Unit 8: Topic 5 - Acid-Base Titrations

Question 1

Explain the steps of an acid-base titration.

Answer 1

An acid of unknown concentration is placed in a flask. A strong base of known concentration is added to the flask 1mL at a time (or any small amount). The acid-base reaction that proceeds in the reaction is HA (aq) + OH- (aq) –> H2O(l) + A- (aq). There are 4 stages to this reaction.
1. No base is added yet: The acid, strong or weak, will have a pH dependent on Ka.
2. Some base is added but the acid is still in excess: There is less concentration of the acid HA, and therefore the pH of the solution increases slowly.
3. Enough base is added to have equal parts of acid and base. If the acid is strong, the resulting solution has pH 7. If the acid is weak, A- is a strong enough base so the resulting solution is basic.
4. Excess base is added: the pH depends on the concentration of the excess OH-.

Question 2

What are some differences between titration of a strong acid with a strong base and titration of a weak acid with a strong base? Justify using titration curves.

Answer 2

Here is a side-by-side of a strong acid with a strong base and a weak acid with a strong base.
1. Initially, the pH of the strong acid is much lower than that of the weak acid due to the Ka values.
2. As titration begins, the weak acid will start to form a buffer solution, a solution with weak acid and conjugate base both present. This does not occur in the titration of the strong acid as the conjugate base has no strength.
3. At the equivalence point, the pH of the strong acid solution is exactly 7, while the pH of the weak acid solution is slightly basic.
4. After the equivalence point, OH- ion dominates both titration curves, so there is little difference here.

Question 3

50 mL of HF is originally added to a container. The Ka of HF is 6.6 * 10^-4. After 10 mL of 0.50M NaOH is added to the container, the pH of the final solution is 8.3, slightly basic. Approximate the concentration of the HF.

Answer 3

A titration of a weak acid with a strong base yields a slightly basic solution at the equivalence point. Since the pH at this point is 8.3, we can assume the titration has reached the equivalence point. We can use the formula M1V1 = M2V2 to calculate the concentration of HF, which is 0.10M.

Question 4

What happens when a weak acid is titrated with strong base, but the weak acid is in excess?

Answer 4

The reaction is HA(aq) + OH-(aq) –> H2O(l) + A-(aq). Since HA is in excess, the compounds affecting pH are HA and A-. The Henderson-Hasselbalch Equation can be used to predict the pH during this buffer solution, most notably when half of the original HA reacts to produce equal amounts of HA and A- ion.
Since pH = pKa + log([A-]/[HA]), when there are equal amounts of weak acid and conjugate base, pH = pKa. This is known as the half-equivalence point.

Question 5

Sulfuric acid is titrated with sodium hydroxide. Explain what occurs and provide a titration curve.

Answer 5

Sulfuric acid, H2SO4, has two protons. The first proton dissociates completely: H2SO4(aq) –> H+(aq) + HSO4-(aq). The second proton dissociates with a Ka of 1.2 * 10^-2. When NaOH is added via titration, the first proton dissociates first: H2SO4(aq) + OH-(aq) –> H2O(l) + HSO4-(aq). Since HSO4- is acidic, the pH of the solution after the first equivalence point is still acidic. However, the second equivalence point is similar to titration of weak acid by strong base: there are two protons, so there will be two equivalence points.
Titration Curve.
Note: polyprotic acids become weaker as a H+ ion is lost.