2. Starter question. Consider propanoic acid (CH 3 CH 2 COOH) reacting with water….. 1. Write a symbol equation for the equilibrium that is established. 2. Identify the conjugate base and acid on the right hand side of the equation. 3. Write a K c expression for the equilibrium.

3. Starter answers CH 3 CH 2 COOH (aq) + H 2 0(l) ↔ H 3 O + (aq) + CH 3 CH 2 COO - (aq) Conjugate base….CH 3 CH 2 COO - Conjugate acid H 3 O + Ka = [H + (aq)][CH 3 CH 2 COO - (aq)] [CH 3 CH 2 COOH (aq)]

4. The pH scale

5. Devised by a Danish chemist called Soren Sorensen. It indicates how much acid or alkali is present in a solution. The ‘p’ in pH stands for ‘potens’, which is latin for power. pH = -log[H + (aq)]

7. Calculating the pH of a strong acid.

8. Calculating the pH of a strong acid The pH of strong acids are relatively simple to calculate. If we assume that the acid molecules fully dissociate then [H + ] = [HA] Therefore pH = -log [HA] E.g. 1 mol HCl, pH = –log[1] = 0 E.g. 0.1 mol HCl, pH = -log[0.1] = 1

9. Calculating the pH of a weak acid.

10. Calculating the pH of a weak acid. An example. What is the pH of 0.1moldm -3 ethanoic acid, which has a Ka of….1.74X10 - 5 moldm -3 ?) You need to… 1. Write out the equilibrium expression (K a ) 2. convert hydrogen concentration [H + ] to pH

11. Step 1 the Ka expression. CH 3 COOH (aq) ↔ H + (aq) + CH 3 COO - (aq) Equilibrium equation. Ka = [CH 3 COO - (aq)][H + (aq)] [CH 3 COOH(aq)] So the general Ka expression will be…… But we need to consider when full equilibrium has been reached and take account of our initial concentration of acid.

12. Step 2 the equilibrium position. CH 3 COOHH+CH3COO- Start Conc.0.10000 At equilibrium. 0.100- [CH 3 COO - ][H + ][CH 3 COO - ] CH 3 COOH ↔ H + + CH 3 COO - CH 3 COOH H+H+ CH 3 COO - For every ethanoic acid molecule that dissociates an ethanoate ion and a H + also forms.

13. Writing the new K a expression at equilibrium. K a = [CH 3 COO - (aq)] eq [H + (aq)] eq 0.100 – [CH 3 COO - (aq)] eq As [CH 3 COO - ] = [H + ] at equilibrium we can write… K a = [H + ] 2 0.100 – [H + ] Initial concentration minus the dissociation concentration of the ions. Because the disassociation of the weak acid is so small we assume 0.100 – [H + ] = 0.100

14. Plugging the numbers in.. Substituting in the values gives…. 1.74 X 10 -5 = [H+] 2 0.100 [H+] 2 = 0.100 X 1.74 X 10 -5 [H+] = √(1.74 X 10 -6 ) [H+]= 1.319 X 10 -3 moldm-3 = pH = 2.88 (using –log)

15. Comparing strong and weak acids FactorsStrong AcidWeak Acid Acid diluted by a factor of 100 [H + (aq)] becomes 100 x smaller [H + (aq)] becomes 10 x smaller pH increases by 2pH increases by 1 ConductivityHigher as more free ionsLower as less free ions. Reaction RateReacts far quicker due to greater dissociation Reacts slower as relies on H+ being removed from equilibrium to completely dissociate acid molecules.

16. Strength vs. Concentration Concentration is a measure of the amount of substance in a given volume of solution. Strength is a measure of the extent to which an acid can donate H +. Measured as pK a values: pK a = -log K a

17. Calculating the pH of a strong base.

18. Calculating pH of a strong base. To begin with we need to understand another term, K w, this is the ionic product of water. For water, K a = [H + (aq)] [OH - (aq)] / [H 2 O(l)] As water is always present in excess we ignore the term, [H 2 O(l)] to give: K w = [H + (aq)] [OH - (aq)] At 298K K w = 1x10 -14 mol 2 dm -6

19. Calculating pH of a strong base. K w = [H + (aq)] [OH - (aq)] At 298K K w = 1x10 -14 mol 2 dm -6 K w = 1x10 -14 = [H + ] 2 Therefore [H + ] = 1x10 -7 =pH7 at 298K, pH will fall as temperature increases.

20. Calculating pH of a strong base. What is the pH of a 0.1 mol solution of sodium hydroxide? K w = 1 x 10 -14 = [H + ] x [OH - ] K w = 1 x 10 -14 = [H + ] x [0.1] [H + ] = 1 x 10 -14 / 0.1 [H + ] = 1 x 10 -13 pH = -log [1 x 10 -13 ] = 13