Acids and Bases: Weak acids

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A weak acid (HA) is an acid that dissociates partially into H+ and A–.

We measure the extent of dissociation with the acidity constant KA. The greater the extent of dissociation, the “stronger” the weak acid is, and the greater KA is.

The dissociation of a weak acid produces H+ and A– in equal amounts, so we can say that [H+] = [A–], and calculate [H+] as follows:

HUGELY IMPORTANT: This assumption ONLY applies if there are no other processes that produce a significant amount of H+ or A–. Think really carefully before using the assumption. One mistake will cost MANY marks as the method will be wrong.

Some examples of where we CAN make this assumption :

  • A weak acid on its own with nothing added
  • e.g., CH3COOH
  • A weak acid diluted with distilled water
  • e.g., CH3COOH + H2O

Some examples of where we CANNOT make this assumption and doing so will always cost you MANY marks:

  • Where the weak acid is reacted with a strong base
  • e.g., CH3COOH + NaOH
  • Where the weak acid is mixed with its soluble salt
  • e.g., CH3COOH + CH3COONa

Here is an example of a calculation where we can make the assumption that [H+] = [A–].

Question:

Calculate the pH of a 0.4 mol dm-3 HA weak acid solution (KA = 2.0 x 10-5 mol dm-3)

Answer:

We write out the KA expression for this acid.

Then, since it is a weak acid on its own with nothing else added, we can safely assume that H+ = A- and do some algebra:

Substituting gives [H+] = 2.8 x 10-3 mol dm-3 and pH 2.55.

Here is another:

Question:

Calculate the pH change of the acid when 900 cm3 of distilled water is added to 100 cm3 0.4 mol dm-3 HA weak acid solution (KA = 2.0 x 10-5 mol dm-3)

Answer:

Inital pH = 2.55 (from previous question)

We can see that the volume of the acid has increased by a factor of 10, so the final [HA] has decreased by a factor of 10 i.e., [HA] = 0.04.

As before, we write out the KA expression for this acid.

Then, since it is a weak acid on its own with just water added, we can still safely assume that H+ = A- and do some algebra (again!):

Substituting gives [H+] = 8.9 x 10-4 mol dm-3 and pH 3.05.

The pH change is therefore +0.50.

NOTE – This is an interesting result because it shows that diluting a weak acid by a factor of 10 increases the pH by only 0.50 unit.

This differs from diluting a strong acid by a factor of 10 which increases the pH by 1.00 unit.