16.2.3 Weak Acids

Question 1

Weak Acids

Answer 1

• The acid-dissociation constant (K a ) is the equilibrium constant for the partial dissociation of a weak acid in aqueous solution.
• K a values can be used to calculate equilibrium concentrations of H+ in acid solutions.
• For a weak acid, percent dissociation is less than 100% and depends on the initial concentration of HA.

Question 2

note

Answer 2

• Weak acids only partially dissociate in aqueous solution. The acid-dissociation constant (K a ) is the equilibrium constant for the partial dissociation of a weak acid in aqueous solution.
• The stronger the acid, the larger the values of [H 3 O + ] and [A – ]. Therefore, stronger acids have higher Ka values.
• K a values are not typically used for strong acids because strong acids essentially dissociate completely, and the Ka values are extremely large.
• pK a = –log (K a ). Since pK a values are usually small positive numbers, pK a values are more easily compared and manipulated.
• K a values can be used to calculate equilibrium concentrations of H + in acid solutions.
• Problem: What is [H + ] and pH for a 0.010 M solution of HCN?
• An ICE diagram (Initial, Change, Equilibrium) is useful for weak acid dissociation calculations. The equilibrium concentration of H + can be found using the same techniques used for other equilibrium problems.
• Note that if this had been a strong acid, the pH would have been 2.00, rather than 5.70.
• Percent dissociation is the concentration of ionized acid divided by the initial acid concentration, multiplied by 100%. Since [ionized acid] = [H+] for a monoprotic acid, percent dissociation = [H + ] / [HA] o · 100%.
• For a strong acid, [H + ] = [HA] o , so percent
  dissociation = 100%.
• For a weak acid, [H + ] < [HA] o , so percent
  dissociation < 100%.
• In this 0.50 M solution of HF, only 3.8% of the HF is
  dissociated into H + and F – . The other 96.2% exists as intact HF molecules in solution.
• For a 0.05 M solution of HF, only 11.2% is dissociated into H + and F – . The other 88.8% exists as intact HF molecules in solution.
• For a weak acid, percent dissociation is less than 100% and depends on the initial concentration of HA.

Question 3

Calculate the percent dissociation of a 0.100 M solution of acetic acid, HC2H3O2. (Ka = 1.8 × 10^−5 )

Answer 3

1.3

Question 4

Which of the following is the formula for pKa?

Answer 4

− log 10 (Ka )

Question 5

As an acid gets weaker, what happens to Ka?

Answer 5

It decreases.

Question 6

Which of the following is TRUE regarding strong acids and weak acids?

Answer 6

• Weak acids have an equilibrium constant less than 1.0.
• Strong acids are considered to be completely ionized.
• The equilibrium point for HA <==> H+ + A- is far to the left for weak acids

Question 7

A 2.5 M solution of a monoprotic acid has a pH of 2.50 at 25°C. What is the equilibrium constant for this acid? Is it a strong or weak acid?

Answer 7

4.1×10^−6, weak

Question 8

What is the pH of 0.180 M solution of carbonic acid (H2CO3 )?
Although it is a diprotic acid, assume that the 2nd dissociation is not significant.
Use Ka = 3.52 × 10^−7, the constant for the first dissociation hydrogen ion.

Answer 8

3.60

Question 9

Which of the following describes a weak acid?

Answer 9

An acid that does not dissociate completely

Question 10

Hydrofluoric acid is a weak acid with an equilibrium constant (at 25°C) of Ka = 7.2 × 10^−4. What is the pH of a 1.00 M solution of this acid?

Answer 10

1.57

Question 11

Suppose a weak monoprotic acid has a Ka = 1.0 × 10^−8. What is the pH of a 1.5 × 10^−4 M solution of this acid?

Answer 11

5.92

Question 12

For water at 25°C, Kw = 1.0 × 10^−14. Is this equivalent to Ka or to Kb?

Answer 12

neither