1. Topic 18.1 Calculations Involving Acids and Bases

2. Assessment Statements 18.1.1 State the expression for the ionic product constant of water (K w ) 18.1.2 Deduce [H + (aq)] and [OH - (aq)] for water at different temperatures given K w values. 18.1.3 Solve problems involving [H + (aq)], [OH - (aq)], pH and pOH. 18.1.4 State the equation for the reaction of any weak acid or weak base with water, and hence, deduce the expressions for Ka and Kb.

3. Assessments Statements 18.1.5 Solve problems involving solutions of weak acids and bases using the expressions: Ka + Kb = Kw pKa + pKb = pKw pH + pOH = pKw 18.1.6 Identify the relative strengths of acids and bases using values of Ka, Kb, pKa and pKb

4. Assessment Statements 18.1.1 State the expression for the ionic product constant of water (K w )

5. When water is purified by repeated distillation its electrical conductivity falls to a very low (and constant) value. This means that pure water dissociates to a small extent to form ions. H 2 O(l) ⇌ H+(aq) + OH-(aq) 2H 2 O(l) ⇌ H 3 O+(aq) + OH-(aq)

6. The Dissociation Constant At 25 o C (298K) the concentration of hydrogen and hydroxide ions are equal at 1.0 x 10 -7 mol dm -3 K w = [H + ][OH - ] = 1.00 x 10 -14 (at 298K) Dissociation Constant for water OR The Ionic Product Constant of Water

7. No matter the value of Kw (as it changes depending on the surrounding conditions), concentrations of [H+] and [OH-] will equal each other in pure water. To understand fully this topic you need to remember Topic 5 (Energetics) and Topic 7 (Equilibrium) particularly Le Chateliers Principle.

8. Assessment Statements 18.1.2 Deduce [H + (aq)] and [OH - (aq)] for water at different temperatures given K w values.

9. Water at Different Temperatures H 2 O (l) ↔ H + (aq) OH - (aq) ∆H = +57 k mol -1 Endothermic What happens to the pH if we increase the temperature? Will it be acidic, basic or neutral?

10. The Effect of Temperature pH Temperature

11. Water, pH and Temperature Water becomes increasingly dissociated and hence acidic as the temperature rises This accounts for the corrosive action of hot pure water on iron pipes: [H+] ions increased The concentrations of ions [H+] and [OH-] both increase equally, so the solution is neutral, But Kw changes

12. Kw Changes with Temperature

13. Dissociation of water is an Endothermic process: Topic 05 – energy is needed to separate oppositely charged ions Topic 07 – Le Châtelier’s principle predicts that increasing the temperature will favor the reaction that absorbs the heat (negates change) in this case, the endothermic process Increasing Kw values also tells that there is an increase in products of the equilibrium (more [H+] and [OH-]) H2O(l) ⇌ H+(aq) + OH-(aq)

14. Practice…….. At 60 o C, the ionic product constant of water is 9.6x10 -14 mol 2 dm -6. Calculate the pH of a neutral solution at this temperature Remember, what does neutral mean? What equations will we use? Calculate the pH of water at the temperatures in the table

15. Assessment Statements 18.1.3 Solve problems involving [H + (aq)], [OH - (aq)], pH and pOH.

16. Calculations The pH of a solution is dependant on the Hydrogen ions. pH = -log [H + ] or [H + ] = 10 -pH pOH = -log [OH - ] or [OH + ] = 10 -pOH

17. Try these……… What is the pH of these solutions when the [H + ] is…….. 1.00 x 10 -5 mol dm -3 5.00 x 10 -4 mol dm -3 What is the [H + ] of these solutions when the pH is…… 11.70 3.30

18. Note: The value pKw is used for –logKw As [H+][A-] = Kw Then: pH = pOH = pKw = 14.00 (at 25 o C)

19. Strong Acids and Bases Calculate the pH of 0.01 mol dm-3NaOH(aq) Calculate the pH of 0.01 mol dm-3Ba(OH)2 Calculate the pH of 0.01 mol dm-3sulfuric acid Hint: think about the strong/weak properties of sulfuric acid and it’s conjugates as it’s deprotonated A solution of HNO3(aq) contains 1.26g of the pure acid in every 100cm3 of aqueous solution. Calculate the pH of the solution A solution of NaOH(aq) contains 0.40g in every 50cm3 of aqueous solution. Calculate the pH of the solution

20. Tricky Ones........(Dilute Solutions) Calculate the pH of a dilute acid solution 1.0x10-8 moldm -3 HCl (aq) What wrong with your answer??????

21. This does not take into account that there are two sources of H+ions (the water and HCl) [H + ] Total = [H + ] H20 + [H+] HCl At higher concentrations of acid, the [H + ] contribution from water is insignificant and can be neglected. But when [H + ] < 1.0x10 -6

22. So for 1.0x10 -8 moldm -3 HCl: [H + ] Total = [H + ] H20 + [H + ] HCl Let [H + ] H20 = y Then OH is also, [OH - ] H20 = y Substitute: [H + ] Total = [H + ] H20 + [H + ] HCl [H + ] x [OH - ] = 1x10 -14 [[H + ] H20 + [H + ] HCl ] x [OH - ] = 1x10 -14 [ y + (1.0x10 -8 ) ] x [y] = 1x10 -14 y 2 + (1.0x10 -8 )y – (1.00x10 -14 ) = 0 Solve for y = 9.5x10 -8 = [H + ] H2O [H + ] total = [H + ] H20 + [H + ] HCl = 9.5x10 -8 + 1.0x10 -8 [H + ] total = 10.5x10 -8 pH = 6.98 so the solution is weakly acidic Although it is good to be aware, PROBABLY you would not be assessed on this……………

23. Also……… The scale is 0-14, but negative and 14+ values are possible 12M HCl, pH = -1.1 (Commercially available HCl) 10M NaOH, pH = 15.0 (Saturated NaOH) Again, it is good to know but your IB questions should be within the 0-14 range

24. Assessment Statements 18.1.4 State the equation for the reaction of any weak acid or weak base with water, and hence, deduce the expressions for Ka and Kb.

25. For Weak Acids HA (aq) ↔ H + (aq) + A - (aq) K a = ( [H+][A-] / [HA] ) Acid Dissociation Constant pKa = -log Ka hence Ka = 10 -pKa Note the greater the pKa value, the weaker the acid

26. Consider the Equilibrium HA (aq) ↔ H + (aq) + A - (aq) Initial Concentrations Final Concentrations a00 a-xxx

27. Substituting K a = ( [H+][A-] / [HA] ) = x.x /a-x = x 2 / a-x Since, in a weak acid, it is only slightly disassociated.. K a ≈ x 2 / a

28. Calculating K a A 0.0100 mol dm -3 solution of a weak acid has a pH of 5.00. What is the dissociation constant of the acid? If pH =5.00 then [H+] = [A-] = 1.00 x 10 -5 mol dm -3 K a ≈ x 2 / a ≈(1.00 x 10 -5 ) 2 / 0.0100 ≈ 1.00 x 10 -8 mol dm -3

29. Calculating pH Benzoic acid has a pKa of 4.20. What is the pH of a 0.100 mol dm -3 solution of acid? If pKa = 4.20, then Ka = 10 -4.20 = 6.31 x 10 -5 mol dm -3 K a ≈ x 2 / a 6.31 x 10 -5 ≈ x 2 / 0.10 ≈ 2.51 x 10 -3 mol dm -3 pH = -log(2.51 x 10 -3 ) = 2.6

30. Calculating Concentration What concentration of hydroflouric acid is required to give a solution of pH 2.00 and what percentage of hydroflouric Acid is dissociated at this pH if the dissociation constant of the acid is 6.76 x 10 -3 mol dm -3 ? If pH = 2.00, [H+] = [F-] = 1.00 x 10 -2 mol dm -3 K a ≈ x 2 / a 6.76 x 10 -3 ≈ (1.00 X 10-2) 2 / a a ≈ 0.148 mol dm -3 Percentage Dissociation = 100 x 1.00 X 10-2 / 0.148 = 6.76%

31. Weak Bases The Same principle holds true for weak bases B (aq) + H 2 O (l) ↔ BH + + OH - (aq) K b = ( [BH+][OH-] / [B] ) = y.y /b-y = y 2 / b Since, in a weak base, it is only slightly disassociated.. pK b = -logK b

32. Calculations involving a Weak Base What is the pH of a 0.0500 mol dm -3 solution of ethyamine ( pKb = 3.40) pKb = 3.40 Therefore Kb = 10 -3.4 = 3.98 x 10 -4 Kb = y 2 / b

33. 3.98 x 10 -4 = y 2 / 0.0500 y = √1.99 x 10 -5 y = [OH - ] = 4.46 x 10 -3 pOH = -log [OH - ] = 2.40 Therefore the pH = 14 – pOH = 11.6

34. Types of Calculations Calculating Ka Calculating pH Calculating [ ] Calculation involving weak base

35. In General HA (aq) ↔ H + (aq) + A - (aq) B (aq) + H 2 O (l) ↔ BH + + OH - (aq) therefore.. K a = ( [H+][A-] / [HA] ) and K b = ( [BH+][OH-] / [B] )

36. Recap Acid ionization constant (Ka) – equilibrium constant for the ionization of an acid. pKa= -log10Ka Base ionization constant (Kb) – equilibrium constant for the ionization of a base. pKb= -log10Kb Ionic product constant for water (Kw) – equilibrium constant for the ionization of water Kw 1.0x10-14 Only @25oC pKw= -log10Kw pKw= 14

37. Ka vs pKa and its Conjugates HCl(aq) + H2O(l)  H3O+(aq) + OH-(aq) Strong acids have weak conjugate bases Weak acids have strong conjugate bases Strong bases have weak conjugate acids Weak bases have strong conjugate acids

38. Summary Ka values, because often small (for weak acids) can be represented as pKa values -log10Ka= pKa Smaller pKa values are stronger Bigger Ka values are stronger Just like the pH scale, a change of 1 is a factor of 10; change of 2, a factor of 100 Ka values for strong acids are not usually quoted as the reactions go to completion and approach infinity in dilute solutions. Only changes in Temp affect the Ka, Kb, or Kc