Acid, bases and buffers

Cards (76)

  • Kw
    ionic product of water
  • Monobasic acid
    Acid that donates one mole of a proton
  • Brønsted-Lowry acid
    Species that donates a proton
  • monobasic acids
    HCl
    HCN
    HNO3
    CH3COOH
  • dibasic acid
    H2SO4
    H2CO3
  • tribasic acids
    H3PO4
    H3BO3
  • conjugate acid-base pair
    pair of two species that can be interconverted by a transfer of protons
    • A1 B2 —> A2 B1
    • acids have a positive charge
    • bases have a negative charge
  • larger Ka means more dissociation and a stronger acid
  • Ka
    acid dissociation constant
  • Ka of weak acids
    only paritally dissociate so theres more molecules of weak acid than of H+ and A-
  • Assumption of Ka of weak acids
    H+ = A-
    • HA dissociates to produce equal concentrations of the two
    • concentrant on of H+ due to the ionisation of water is negligible compared to the H+ concentration of the acid
  • Brønsted-Lowry base
    species that accepts a proton
  • Dibasic acid
    one mole of an acid can donate/release 2 protons
  • tribasic acid
    one mole of an acid can release/ donate 3 protons
  • acids dissociate and release H+ ions in aqueous solutions and alkalis dissociate and release OH- in aqueous solution
  • Calculating pH of strong Monobasic acids
    pH = -log[H+]
  • calculating the pH of strong bases
    Kw = [OH-][H+]
  • calculating the pH of weak acids 

    Ka= [H+]²/[HA]
  • Kw increases with increasing temperature as more bonds are broken (endothermic process)
  • assumptions for calculating pH of weak acids
    [HA] at the start is the same as [HA] in equilibrium
    • remains constant as dissociation is small
    this doesn’t apply when there’s a stronger weak acid
  • assumption for pH of weak acids do not always work
    -[H+] doesnt equal [A-] if the acid is very dilute
    -[HA] is not the same throughout if there’s a stronger weak acid
  • range of weak acids lies between pH 3 and 6
  • pH of strong monobasic acids
    complete dissociation in solution
    • [H+] = [HA]
    pH of strong acid can be calculated directly from the concentration of an acid
  • Kw
    if [H+(aq)] > [OH- (aq)] then the solution is acidic
  • Kw
    if [H+(aq)] < [OH- (aq)] then the solution is alkaline
  • pH of a buffer solution

    [H+] = Ka x [HA] / [A-]
  • what is a buffer solution?
    solution that minimises pH changes on the addition of small amounts of bases or acids
  • buffer
    [H+] / Ka = [HA] / [A-]
    • ratio of acid to its conjugate base
  • Kw
    if temperature increases, acid dissociates to a greater extent and so the Kw is higher
  • acid + base —> salt + water
  • acid + metal oxide —>salt + hydrogen (gas)
  • acid + alkali —> salt + water
  • How are buffers formed?
    weak acids and the salt of a weak acid
    • weak acid partially dissociates in solution and forms a relatively low concentration of A- ion
    • salt fully ionises in solution
    reserve supplies of acid and its conjugate base
  • formation of buffers example
    weak acid and salt of weak acid
    • ethanoic acid and sodium ethanoate
    • CH3COOH / CH3COONa
  • How are buffers formed?
    partial neutralisation of the weak acid
    • add aq alkali to excess weak acid
    • weak acid is partially neutralised by the alkali forming the conjugate base
    • some of the weak acid is left over unreacted so a mixture of salt and unreacted acid forms
  • buffer formation
    partial neutralisation e.g CH3COOH
    CH3COOH + NaOH —> CH3COONa + H2O
    CH3COOHCH3COO- + H+
  • action of a buffer
    Addition of an acid:
    • equilibrium shifts to the left
    • added H+ react with conjugate base
    • [H+] increases and pH is lowered
  • action of a buffer
    Addition of an alkali:
    • equilibrium shifts to the right (to replenish lost H+)
    • H+ (aq) + OH- (aq) —> H2O (l)
    • HA dissociates
    • [OH-] increases and pH increaases
  • buffer is most effective when there are equal concentrations of weak acid and its conjugate base
  • calculation of the pH of a buffer when its partial neutralisation
    n(acid used)
    n(alkali) = n(conjugate base formed)
    n(acid remaining) = n(acid used) - n(alkali added)