x =.0045. Total [H + ] = 0.01(from 1 st dissociation step) +.0045 = 0.0145. pH = -log(0.0145) = 1.84 RHSO 4 - H+H+ SO 4 2- I0.01 M0.010 C-x+x E0.01 – x0.01 + xx From 1 st dissociation step

22. Mullis22 Salts of Weak Acids or Bases Cations of strong bases do not change pH of a solution. –(Ex: Na + will not attract or contribute H + ) Salts in which cations do not change pH and the anion is the conjugate base of a weak acid will produce basic solutions. –Ex. KC 2 H 3 O 2 dissociates to produce C 2 H 3 O 2 - and OH - Salts in which anions are not a base and the cation is a conjugate acid of a weak base will produce acidic solutions. –Ex. NH 4 Cl dissociates to produce NH 3 and H +.

23. Mullis23 Finding K b for Conjugates Look up the K a for the weak acid that makes the conjugate base. Ex: NaC 2 H 3 O 2 (aq) C 2 H 3 O 2 - + H 2 O  HC 2 H 3 O 2 + OH - K a(acetic acid) = 1.8 x 10 -5 K a K b = K w K b = 1x10 -14 = 5.6 x 10 -10 1.8 x 10 -5

24. Mullis24 Salt as a Weak Base Find pH of a 0.30 M NaF solution. Na + does not change pH. F - is the conjugate base of HF. Water has to be the proton donor for this solution. F - (aq) + H 2 O (aq)   HF (aq) + OH - (aq) K a for HF is 7.2 x 10 -4. K b = 1.0 x 10 -14 = 1.4 x 10 -11 7.2 x 10 -4

25. Mullis25 Salt as a Weak Base K b = 1.4 x 10 -11 = [HF][OH - ] = x 2 [F - ].30-x x 2 =.30(1.4 x 10 -11 )x = 2.0 x 10 -6 (<5% of.30) [OH - ] = 2.0x10 -6 pOH = 5.69pH= 14.00 – 5.69 = 8.31 RF-F- H 2 O ( l )HFOH - I0.30--0≈ 0 C-x--+x E0.30 - x-- + xx

26. Mullis26 Salt as a Weak Acid Find pH of a 0.10 M NH 4 Cl solution. Cl - does not change pH. NH 4 + produces H + in water. It is the conjugate acid of NH 3. NH 4 + (aq)   NH 3 (aq) + H + (aq) K b for NH 3 is 1.8 x 10 -5. K a for NH+ = 1.0 x 10 -14 = 5.6 x 10 -10 1.8 x 10 -5

27. Mullis27 Salt as a Weak Acid K b = 5.6 x 10 -10 = [NH 3 ][H + ] = x 2 [NH 4 +].10-x x 2 =.10(5.6 x 10 -10 )x = 7.5 x 10 -6 (<5% of.10) [H + ] = 7.5 x 10 -6 pH =-log(7.5 x 10 -6 ) = 5.13 RNH 4 + NH 3 +H+H+ I0.100≈0≈0 C-x+x E0.10 - x + xx

28. Mullis28 Acid-Base Properties of Salts Salt Example pH + from strong base - from strong acid NaClneutral + from strong base - from weak acid NaC 2 H 3 O 2 Anion acts as base basic + conj. acid fm. weak base - from strong acid NH 4 Cl Cation acts as acid acidic + conj. acid fm. weak base - conj. base fm. weak acid NH 4 C 2 H 3 O 2 Anion acts as base, cation as acid Acidic if K a > K b + from metal ion - from strong acid FeCl 3 Hydrated cation acts as acid acidic

29. Mullis29 Practice 1.A 0.15 M solution of a weak acid is 3.0% dissociated. Calculate K a. 2.What are the major species present in a 0.150 M NH 3 solution? Calculate the [OH - ] and the pH of this soln. 3.Calculate the pH of a 5.0 x 10 -3 M solution of H 2 SO 4. 4.Sodium azide is sometimes added to water to kill bacteria. Calculate the concentration of all species in a 0.010 M solution of NaN 3. The K a value for HN 3 (hydrazoic acid) is 1.9 x 10 -5. 5.Calculate the pH of 0.10 M CH 3 NH 3 Cl. K b of CH 3 NH 3 + is 4.38 x 10 -4.