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Weak acid and base calculations

Weak acid and base calculations. 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.

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Weak acid and base calculations

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  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 [H3O+] = 1 molL-1 For a weak acid e.g. CH3COOH, 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. Ka = 1.74 x 10-5 Step 1: Write a Ka expression for the dissociation of the weak acid. CH3COOH + H2O  CH3COO- + H3O+ Ka = [H3O+][CH3COO-] [CH3COOH] Step 2: Make 2 assumptions 1. That the amount of dissociation is so small it is insignificant. Therefore [CH3COOH] = 0.1 molL-1 in this example 2. That [H3O+] = [CH3COO-] (we will call this X)

  4. Example ctd… Step 3: Substitute the values in and solve for X Ka = [H3O+][CH3COO-] [CH3COOH] 1.74 x 10-5 = [X][X] [0.1 molL-1] X2 = 1.74 x 10-5 x 0.1 molL-1 X = √ 1.74 x 10-5 x 0.1 molL-1 [H3O+] = 1.32 x 10-3 molL-1 Step 4: Calculate the pH in the usual way pH = -log [H3O+] = -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 Ka (propanoic acid) = 1.4 x 10-8 Answer = 4.43 2. Calculate the pH of an 0.02 molL-1 ammonium chloride solution given pKa (ammonium ion) = 9.24 Answer = 5.46 3. Calculate the pH of a 1 x 10-3 molL-1 hydrogen fluoride solution given Ka (HF) = 6.76 x 10-4 Answer = 3.09

  6. Now try a weak base! Calculate the pH of a 0.25 molL-1 ammonia solution. Ka (ammonium ion) = 1 x 10-9 1. For weak base calculations, Kb will never be given so you will have to calculate it Ka x Kb = 1 x 10 -14 1 x 10-14 / 1 x 10-9 = 1 x 10-5 2. Write a Kb expression as before NH3 + H2O  NH4+ + OH- Kb = [OH-][NH4+] [NH4]

  7. Weak base ctd…… Step 2: Make 2 assumptions 1. That the amount of dissociation is so small it is insignificant. Therefore [NH3] = 0.25 molL-1 in this example 2. That [OH-] = [NH4+] (we will call this X) Step 3: Substitute the values in and solve for X Kb = [OH-][NH4+] [NH3] 1 x 10-5 = [X][X] [0.25 molL-1] X2 = 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 H3O+ we need to use pH + pOH = 14 (or [OH-][H3O+]=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 Ka for ethanoic acid = 1.74 x 10-5

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