1. Acids, bases and pH

2. The Brønsted–Lowry theory of acids and bases involves the transfer of protons (hydrogen ions). Acids are proton donors.

3. Strong acids such as sulphuric acid are fully ionised in solution.

4. Weak acids such as ethanoic acid are partially ionised in solution.

5. In aqueous solution acids, shown as HA, ionise and donate protons to water molecules. The H 3 O + ion is called the oxonium ion.

6. The equation may be simplified like this.

7. Hydrochloric acid and nitric acid are strong acids and ionise fully.

8. They are monoprotic acids because they can donate one proton per molecule.

9. Sulphuric acid is also a strong acid and ionises fully. It is a diprotic acid because it can donate two protons per molecule.

10. We can calculate the pH of a solution if its hydrogen ion concentration is known.

11. We can also calculate the hydrogen ion concentration of a solution if its pH is known.

12. For a strong monoprotic acid like hydrochloric acid, the hydrogen ion concentration is equal to the concentration of the acid itself.

13. What is the pH of 0.10 mol dm –3 hydrochloric acid?

14. The hydrogen ion concentration will also be 0.10 mol dm –3.

15. The pH will be 1.0 for 0.10 mol dm –3 hydrochloric acid.

16. For a strong diprotic acid like hydrochloric acid, the hydrogen ion concentration is equal to twice the concentration of the acid itself.

17. What is the pH of 0.10 mol dm –3 sulphuric acid? The hydrogen ion concentration will be 0.20 mol dm –3.

18. The pH will be 0.69 for 0.10 mol dm –3 sulphuric acid.

19. What is the hydrogen ion concentration in nitric acid with a pH of 1.7?

20. Substitute the pH into the equation. Remember to enter ‘minus pH’ when you use a calculator.

21. The hydrogen ion concentration for nitric acid with a pH of 1.7 will be 0.02 mol dm –3. As it is a strong monoprotic acid, this will also be the concentration of the nitric acid itself.

22. The Brønsted–Lowry theory of acids and bases involves the transfer of protons, hydrogen ions. Acids are proton donors.

23. Strong acids such as sulphuric acid are fully ionised in solution.

24. Weak acids such as ethanoic acid are partially ionised in solution.

25. Weak acids are only partially ionised in solution, unlike strong acids (which are fully ionised).

26. In a weak acid like ethanoic acid, the hydrogen ion concentration is much less than the concentration of the acid.

27. We need to use the acid dissociation constant, K a, of the weak acid to calculate its hydrogen ion concentration.

28. This equation is often simplified using two assumptions: that [H + ]=[A – ] and that [HA] at equilibrium is equal to [HA] at the start. This gives a very close approximation.

30. The equation can be usefully rearranged to let us find the hydrogen ion concentration of a weak acid.

31. For example, what is the hydrogen ion concentration of 0.10 mol dm –3 ethanoic acid?

32. The acid dissociation constant for ethanoic acid is 1.7 x 10 –5 mol dm –3 at 298 K.

33. Substitute the acid concentration into the equation.

34. The hydrogen ion concentration of 0.10 mol dm –3 ethanoic acid is 1.3 x 10 –3 mol dm –3. (Using pH = –log 10 [H + (aq)], we find that the pH is 2.9.)

35. Water partially ionises to form hydrogen ions and hydroxide ions.

36. We can write an expression for the equilibrium constant, K c.

37. The concentration of water is effectively constant, so we can simplify the expression.

38. The resulting equilibrium constant is called the ionic product of water, and it has the symbol K w.

39. K w is 1 x 10 –14 mol 2 dm –6 at 298 K. The ionisation of water is an endothermic process, so K w increases as the temperature increases.

41. The Brønsted–Lowry theory of acids and bases involves the transfer of protons (hydrogen ions). Bases are proton acceptors.

42. Strong bases such as sodium hydroxide are fully ionised in solution.

43. Weak bases such as ammonia are partially ionised in solution.

44. In aqueous solution bases, shown as B, accept protons from water molecules.

45. The equation may be simplified like this.

46. Sodium hydroxide and potassium hydroxide are strong bases and ionise fully.

47. They are monoprotic bases because they can accept one proton per molecule.

48. To calculate the pH of a solution of a strong base we need to know the hydrogen ion concentration. We find this using the ionic product of water, K w.

50. Once we have worked out the hydrogen ion concentration we can work out the pH.

51. What is the pH of 0.10 mol dm –3 potassium hydroxide?

52. For a strong monoprotic base like potassium hydroxide, the hydroxide ion concentration is equal to the concentration of the base itself.

53. We need to use the ionic product of water to find the hydrogen ion concentration.

54. K w is 1.0 x 10 –14 mol 2 dm – 6 at 298 K, so we substitute this value and the hydroxide ion concentration into the equation.

55. This gives us a hydrogen ion concentration of 1.0 x 10 –13 mol dm –3.

56. The pH will be 13 for 0.10 mol dm –3 potassium hydroxide.