20.4 - pH of weak acids

Question 1

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

Answer 1

Strong

They are equivalent

Weak

They are not equal

Question 2

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

Answer 2

Approximation 1

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

Approximation 2

• Since the dissociation is in aqeuous conditions, water also dissociates
  
  • H₂O_(l) ⇔ H⁺_(aq) + OH^-_(aq)
• In the K_a expression, the [H⁺_(aq)] should also include protons from water dissociation
• However, H₂O** **neglibibly 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^-]

Question 3

How is Ka simplified?

Answer 3

= { [H⁺]²}/[HA]

Question 4

What are the limitations of the approximations?

Answer 4

Essentially, when there is a stronger weak acid (with K_a > 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