1. Weak acid and base calculations

2. What’s so hard? Unlike strong acids and bases, weak examples do not dissociate fully. For example, a 1 molL -1 HCl solution dissociates fully, so we make the approximation that the [H 3 O + ] = 1 molL -1 For a weak acid e.g. CH 3 COOH, dissociation is only partial, so we cannot make the same approximation.

3. Example Calculate the pH of a 0.1 molL -1 ethanoic acid solution. K a = 1.74 x 10 -5 Step 1: Write a K a expression for the dissociation of the weak acid.  CH 3 COO - + H 3 O + CH 3 COOH + H 2 O  CH 3 COO - + H 3 O + Ka = Ka = [H 3 O + ][CH 3 COO - ] [CH 3 COOH] 1. That the amount of dissociation is so small it is insignificant. Therefore [CH 3 COOH] = 0.1 molL -1 in this example 2. That [H 3 O + ] = [CH 3 COO - ] (we will call this X) Step 2: Make 2 assumptions

4. Example ctd… Step 3: Substitute the values in and solve for X Ka = Ka = [H 3 O + ][CH 3 COO - ] [CH 3 COOH] 1.74 x 10 -5 = 1.74 x 10 -5 = [X][X] [0.1 molL -1 ] X 2 = 1.74 x 10 -5 x 0.1 molL -1 X = √ 1.74 x 10 -5 x 0.1 molL -1 [H 3 O + ] = 1.32 x 10 -3 molL -1 Step 4: Calculate the pH in the usual way pH = -log [H 3 O + ] = -log (1.32 x 10 -3 ) = 2.88

5. Now try these: 1. Calculate the pH of an 0.1 molL -1 propanoic acid solution given K a (propanoic acid) = 1.4 x 10 -8 Answer = 4.43 2. Calculate the pH of an 0.02 molL -1 ammonium chloride solution given pK a (ammonium ion) = 9.24 3. Calculate the pH of a 1 x 10 -3 molL -1 hydrogen fluoride solution given K a (HF) = 6.76 x 10 -4 Answer = 5.46 Answer = 3.09

6. Now try a weak base! Calculate the pH of a 0.25 molL -1 ammonia solution. K a (ammonium ion) = 1 x 10 -9 1. For weak base calculations, K b will never be given so you will have to calculate it K a x K b = 1 x 10 -14 1 x 10 -14 / 1 x 10 -9 = 1 x 10 -5 2. Write a K b expression as before  NH 4 + + OH - NH 3 + H 2 O  NH 4 + + OH - K b = K b = [OH - ][NH 4 + ] [NH 4 ]

7. Weak base ctd…… Step 2: Make 2 assumptions 1. That the amount of dissociation is so small it is insignificant. Therefore [NH 3 ] = 0.25 molL -1 in this example 2. That [OH - ] = [NH 4 + ] (we will call this X) Step 3: Substitute the values in and solve for X Kb = Kb = [OH - ][NH 4 + ] [NH 3 ] 1 x 10 -5 = 1 x 10 -5 = [X][X] [0.25 molL -1 ] X 2 = 1 x 10 -5 x 0.25 molL -1 X = √ 1 x 10 -5 x 0.25 molL -1 [OH - ] = 1.58 x 10 -3 molL -1

8. Weak base ctd…. Step 4: Calculate the pH in the usual way To calculate H 3 O + we need to use pH + pOH = 14 (or [OH - ][H 3 O + ]=1 x 10 -14 ) pOH = - log [OH - ] pH = 14 – pOH pH = 14 – 2.80 = 11.2

9. Now try these! 1. Calculate the pH of an 0.55 molL -1 solution of sodium ethanoate, given the K a for ethanoic acid = 1.74 x 10 -5