Notes on Acid / Base titrations and Buffer solutions

Acid/Base notes:

The equivalence point is when the acid base reaction is complete. For example if 25cm³ of 0.1M  of sodium hydroxide is added to 25cm³ of  0.1M of ethanoic acid the reaction will be complete. Adding more base will cause the pH to shoot up to the pH of the base.

The pH at the equivalence point is not necessarily 7. For strong bases with strong acids it is. This is because the salts formed are neutral and have no reaction with water. For example sodium chloride (NaCl) is a neutral salt and is formed from the reaction of sodium hydroxide with hydrochloric acid.

For weak acids and a strong base the pH at the equivalence point will be above 7, in other words it will be an alkaline solution.  This is because the salt that is formed is a conjugate base of the weak acid. For example the salt of the reaction with ethanoic acid and sodium hydroxide is sodium ethanoate (CH₃COONa). In solution the ethanoate ions (CH₃COO^-) act as a base, and accept a proton from water, therefore, releasing OH^- ions.

CH₃COO^-   +  H₂O  ↔  CH₃COOH  +  OH^-

For weak bases and strong acids the pH at the equivalence point will be acidic. This is because the salt formed is a conjugate acid of the weak base. For example the salt formed when a solution of ammonia reacts with hydrochloric acid is ammonium chloride (NH₄Cl). In solution the NH₄⁺ ions will donate a proton and therefore, the pH of the solution will be below 7.

NH₄⁺  ↔ NH₃ + H⁺

Or more truly:   NH₄⁺   + H₂0  ↔ NH₃  +  H₃O⁺

The buffer region in titrations:

When a strong base is added to a weak acid or a strong acid is added to a weak base, around the half equivalence point (if the equivalence point is when 25cm³ has been added, the half point is when 12.5cm³ has been added). At this point and around this point there is the unreacted weak acid and the salt formed which is the conjugate base. Or the unreacted base and the salt formed which is the conjugate acid. Let’s take the weak acid strong base example first:

 At the half equivalence point, half of the weak acid remains, E.g. CH₃COOH, in addition there is the same amount of salt in the solution. This means that there is the acid to react with small amounts of base that is added.  Therefore, pH remains relatively constant around the half equivalence zone. There is also the conjugate base which can react with small amounts of acid, if this was added. Buffer solutions can resist changes in pH upon small amounts of acid or alkali added.

For a weak base and a strong acid which is added, at the half equivalence point, there are equal amounts of the weak base and its salt (a conjugate acid). This means that the pH stays relatively constant around this area as there is some weak base to react with the added acid. It would also remain relatively constant if a small amount of base was added as there is the conjugate acid to react with it.

Buffer solutions

You have seen that a buffer solution is formed when a weak acid is partially reacted with a strong  base or a weak base is partially reacted with a strong acid. A buffer solution could also be formed by adding the salt of a weak acid to the weak acid or adding the salt of a weak base to the weak base. Effectively you end up with the same thing.

Weak acid buffer: E.g.  ethanoic acid and sodium ethanoate

The acid can react with small amounts of alkali added: CH₃COOH + OH^- → CH₃COO^-  + H₂O

The conjugate base can react with small amounts of acid added: CH₃COO^-  + H⁺  → CH₃COOH

Or CH₃COO^- + H₃O⁺  →  CH₃COOH  + H₂O

Note that the weak acid is even less ionised due to the presence of the conjugate base and the salt is therefore, pretty much totally from the salt. So we could put the following equation:

CH₃COOH  ↔  CH₃COO^-  +  H⁺

As:  Acid ↔ Salt + H⁺

Therfore Ka can be expressed as: Ka = [salt] [H⁺]/[acid]

[H⁺]= Ka [acid]/[salt]

Or pH = pKa + log [salt]/[acid]

And with this  equation you can see clearly that at the half equivalence point the pH = pKa. You can therefore, use this to obtain the pka value of a weak acid.

Now follow the same logic for a weak base and its salt.

The relationship between Ka and Kb for an acid base conjugate pair

Pick and example or just use HA and A^-  to represent a weak acid and its conjugate base, and then write the appropriate equations and see if you can work out the relationship between them. This can be useful if you are given one value and you need to use the other.