1.12 ACID BASE EQUILIBRIA

• Bronsted-Lowry acid


  A substance that can donate a proton
• Bronsted-Lowry base


  A substance that can accept a proton
• HCl (g) + H2O (l)

  H3O+ (aq) + Cl- (aq)
• Each acid is linked to a conjugate base


  On the other side of the equation
• pH


  • log [H+]
• Calculating pH of strong acids

  1. Strong acids completely dissociate

  2. The concentration of hydrogen ions will be the same as the concentration of the acid
• Always give pH values to 2d.p. in the exam
• Finding [H+] from pH

  [H+] = 1 x 10-pH
• In all aqueous solutions and pure water the following equilibrium occurs: H2O (l) H+ (aq) + OH- (aq)
• Ionic product for water

  Kc = [H+ (aq)][OH- (aq)] / [H2O (l)]
• Kw


  Kw = [H+ (aq)][OH- (aq)]
• At 25oC the value of Kw for all aqueous solutions is 1x10-14 mol2dm-6
• Finding pH of pure water

  [H+ (aq)] = [OH- (aq)] = √ Kw = 1x10-7 so pH = 7
• Increasing temperature


  Pushes the equilibrium to the right giving a bigger concentration of H+ ions and a lower pH
• Strong bases


  Completely dissociate into their ions
• Weak acids


  Only slightly dissociate when dissolved in water, giving an equilibrium mixture
• Weak acids dissociation expression

  Ka = [H+ (aq)][A- (aq)] / [HA (aq)]
• Ka


  The larger Ka the stronger the acid
• Calculating pH of a Weak Acid

  Assume [H+ (aq)]eqm = [A- (aq)] eqm and [HA (aq) ] eqm = [HA(aq) ] initial, Ka = [H+ (aq)]2 / [HA (aq)]initial
• pKa


  pKa = -log Ka, Ka = 10-pKa
• pH calculations involving neutralisation reactions

  1. Work out moles of original acid and hence moles H+

  2. Work out moles of base added and hence moles OH-

  3. Work out which one is in excess

  4. Work out new concentration of excess H+ or OH- ions

  5. Calculate pH
• Compounds


  • Ba(OH)2
• Reaction of HCl and Ba(OH)2

  1. 10 15cm3 of 0.5mol dm-3 HCl is reacted with

  2. 35cm3 of 0.45 mol dm-3 Ba(OH)2
• Reaction of H2SO4 and NaOH

  1. 35cm3 of 0.5 mol dm-3 H2SO4 is reacted with

  2. 30cm3 of 0.55 mol dm-3 NaOH
• Weak Acid and Strong Base Neutralisation

  Work out moles of original acid

  2. Work out moles of base added

  3. Work out which one is in excess

  4. Work out new concentration of excess HA

  [HA] = (initial moles HA – moles OH-) / total volume (dm3)

  5. Work out concentration of salt formed [A-]

  [A-] = moles OH- added / total volume (dm3)

  6. Rearrange Ka = [H+] [A-] / [HA] to get [H+]

  7. pH = – log [H+]
• Half neutralisation


  When a weak acid has been reacted with exactly half the neutralisation volume of alkali
• At half neutralisation we can make the assumption that [HA] = [A-]