Percent dissociation of weak acids

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

2. Dissociation increases with dilution

3. Polyprotic acids • Dissociations are stepwise H2CO3 + H2O = H3O+(aq) + HCO3-(aq) HCO3-(aq) + H2O = H3O+(aq) + CO32-(aq)

4. Ka 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 HnX

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 H2CO3 and H2O • Step 2: Ka1» Kw principal process is ionization of H2CO3

6. ICE Age Two

7. Step 5: Obtain x from Ka X <<0.02

8. Step 6: Big concentrations • [H3O+] = [HCO3-] = x = 9.3 x 10-5 M • [H2CO3] = 0.020 – x = 0.020 M

9. Step 7: small concentrations • Obtained from subsidiary equilibria: HCO3-(aq) + H2O = H3O+(aq) + CO32-(aq) [CO32-] = Ka2 = 5.6 x 10-11 M In general, [A2-] = Ka2 Ignore H3O+ generated in second ionization

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

11. Step 8: Calculate pH pH = -log10[H3O+] = -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 + H2O = CodH+(aq) + OH-(aq) Kb = 1.6x10-6 H2O + H2O = H3O+(aq) + OH-(aq) Kw = 1.0 x 10-14

14. Principal reaction is protonation of codeine

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 • [H3O+] from dissociation of water

18. Step 8: Calculate pH pH = -log10[H3O+] = -log(2.3x10-10) = 9.64

19. Relationship between Ka and Kb for conjugate acid-base pair • Acid HA + H2O = H3O+(aq) + A-(aq) • Base A- + H2O = HA(aq) + OH-(aq)

21. K for the overall reaction is product of K’s for individual reactions • In general Knet = K1 x K2 x K3 x… • For conjugate acid-base pairs Ka x Kb = Kw

22. Salts • Products of acid-base neutralization HCl + NaOH = NaCl + H2O HCl + KOH = KCl + H2O HNO3 + KOH = KNO3 + H2O 2HCl + Ca(OH)2 = CaCl2 + 2H2O HCN + NaOH = NaCN + H2O

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 Be2+) • Anions from strong monoprotic acids (Cl-, Br-, I-, NO3-, ClO4-)

25. Salts that yield acid solutions • Salts of weak bases NH4+(aq) + H2O = NH3 + H3O+(aq) • Hydrated small polarizing metal cations

26. What is pH of a 0.10 M solution of AlCl3 Ka = 1.4 x 10-5 • Using prior strategy: • Al(H2O)63+, Cl- and H2O are initial species • Al(H2O)63+ is stronger acid than H2O

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) + H2O = HCN(aq) + OH-(aq) • Calculation of pH follows same strategy as for the salt of strong acid and weak base except use expression for Kb

29. Weak acid and base: what then? • Competition between relative acid strength of cation and base strength of anion • Consider (NH4)2CO3 NH4+(aq) + H2O = H3O+(aq) + NH3(aq) Ka CO32-(aq) + H2O = HCO3-(aq) + OH-(aq) Kb • If Ka > Kb then pH < 7 • If Ka < Kb then pH > 7 • If Ka≈ Kb then pH ≈ 7

30. Summing up

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

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

34. Summing up

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

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

38. Lewis acids – no protons required • Lewis acid: an electron pair acceptor (BF3) • Lewis base: an electron pair donor (NH3) BF3 NH3 DONOR ACCEPTOR

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 • Al3+(acid) H2O (base) • Cu2+(acid) NH3 (base) • SO3 (acid) H2O (base)