acids & bases

• 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)
• pH


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

  1. Strong acids completely dissociate

  2. The concentration of hydrogen ions in a monoprotic strong acid will be the same as the concentration of the acid
• Always give pH values to 2 decimal places in the exam
• Finding [H+] from pH

  [H+] = 1 x 10-pH
• Ionic product for water

  Kc = [H+ (aq)][OH- (aq)] / [H2O (l)]
• At 25°C the value of Kw for all aqueous solutions is 1x10-14 mol2 dm-6
• Finding pH of pure water

  1. Pure water/ neutral solutions are neutral because [H+ (aq)] = [OH- (aq)]

  2. Using Kw = [H+ (aq)][OH- (aq)], when neutral Kw = [H+ (aq)]2

  3. At 25°C [H+ (aq)] = √1x10-14 = 1x10-7 so pH = 7
• As temperature increases


  The dissociation of water is endothermic so increasing the temperature would push the equilibrium to the right giving a bigger concentration of H+ ions and a lower pH
• Calculating pH of a strong base

  1. Assume complete dissociation

  2. Use Kw = [H+ (aq)][OH- (aq)] to calculate [H+ (aq)]
• Weak acids

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

  Ka = [H+ (aq)][A- (aq)] / [HA (aq)]
• The larger Ka the stronger the acid
• Calculating pH of a Weak Acid

  Assume [H+ (aq)]eqm = [A- (aq)]eqm

  Assume [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

  Work out moles of original acid and hence moles H+

  Work out moles of base added and hence moles OH-

  Work out which one is in excess

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

  Calculate pH from [H+] or [OH-]
• 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
• Moles HCl = mol H+ = conc x vol = 0.5 x 0.015 = 0.0075mol
• Moles Ba(OH)2 = conc x vol = 0.45 x 0.035 = 0.01575
• Moles OH- = 0.01575 x2 = 0.0315