20.4 - pH of weak acids Flashcards Preview

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How does [H+] compare to [HA] in strong and weak acids?


They are equivalent


They are not equal


What are the approximations involving weak acid calculations (in Ka)? 

Approximation 1

  • HA(aq) ⇔ H+(aq) + A-(aq) (everything in 1:1)
    • (Note: Ka expression the concentrations are @ eqm​)
  • Dissociation of HA is very small (<1%), so you can approximate [HA]eqm ~ [HA]undissociated
  • Therefore, in Ka you can use the original [HA] 

Approximation 2

  • Since the dissociation is in aqeuous conditionswater also dissociates
    • H2O(l) ⇔ H+(aq) + OH-(aq)
  • In the Ka expression, the [H+(aq)] should also include protons from water dissociation
  • However, H2neglibibly dissociates (even less than HA), so you can approximate [H+]eqm ~ [H+]acid only 
  • Therefore, when [HA] dissociates, you are left with equal concentrations of [H+] = [A-]


How is Ka simplified?

= { [H+]2}/[HA]


What are the limitations of the approximations?

Essentially, when there is a stronger weak acid (with K> 10-2) that dissociates more

  • If HA dissociates >5%, the approximations are not valid
    • [HA]eqm not equal to [HA]undissociated
  • Also if pH > 6,  [H+] from water becomes significant
    • Therefore, [H+]eqm not equal to [A-]eqm