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8 Acid Base Equilibria
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8 Acid Base Equilibria

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  • 1. 8 Weak Acid Base Equilibria
  • 2. Bronsted-Lowry Theory
    • An acid is a species that can donate a proton to a base.
    • A base is a species that can accept a proton.
    • A  B + H +
    • acid base
    • (proton donator) (proton acceptor)
    • Note: A free proton does not exist. It is always attached by a coordinate bond to a lone pair of electrons on the solvent molecule
  • 3.
    • Conjugate acid base pairs
    • CH 3 COOH  CH 3 COO - + H +
    • acid conjugate base
    • H 2 O + H +  H 3 O +
    • base conjugate acid
    • Add 1) and 2)
    • CH 3 COOH + H 2 O  H 3 O + + CH 3 COO -
    • acid 1 base 2 acid 2 base 1
    • There are 2 conjugate acid/base pairs in this reaction
  • 4.
    • Strong and Weak Acids
    • A strong acid completely dissociates in water
    • A weak acid only partially dissociates
    • A strong base completely dissociates in water
    • A weak base only partially dissociates
    • Strong Acids
    • HCl H 2 SO 4 HNO 3
    • Weak Acids
    • All organic acids e.g. Ethanoic acid (acetic acid)
    • CH 3 COOH
    • Strong Bases
    • NaOH KOH
    • Weak Bases
    • NH 3
  • 5.
    • Ionic Product of Water
    • Water only slightly dissociates into ions
    • H 2 O  H + + OH -
    • K c = [H + ][OH - ]
    • [H 2 O]
    • As water is only slightly ionised [H 2 O] is much bigger then [H + ] or [OH - ]
    •  we can regard [H 2 O] as constant
    • K c [H 2 O] = [H + ][OH - ]
    • As K c is a constant and [H 2 O] is a constant we can replace both with a new constant called K w the ionic product of water
    •  K w = [H + ][OH - ]
  • 6.
    • K w = [H + ][OH - ]
    • At 25 0 C K w = 1 x 10 -14 mol 2 /L 2
    • and pK w = pH + pOH
    • p K w = 14
    •  pH + pOH = 14
    • So if we know pOH we can find pH
  • 7.
    • Weak Acids and Bases
    • General Equation for a Weak Acid
    • HA  A - + H +
    • acid conjugate proton
    • base
    • K c = [A - ][H + ] for weak acids we use K a (acid dissociation constant)
    • [HA]
    •  K a = [A - ][H + ]
    • [HA]
  • 8.
    • Some examples of weak acids
    • Phenol K a = 1 x 10 -10 mol/L
    • Ethanoic acid K a = 2.1 x 10 -4 mol/L
    • pK a = -log K a
    •  pK a phenol = 10
    • pK a ethanoic acid = 3.67
    • Acid strength increases as pK a decreases (compare with pH)
    Increasing strength Increasing strength
  • 9.
    • Finding the pH of weak acids
    • HA  A - + H +
    • K a = [A - ][H + ]
    • [HA]
    • As there is very little dissociation of the weak acid we can assume that [HA] at equilibrium = initial concentration of the acid.
    • Let this conc = c
    • Then K a = [A - ][H + ]
    • c
    • Also A - = H +  K a = [ H + ] 2
    • c
    • Then H + = √K a C
  • 10. Example Find the pH of 0.1M ethanoic acid. K a = 1.8 x 10 -5 mol/L H + = √KaC = √ 1.8 x 10 -5 x 0.1 = 1.34 x 10 -3 mol/L pH = - log H +  pH = 2.87
  • 11.
    • Weak Bases
    • BOH  B + + OH -
    • K b = [B + ][OH - ]
    • [BOH]
    • We can assume [BOH] = c and [B + ] = [OH - ]
    • K b = [OH - ] 2
    • c
    • And [OH - ] = √ K b x c
  • 12.
    • Example
    • Find the pH of a solution of 0.1m ammonium hydroxide
    • if K b = 1.7 x 10 -5 mol/L
    • [OH - ] = √ K b x c
    • = √ 1.7 x 10 -5 x 0.1mol/L
    • [OH - ] = 1.3 x 10 -3 mol/L
    • pOH = 2.88
    •  pH = 14 – 2.88 = 11.1