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+]
• [H+]


  Concentration of hydrogen ions in the solution
• 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
• Finding [H+] from pH

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

  [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

  [H+ (aq)] = [OH- (aq)] = √Kw = 1x10-7 mol dm-3, so pH = 7
• Increasing temperature


  Pushes the water dissociation equilibrium to the right, increasing [H+] and decreasing pH
• Calculating pH of a strong base


  Use Kw = [H+ (aq)][OH- (aq)] to find [H+ (aq)] from [OH- (aq)]
• Weak acids


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

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


  Acid dissociation constant, the larger the value the stronger the acid
• Calculating pH of a weak acid

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


  • log Ka
• pH calculations involving neutralisation reactions

  Determine moles of acid and base, find which is in excess, calculate concentration of excess H+ or OH-
• Such as Ba(OH)2
• Calculating pH


  1. Moles HCl = 0.0075mol

  2. Moles Ba(OH)2 = 0.01575

  3. Moles OH- = 0.0315

  4. Moles of OH- in excess = 0.024

  5. [OH-] = 0.48 mol dm-3

  6. [H+] = 2.08x10-14 mol dm-3

  7. pH = 13.68
• Calculating pH


  1. Moles H2SO4 = 0.0175mol

  2. Moles H+ = 0.035

  3. Moles NaOH = 0.0165

  4. Moles of H+ in excess = 0.0185

  5. [H+] = 0.28 mol dm-3

  6. pH = 0.55
• Weak Acid and Strong Base Neutralisations

  • Work out moles of original acid
  • Work out moles of base added
  • Work out which one is in excess
  • Work out new concentration of excess HA
  • Work out concentration of salt formed [A-]
  • Rearrange Ka = [H+] [A-] / [HA] to get [H+]
  • pH = -log [H+]
• Calculating pH of weak acid-strong base neutralisation

  1. Moles CH3CO2H = 0.0275mol

  2. Moles NaOH = 0.00875

  3. Moles of CH3CO2H in excess = 0.01875

  4. [CH3CO2H] = 0.234M

  5. pH = 4.44

  6. [CH3CO2-] = 0.109M

  7. [H+] = 3.64 x 10-5 M
• Weak Acid at Half Equivalence

  • At half neutralisation [HA] = [A-]
  • pH = pKa
• pH = pKa = 4.77
• Diluting an acid or alkali

  • pH of diluted strong acid: [H+] = [H+]old x old volume / new volume, pH = -log [H+]
  • pH of diluted base: [OH-] = [OH-]old x old volume / new volume, [H+] = Kw / [OH-], pH = -log [H+]
• Calculating pH of diluted strong acid

  1. [H+] = 0.0136 mol dm-3

  2. pH = 1.87
• Buffer Solutions


  • Acidic buffer: weak acid + salt of weak acid
  • Basic buffer: weak base + salt of weak base
  • Buffer solutions resist changes in pH when small amounts of acid or alkali are added
• In a buffer solution there is a much higher concentration of the salt conjugate base ion than in the pure acid
• How buffer solutions work

  • If acid is added, the conjugate base reacts with it to remove the H+ ions
  • If alkali is added, the equilibrium shifts to produce more H+ ions
  • The ratio [HA]/[A-] stays almost constant, so pH stays constant
• Calculating pH of buffer made by adding salt solution

  1. Moles ethanoic acid = 0.0045 mol

  2. Moles sodium ethanoate = 0.0075 mol

  3. [H+] = 1.02x 10-5 mol dm-3

  4. pH = 4.99
• Calculating pH of buffer made by adding solid salt

  1. Moles ethanoic acid = 0.04 mol

  2. Moles sodium ethanoate = 0.0134 mol

  3. [H+] = 5.07x 10-5 mol dm-3

  4. pH = 4.29
• Calculating pH of buffer made by partially neutralising a weak acid

  1. Moles CH3CO2H = 0.0275 mol

  2. Moles NaOH = 0.00875 mol

  3. Moles excess CH3CO2H = 0.01875 mol

  4. [CH3CO2H] = 0.234 mol dm-3

  5. [CH3CO2-] = 0.109 mol dm-3

  6. [H+] = 3.64 x 10-5 mol dm-3

  7. pH = 4.44
• Calculating the pH of Buffer Solutions


  • Use the weak acid dissociation expression: [H+][A-]/[HA] = Ka
  • Assume [A-] is due to the added salt only
  • Assume [HA] has remained constant
• The salt content can be added in several ways: a salt solution, some solid salt, or by partially neutralising a weak acid with alkali
• pH


  • log[H+]
• Calculating pH of Buffer Solutions


  [H+

  (aq)][A-

  (aq)]/[HA (aq)] = Ka
• We still use the weak acids dissociation expression
• We assume the [A-] concentration is due to the added salt only