acids, bases and buffers

• Bronsted-Lowry acid


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


  A substance that can accept a proton
• Bronsted-Lowry acid-base reaction


  Acid 1 + Base 2 ⇌ Base 1 + Acid 2
• Each acid is linked to a conjugate base on the other side of the equation
• Bronsted-Lowry acid-base reactions


  • HCl (g) + H2O (l) ⇌ H3O+ (aq) + Cl- (aq)
  • HNO3 + HNO2 ⇌ NO3- + H2NO2+
  • HCOOH + CH3(CH2)2COOH ⇌ HCOO- + CH3(CH2)2COOH2+
• The substance with the bigger Ka will act as the acid in the reaction
• Reactions of acids with metals, carbonates, bases and alkalis

  1. Acid + metal → salt + hydrogen

  2. Acid + alkali (NaOH) → salt + water

  3. Acid + carbonate (Na2CO3) → salt + water + CO2
• Ionic equation


  • H+ + OH- ⇌ H2O
  • 2H+ + CO32- ⇌ H2O + CO2
• pH


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

  pH = -log[H+]
• Finding [H+] from pH

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

  At 25°C, Kw = 1x10-14 mol2 dm-6
• Calculating pH of pure water

  At 25°C, [H+] = [OH-] = √(1x10-14) = 1x10-7 mol dm-3, pH = 7
• Calculating pH of strong bases


  Kw = [H+][OH-]

  [H+] = Kw / [OH-]

  pH = -log[H+]
• Weak acid


  Acid that only slightly dissociates in water, giving an equilibrium mixture
• Calculating pH of weak acids


  [H+]2 = Ka[HA]

  pH = -log[H+]
• pKa


  • log Ka
• Weak acids


  • Ethanoic acid (Ka = 1.7x10-5 mol dm-3)

  Propanoic acid (Ka = 1.35x10-5 mol dm-3)
• Calculating pH of buffer solutions

  Buffer made from weak acid and salt of weak acid:

  HA + A- ⇌ A- + H+

  If acid added, equilibrium shifts left to remove H+

  If base added, equilibrium shifts right to produce more H+
• Diluting strong acids and bases

  For acids: [H+]new = [H+]old × (old volume / new volume)

  For bases: [OH-]new = [OH-]old × (old volume / new volume)

  pH = -log[H+]
• Making buffer solutions


  • Adding solid salt to weak acid

  Partially neutralising weak acid with alkali
• 1/82


  0.0134 mol
• [HA(aq)]


  [A-(aq)]
• [H+(aq)]


  Ka
• Entering moles of acid and salt

  1. Straight into the equation

  2. As they both have the same new final volume
• 0.04


  0.0134
• [H+(aq)]


  1.7 x 10-5 x [H+(aq)] = 5.07x 10-5
• pH


  1. -log [H+]

  2. = -log 5.07x 10-5

  3. = 4.29
• Buffer


  Made by adding sodium hydroxide to partially neutralise a weak acid
• Example 10


  1. 55cm3 of 0.50 mol dm-3 CH3CO2H is reacted with 25cm3 of 0.35 mol dm-3 NaOH

  2. Calculate the pH of the resulting buffer solution
• Moles CH3CO2H


  conc x vol =0.50x 0.055 = 0.0275mol
• Moles NaOH


  conc x vol = 0.35 x 0.025 = 0.00875 mol
• Moles of CH3CO2H in excess

  0.0275-0.00875 = 0.01875 (as 1:1 ratio)
• [CH3CO2H]


  moles excess CH3CO2H / total volume (dm3) = 0.01875/ 0.08 = 0.234 mol dm-3
• Reaction


  CH3CO2H+ NaOH CH3CO2Na + H2O
• pH


  1. -log [H+]

  2. = -log 3.64 x 10-5

  3. = 4.44
• [CH3CO2-]


  moles OH- added / total volume (dm3) = 0.00875/ 0.08 = 0.109 mol dm-3
• Ka


  [H+] [CH3CO2-] / [CH3CO2H] = 1.7 x 10-5 mol dm-3
• [H+]


  Ka x [CH3CO2H] / [CH3CO2-] = 1.7 x 10-5 x 0.234 / 0.109 = 3.64 x 10-5