1. Acids and bases Different concepts Calculations and scales

2. Learning objectives You will be able to: –Identify acids and bases according to Arrhenius and Bronsted definitions –Identify conjugate acid base pairs –Calculate pH for solutions –Write dissociation constant expression K a –Determine pH of weak acid/base solutions –Predict pH of salt solutions –Identify factors that influence acid strength

3. ACIDS AND BASES …for it cannot be But I am pigeon-liver’d and lack gall To make oppression bitter… Hamlet Definitions of acids and bases pH scale Weak acids and equilibria Lewis acids and bases

4. ACIDS AND BASES The meaning of acid and base has evolved Arrhenius acid is one that generates protons when dissolved in water HA(aq) = H + (aq) + A - (aq) Arrhenius base is one that generates hydroxide ions when dissolved in water MOH(aq) = M + (aq) + OH - (aq)

5. Brønsted and Lowry A broader definition of acids and bases In the reaction NH 3 + HCl = NH 4 Cl has all the elements of acid-base neutralization but no H 2 O as would be required in the Arrhenius definition Brønsted acid donates a proton Brønsted base accepts a proton

6. H 3 O +, H 2 O and OH - all have the same number of valence electrons – they are isoelectronic

7. Neutralization The mixing of an acid with a base: ACID + BASE = SALT + WATER The reaction of carbonic acid (CO 2 in H 2 O) to give limestone: H 2 CO 3 + Ca(OH) 2 = CaCO 3 + 2H 2 O

8. The essence of neutralization Elimination of the components of acid and base by combination to give H 2 O H + + OH -  H 2 O ACID BASE

9. Brønsted acid HCl + H 2 O = H 3 O + + Cl - DONOR ACCEPTOR

10. Brønsted base H 2 O + NH 3 = NH 4 + + OH - water NH 3 + HCl = NH 4 + Cl - No water DONOR ACCEPTOR

11. The products are themselves acids and bases HA + B ↔ BH + + A - DONOR ACCEPTOR

12. Equilibrium: solution contains mixture of all components ACCEPTORDONOR ACCEPTOR

13. Conjugate acids and bases: follow the proton The difference is a proton HA + B ↔ A - + HB + Conjugate acid-base pair Conjugate base Conjugate acid base acid

14. Substances can be both acids and bases – depends on environment Note that in one instance H 2 O behaves like a base – accepting protons, and in another, behaves like an acid – donating protons HCl + H 2 O = H 3 O + + Cl - In presence of an acid H 2 O is a base NH 3 + H 2 O = NH 4 + + OH - In presence of a base H 2 O is an acid

15. It’s a competition for protons The substance that is a stronger proton donor becomes the acid HCl + H 2 O = H 3 O + + Cl - HCl + H 2 O = H 3 O + + Cl - The substance that is the stronger proton acceptor becomes the base NH 3 + H 2 O = NH 4 + + OH - NH 3 + H 2 O = NH 4 + + OH -

16. Strength and concentration Not all acids completely donate the protons to water molecules in solution HA + H 2 O  A - + H 3 O + The degree of ionization is described by strength The total number of moles per unit volume is described by concentration

17. Changing concentration does not change strength Strength refers to degree of ionization: –Strong is completely ionized (100 %) –Weak is partly ionized (1 % - 1:10 6 ) Concentration refers to number of moles per unit volume An acid (or base) can be strong and concentrated, weak and concentrated, strong and dilute, weak and dilute

18. Hydronium ion is the active ingredient of an acid in aqueous solution Protons do not exist in solution CH 3 CO 2 H + H 2 O = H 3 O + + CH 3 CO 2 - Vinegar in water produces hydronium ions

19. Hydroxide ion is the active ingredient of a base – in aqueous solution NH 3 + H 2 O = NH 4 + + OH - Ammonia, a base, dissolves in water and produces hydroxide ions

20. Amphotericity A substance that behaves as an acid and a base is amphoteric (amphiprotic). Water is a good example Acid Base

21. Ionization of water Even in pure water, a fraction of the molecules are ionized and the concentrations of OH - and H 3 O + are equal H 2 O + H 2 O = H 3 O + + OH - [H 3 O + ] = [OH - ] Concentration

22. In all aqueous solutions, product of concentrations is a constant [H 3 O + ][OH - ] = K w At 25°C, [H 3 O + ] = 1 x 10 -7 M Increasing [H 3 O + ] decreases [OH - ] (acidic conditions) Increasing [OH - ] decreases [H 3 O + ] (basic conditions)

23. Calculating [OH - ] from [H 3 O + ] [H 3 O + ][OH - ] = K w [OH - ] = K w / [H 3 O + ] Example: if [H 3 O + ] = 1 x 10 -3 M Example: if [H 3 O + ] = 1 x 10 -3 M Then [OH - ] = 10 -14 /10 -3 M = 10 -11 M = 10 -11 M

24. The pH scale – reduces large range of numbers to small In fact, [H 3 O + ][OH - ] = 10 -14 M 2 pH = - log 10 [H 3 O + ] –Range of [H 3 O + ] from 10 – 10 -14 M –Range of pH from -1 - 14 Low pH = acid; high pH = basic For example, 10 -1 M HCl has pH = 1 Pure water has [H 3 O + ] = 10 -7 M, pH = 7 Ammonia has [H 3 O + ] =10 -11 M, pH = 11 Note: change of 1 unit in pH is factor of ten in M

25. Deconstructing the pH pH = - log 10 [H 3 O + ] = -log 10 (a x 10 -b ) = -log 10 a + b = -log 10 a + b pH = 3.00 [H 3 O + ] = 1.0 x 10 -3 M pH = 2.595 [H 3 O + ] = 2.55 x 10 -3 M Exponent Prefix – 2 S.F. Exponent Prefix – 3 S.F.

26. Indicating pH All indicators involve a conjugate acid-base pair which have different colours HIn(aq) + H 2 O = H 3 O + (aq) + In - (aq) Colour A Colour B Colour A Colour B

27. Strong acids and bases and pH Monoprotic acids are completely ionized HCl + H 2 O → H 3 O + (aq) + Cl - (aq) Polyprotic acids are not completely ionized even if strong (stay tuned) Strong bases are completely dissociated NaOH + H 2 O → Na + (aq) + OH - (aq) CaO + H 2 O → Ca 2+ (aq) + 2OH - (aq)

28. Weak acids and equilibria HA(aq) + H 2 O ↔ A - (aq) + H 3 O + (aq) pK a = -log 10 K a

29. If the pH of 0.250 M HF is 2.036, what is K a ? [ H 3 O + ] = 9.20 x 10 -3 M (antilog -2.036) [F - ] = [ H 3 O + ] = 9.20 x 10 -3 M [HF] = 0.250 – [ H 3 O + ] = 0.250 - 9.20 x 10 -3 M = 0.241 M

30. Pathway to pH in a weak acid

31. Step 1: species present before dissocation HCN and H 2 O

32. Step 2: possible proton transfer processes 1. HCN(aq) + H 2 O = H 3 O + (aq) + CN - (aq) K a = 4.9 x 10 -10 2. H 2 O + H 2 O = H 3 O + (aq) + OH - (aq) K w = 1.0 x 10 -14

33. Step 3: identify principal process Largest K a is principal process, the rest are subsidiary Assume all H 3 O + (aq) is derived from principal process (in this case dissociation of HCN) OH - is derived from subsidiary process (in this case dissociation of water)

34. Step 4: ICE Age Principal process HCN(aq) + H 2 O = H 3 O + (aq) + CN - (aq) Initial conc 0.1000 Change -xxx Equilibrium conc 0.10 - x xx

35. Step 5: Substitute for x and solve Obtain quadratic expression in x But…x « 0.01, so 0.10 – x ≈ 0.10 Assumption x << 0.1 simplifies solution x 2 = 4.9 x 10 -11, x = 7.0 x 10 -6 Need to check assumption: 7.0x10 -6 <<0.1 X << 0.1

36. Step 6: from x calculate concentrations [ H 3 O + ] = [CN - ] = 7.0 x 10 -6 M [HCN] = 0.10 – x = 0.10 - 7.0 x 10 -6 = 0.10 M

37. Step 7: concentration of species from subsidiary processes OH - is derived from dissociation of H 2 O [OH - ] = K w / [ H 3 O + ] = 1.0x10 -14 / 7.0 x 10 -6 = 1.4 x 10 -9 M So, if [OH - ] = 1.4 x 10 -9 M, [ H 3 O + ] from dissociation of H 2 O = 1.4 x 10 -9 M Check: [ H 3 O + ] from HCN > [ H 3 O + ] from H 2 O ( 7.0 x 10 -6 > 1.4 x 10 -9 )

38. Step 8: Calculate pH pH = -log 10 [ H 3 O + ] = -log 10 (7.0 x 10 -6 ) = 5.15