1. Percent dissociation of weak acids Percent dissociation = [HA] dissociated x 100% [HA] dissociated x 100% [HA] initial [HA] initial Increases as K a increases For given acid, increases with dilution

2. Dissociation increases with dilution

3. Polyprotic acids Dissociations are stepwise H 2 CO 3 + H 2 O = H 3 O + (aq) + HCO 3 - (aq) HCO 3 - (aq) + H 2 O = H 3 O + (aq) + CO 3 2- (aq)

4. K a decreases with each step It is harder to remove a proton from a negative ion than from a neutral molecule Solution contains mixture of all species Strongest acid is the unionized molecule H n X

5. Use the same strategy described earlier What are concentrations of species present in 0.020 M solution of carbonic acid? Step 1: initial species are H 2 CO 3 and H 2 O Step 2: K a1 » K w principal process is ionization of H 2 CO 3

6. ICE Age Two Principal process Cod(aq) + H 2 O = CodH + (aq) + OH - (aq) Initial conc 0.001200 Change-xxx Equilibrium conc 0.0012 - x xx

7. Step 5: Obtain x from K a X <<0.02

8. Step 6: Big concentrations [ H 3 O + ] = [HCO 3 - ] = x = 9.3 x 10 -5 M [ H 2 CO 3 ] = 0.020 – x = 0.020 M

9. Step 7: small concentrations Obtained from subsidiary equilibria: HCO 3 - (aq) + H 2 O = H 3 O + (aq) + CO 3 2- (aq) [CO 3 2- ] = K a2 = 5.6 x 10 -11 M In general, [A 2- ] = K a2 Ignore H 3 O + generated in second ionization

10. Step 7 contd. [OH - ] from dissociation of water:

11. Step 8: Calculate pH pH = -log 10 [H 3 O + ] = -log(9.3x10 -5 ) = 4.03

12. Weak base equilibria Treated in an analogous way to weak acids

13. Worked example Calculate pH and concentrations of species in 0.0012 M solution of codeine Cod + H 2 O = CodH + (aq) + OH - (aq) K b = 1.6x10 -6 H 2 O + H 2 O = H 3 O + (aq) + OH - (aq) K w = 1.0 x 10 -14

14. Principal reaction is protonation of codeine Principal process Cod(aq) + H 2 O = CodH + (aq) + OH - (aq) Initial conc 0.001200 Change-xxx Equilibrium conc 0.0012 - x xx

15. Step 5: dissociation X <<0.0012

16. Step 6: Big concentrations [CodH + ] = [OH - ] = x = 4.4x10 -5 M [Cod] = 0.0012 – x = 0.0012 M (x«0.0012)

17. Step 7: small concentrations [H 3 O + ] from dissociation of water

18. Step 8: Calculate pH pH = -log 10 [H 3 O + ] = -log(2.3x10 -10 ) = 9.64

19. Relationship between K a and K b for conjugate acid-base pair Acid HA + H 2 O = H 3 O + (aq) + A - (aq) Base A - + H 2 O = HA(aq) + OH - (aq)

21. K for the overall reaction is product of K’s for individual reactions In general K net = K 1 x K 2 x K 3 x… For conjugate acid-base pairs K a x K b = K w

22. Salts Products of acid-base neutralization HCl + NaOH = NaCl + H 2 O HCl + KOH = KCl + H 2 O HNO 3 + KOH = KNO 3 + H 2 O 2HCl + Ca(OH) 2 = CaCl 2 + 2H 2 O HCN + NaOH = NaCN + H 2 O

23. Unequally yoked Consequences for “neutralized” solutions of the following combinations Strong acid + strong base NEUTRAL Strong acid + weak base ACIDIC Weak acid + strong base BASIC

24. Salts that yield neutral solutions Group 1A cations Group 2A cations (except Be 2+ ) Anions from strong monoprotic acids (Cl -, Br -, I -, NO 3 -, ClO 4 - ) (Cl -, Br -, I -, NO 3 -, ClO 4 - )

25. Salts that yield acid solutions Salts of weak bases NH 4 + (aq) + H 2 O = NH 3 + H 3 O + (aq) Hydrated small polarizing metal cations

26. What is pH of a 0.10 M solution of AlCl 3 K a = 1.4 x 10 -5 Using prior strategy: –Al(H 2 O) 6 3+, Cl - and H 2 O are initial species –Al(H 2 O) 6 3+ is stronger acid than H 2 O Principal process Al(H 2 O) 6 + (aq) + H 2 O(l) = H 3 O + (aq) + Al(H 2 O) 5 OH 2+ (aq) Initial conc 0.1000 Change-xxx Equilibrium conc 0.10 – x xx

27. pH = -log(1.2 x 10 -3 ) = 2.92 X << 1

28. Salts that yield basic solutions Salts of weak acid and strong base CN - (aq) + H 2 O = HCN(aq) + OH - (aq) Calculation of pH follows same strategy as for the salt of strong acid and weak base except use expression for K b

29. Weak acid and base: what then? Competition between relative acid strength of cation and base strength of anion Consider (NH 4 ) 2 CO 3 NH 4 + (aq) + H 2 O = H 3 O + (aq) + NH 3 (aq) K a CO 3 2- (aq) + H 2 O = HCO 3 - (aq) + OH - (aq) K b If K a > K b then pH K b then pH < 7 If K a 7 If K a ≈ K b then pH ≈ 7

30. Summing up Types of salt Examples Ions that react with water pH of solution Cation from strong base, anion from strong acid NaCl, Ba(NO 3 ) 2, None7 Cation from weak base, anion from strong acid NH 4 Cl, (CH 3 ) 4 CCl Cation <7 Small highly charged cation, anion from strong acid AlCl 3, Cr(NO 3 ) 3 Hydrated cation < 7 Cation from strong base, anion from weak acid NaCN, KF, Na 2 CO 3 Anion 7 Cation from weak base, anion from weak acid NH 4 CN, NH 4 F Both K b >7 if K a < K b ≈7 if K a ≈ K b

31. Factors affecting acid strength How easily is the H – X bond broken? –the less easily dissociated, the weaker the acid. Bond strength –Bond strength increases, acid strength decreases Polarity –For a given bond strength, the more polar the more easily dissociated – closer to H +

32. In a group, bond strength is the more important factor Polarity is not a good predictor: H-X polarity decreases down the group as acid strength increases Bond strength decreases more strongly down the group –Bond length increases sharply FClBrI Bond strength/kJmol -1 567431366299 H – X bond length/Å 0.9171.2741.4081.608 Electronegativity4.03.02.82.5

33. Electronegativity more reliable indicator in a period H-X polarity increases across period Acid strength increases across period Bond strength shows weaker trend –Bond lengths vary little CNOF Bond strength/kJmol -1 410390460570 H – X bond length/Å 1.0911.0080.9580.917 Electronegativity2.53.03.54.0

34. Summing up Electronegativity increases Acid strength increases Bond strength decreases Acid strength increases NH 3 H2OH2O HF PH 3 H2SH2S HCl AsH 3 H 2 Se HBr

35. Oxoacids and strength – where are the electrons going How to weaken the OH bond and increase its polarity? –Increase electronegativity of Y –Increase oxidation number of Y (number of O atoms around Y Both serve to withdraw electrons from the H making it more positive – closer to H +

36. Increasing electronegativity of Y In a homologous series, increasing electronegativity of Y increases strength –Y draws charge away from the H atom, more easily ionized NOTE: Opposite dependence to series HA, where A = F, Cl, Br, I Acid strength Acid HOIHOBrHOCl KaKaKaKa 2.3 x 10 -11 2.0 x 10 -9 3.5 x 10 -8 Electronegativity2.52.83.0

37. Increasing oxidation number (O atoms) More O atoms around Y, more acidic –The O atoms draw charge away from the H atom, more easily ionized Also think in terms of relative stabilities of the oxoanions formed in the process Acid strength Acid HOClHOClOHOClO 2 HOClO 3 Hypochlorous ChlorousChloricPerchloric KaKaKaKa 3.5 x 10 -8 1.2 x 10 -2 Large Oxidation number of Cl 1357

38. Lewis acids – no protons required Lewis acid: an electron pair acceptor (BF 3 ) Lewis base: an electron pair donor (NH 3 ) ACCEPTOR DONOR BF 3 NH 3

39. Lewis definition casts the net further All Arrhenius acids are Brønsted acids All Brønsted acids are Lewis acids The converses are not true Examples –Al 3+ (acid) H 2 O (base) –Cu 2+ (acid) NH 3 (base) –SO 3 (acid) H 2 O (base)