1. Calculating pH of strong acids and bases

2. Strong acids or bases are those which dissociate completely. HCl(aq) + H 2 O(l) → H 3 O + (aq) + Cl - (aq) So in a 0.5 mol L -1 solution of HCl there will be 0.5 mol L -1 of H 3 O + and 0.5 mol L -1 of Cl -. Similarly, NaOH is a strong base because 0.1 mol L -1 of NaOH(aq) contains 0.1 mol L -1 of Na + and 0.1 mol L -1 of OH -. There are only four strong acids that we normally deal with: HCl, HBr, HNO 3 and H 2 SO 4. All other acids should be considered as weak acids unless you are told otherwise. Likewise, the only strong bases you are likely to meet are NaOH, KOH and Ca(OH) 2.

3. Weak acids or bases differ from strong ones in that they only partially dissociate. Their reactions with water are equilibrium reactions. Ethanoic acid (in vinegar) is a weak acid. Most ethanoic acid molecules remain as molecules in water: only a few react to form CH 3 COO - and H 3 O +. Later on in this unit you will find out how to calculate exactly how much that ‘most’ is.

4. Example 1: Calculating the pH of a strong acid. Calculate the pH of a 0.0672 mol L -1 solution of nitric acid. Since nitric acid is a strong acid: HNO 3 (aq) + H 2 O(l) → H 3 O + (aq) + NO 3 - (aq) c (HNO 3 ) = 0.0672 mol L -1 so [H 3 O + ] = 0.0672 mol L -1 pH = -log[H 3 O + ] = -log(0.0672) = 1.17 Remember that [x] means ‘the concentration of x’. Do this problem on your calculator and check that you can get this answer.

5. Example 2: Calculating the pH of a strong acid. Calculate the pH of a 2.41 × 10 -3 mol L -1 solution of sulfuric acid. Each molecule of H 2 SO 4 will donate two protons: H 2 SO 4 (aq) + H 2 O(l) → 2H 3 O + (aq) + SO 4 2- (aq) Therefore c (H 2 SO 4 ) = 2.41 × 10 -3 mol L -1 But [H 3 O + ] = 2 × 2.41 × 10 -3 mol L -1 = 4.82 × 10 -3 mol L -1 pH = -log[H 3 O + ] = -log(4.82 × 10 -3 ) = 2.32 Do this problem on your calculator and check that you can get this answer.

6. Example 3: Calculating the pH of a strong base. Calculate the pH of a 0.00250 mol L -1 solution of potassium hydroxide. KOH is a strong base, therefore c (KOH) = 0.00250 mol L -1 = [OH - ] Before we can calculate the pH we need to know [H 3 O + ]. In any aqueous solution at room temperature [H 3 O + ][OH - ] = 10 -14 Therefore

7. So for our 0.00250 mol L -1 solution of potassium hydroxide: [OH - ] = 0.00250 mol L -1

8. An alternative method for calculating pH from [OH – ] This method is shorter than converting [OH – ] to [H 3 O + ], but may be harder to understand. Just as we can write pH = –log[H 3 O + ] so we can write pOH = –log[OH – ] And since 1  10 –14 = [H 3 O + ][OH – ] –log(1  10 –14 ) = (–log[H 3 O + ]) + (–log[OH – ]) 14 = pH + pOH If you don’t understand the mathematics of logs, don’t worry – just learn the bottom formula.

9. Calculate the pH of a 0.00250 mol L -1 solution of potassium hydroxide.