W4.1_pKa and Ionisation of Molecules

Question 1

Define what ionisation is and explain how it impacts different drug properties.

Answer 1

• Ionisation: transfer of H+ ions (or others) to alter the development of charge on molecules
• Direct impact on solubility, permeability, excretion, distribution, on-/off-target protein binding
• Impacts lipophilicity (logD) -> indirectly affects other drug properties

Question 2

Explain what pKa is and how it is defined in water.

Answer 2

• [HA] ⇌ [A-][H+]
• Ka=[H+][A-]/[HA] (equilibrium constant for dissociation of acid at aqueous state)
• pKa = -log10(Ka) (50% ionisation rate when pH = pKa)
• In water: H+ ions do not exist as free species, but will interact with O atoms
• ∴ H+(aq) is interchangeable with H3O+(aq)

Question 3

Explain the definite pKa and pH conditions under standard conditions in water.

Answer 3

• Under standard conditions: [H+] = [-OH] = 10^-7M in water
• Ka = [H+][-OH] = 10^-14 ([H2O] is not a variable, as its concentration has no impact on rate, thus negligible)
• pKa = -log10(Ka) = 14
• pH + pOH ≡ 14

Question 4

Explain how pKb arises from dissociation of acid and how it relates to pKa.

Answer 4

• Dissociation of acid -> formation of conjugate base
  Basicity of conjugate base: pKb (≠reversed equilibrium of pKa)
• [B] ⇌ [-OH][BH+] -> Kb=[BH+][-OH]/[B] -> pKb = -log10(Kb)
• pKb: less useful as it directly relates to [-OH]/pOH instead of [H+]/pH
• Ka=[H+][B]/[BH+] x Kb=[BH+][-OH]/[B] =[H+][-OH]
• ∴pKa + pKb = 14 (under the rule of logab = log a + log b)

Question 5

Explain the relationship between ionisation and acids and bases. How does pKa affect the eqilibrium position in ionisation quantitatively?

Answer 5

• Ionisation = Acid dissociation
• Ionisation = Base association
• Higher pKa: eqm lies to the left vs Lower pKa: eqm lies to the right
• Acid:
• pH ≤ pKa - 2: <1% ionised
• pH = pKa - 1: ≈10% ionised
• pH = pKa: =50% ionised
• pH = pKa + 1: ≈90% ionised
• pH ≥ pKa + 2: >99% ionised
• Alkali:
• pH ≤ pKa - 2: >99% ionised
• pH = pKa - 1: ≈90% ionised
• pH = pKa: =50% ionised
• pH = pKa + 1: ≈10% ionised
• pH ≥ pKa + 2: <1% ionised

Question 6

Explain how ionisation is impacted in acid and bases in low and high pH environment.

Answer 6

• Acid:
• Become ionised by losing H+
• Low pH: solution is rich in H+ -> eqm lies to the left -> less ionisation
• High pH: solution is deficient in H+ -> eqm lies to the right -> more ionisation
• Alkali:
• Become ionised by gaining H+
• Low pH: solution is rich in H+ -> eqm lies to the left -> more ionisation
• High pH: solution is deficient in H+ -> eqm lies to the right -> less ionisation

Question 7

Explain the Henderson-Hasselbach equation.

Answer 7

• % compound ionised=100/1+10^(pH-pKa)
• (- for acids, + for bases)