Topic 12 - Acid-Base Equilibria

• Equivalence point is the point where half of both reactants have been used up
• A strong acid and a strong base with have an equivalence point with pH 7
• A strong acid and a weak base will have an equivalence point of pH below 7
• A weak acid and a strong base will have an equivalence point of above 7
• A strong acid fully disassociates in solution to give H+ ions
• A weak acid slightly disassociates in solution (around 1 per 1000)
• A buffer solution is a solution that resists change in pH when small volumes of strong acid or base are added
• A buffer solution is made from a weak acid and it's salt
• A weak acid's conjugate base must be strong
• A weak acid's conjugate base must be strong
• Adding OH- to buffer solution is the equivalent of taking away H+ ions
• Apply equilibrium principles when working with buffers
• pH=-log10[H+]

  pH formular
• pKa + log([A-]/[HA]) = pH
• pKa + log([A-]/[HA]) = pH
• Breathing is used to maintain equilibrium
• carbonic acid equlibrium - $H^+$$+$$HCO^-_3⇋H_2CO_3⇋CO_2aq+$$H_2O⇋CO_2g+$$H_2O$
• pH = pKa at 1/2 equivalnce point
• pH = pKa at 1/2 equivalnce point
• The basic enthalpy of neutralisation is -57.6kj/mol
• In practical situations, enthalpy of neutralisation is never exact as H+ and OH- cannot exist on their own
• Breaking bonds in acids such as HCN can alter the enthalpy of neutralisation
• A Bronsted-Lowry acid is a substance that can donate a proton
• A Bronsted-Lowry base is a substance that can accept a proton
• Kw = [H+][OH-]
• The Kw expression can be used to calculate [H+] if we know the [OH-] and vice versa
• Kw is 1x10^-14 so pKw is 14
• p infront of something mean -log10 of that thing
• When finding the pH of water, square root Kw as the concentration of H+ and OH- is equal
• When calculating pH of strong bases, use Kw and rearrange for [H+]
• When calculating pH of weak acids, make 2 assumptions:

  • [H+] = [A-] because they have disassociated at a 1:1 ration
  • [HA] at equilibrium = [HA] initial as amount of acid disassociation is small
• When calculating pH of weak acids, at half the neutralisation volume [HA]=[A-] and pH=pKa
• pH of a diluted strong acid:

  [H+]=[H+](old) x old volume/new volume
• A basic buffer is made from a weak base and the salt of that base
• A weak base must have a strong conjugate acid
• Buffer solutions contain a resevoir of HA and A- ions
• If small amounts of alkali are added to the buffer, the OH- ions react with H+ ions (from the weak acid) to form water. Because there is only a small amount of H+ ions (since the acid only slightly disassociates), equilibrium will shift to produce more H+ ions, lowering the pH
• If a small amount of acid is added to the buffer solution, equilibrium will shift to produce more salt and the H+ ions will react with the reservoir A- ions
• A buffer can be made from partially neutralising a weak acid
• Calculating pH of a buffer:

  1. Rearrange Ka for [H+]
  2. Calculate moles of salt and weak acid
  3. Using ratio, find which reactant is in excess and by how much
  4. [acid]=excess/total volume, [salt]=moles/total volume
  5. Plug into rearranged equation
  6. Calculate pH