Weak Acid-Strong Base Titrations

A weak acid-strong base titration is a common analytical method used to determine the concentration of a weak acid by reacting it with a strong base. Unlike strong acids, weak acids do not completely dissociate in water, which influences how the pH changes throughout the titration. As the strong base is gradually added, it neutralizes the weak acid, producing water and a salt. The process is typically monitored using a pH indicator or pH meter, providing insight into acid-base equilibria and buffer behavior.

The titration curve for this reaction differs notably from that of a strong acid-strong base titration. It shows a buffer region, where the weak acid and its conjugate base coexist and resist sharp changes in pH. At the equivalence point, the solution becomes slightly basic due to the hydrolysis of the conjugate base formed during the reaction.

In this system, the strong base completely dissociates in the aqueous medium, while the weak acid only partially ionizes, releasing hydrogen ions (H⁺) into the solution. The reaction between them yields a basic salt and water, giving the solution a pH above 7 at equivalence.

Solving problems involving weak acid-strong base titrations

Let’s take an instance where the amount of 25 cm³ of 0.1 mol dm^-3 CH₃COOH (acetic acid) solution is titrated with 0.1 mol dm^-3 NaOH solution. CH₃COOH solution is taken to the titration flask, and NaOH is added to the burette. The titration is carried out at 25 °C. At 25 °C, the dissociation constant of CH₃COOH is 1.5 × 10^-5 mol dm^-3.  In this titration, the following reaction occurs.

Let’s find out the change in pH of the solution with respect to the volume added of NaOH. Then the pH titration curve is plotted between the pH vs the volume of added NaOH.

Q 01: What is the initial pH of the solution?

Initially, there is only a dilute solution of CH₃COOH in the titration flask. Since CH₃COOH is a weak acid, it will dissociate partially into CH₃COO^- ions and H⁺ ions. The H⁺ ions can be calculated using the dissociation constant of the CH₃COOH. If the dissociation amount of the CH₃COOH is x, the concentration of each component at the equilibrium can be calculated as follows.

In this calculation, the H⁺ ion concentration from the dissociation of water has been neglected.

Q 02: Find the pH of the solution when 5 cm³ of NaOH is added. Ka of the acetic acid is 1.5 × 10^-5 mol dm^-3.

According to the following equation, the stoichiometric ratio between NaOH and CH₃COOH is 1:1. Therefore, it forms an equal number of CH₃COONa mols as the NaOH. Since there is a weak acid and its conjugate base in the system, this solution can act as a buffer solution.

pH of a buffer solution is given by the Henderson-Hasselbalch equation.

In this calculation, the H⁺ ion concentration from the dissociation of water has been neglected.

Q 03: Find the pH of the solution when 10 cm³ of NaOH is added. Ka of the acetic acid is 1.5 × 10^-5 mol dm^-3.

According to the following equation, the stoichiometric ratio between NaOH and CH₃COOH is 1:1. Therefore, it forms an equal number of CH₃COONa mols as the NaOH. Since there is a weak acid and its conjugate base in the system, this solution can act as a buffer solution.

pH of a buffer solution is given by the Henderson-Hasselbalch equation.

In this calculation, the H⁺ ion concentration from the dissociation of water has been neglected.

Q 04: Find the pH of the solution when 20 cm³ of NaOH is added to the solution.

Since the stoichiometric ratio between CH₃COOH and NaOH is 1:1, it forms an equal number of CH₃COONa mols as the NaOH. Since there is a weak acid and its conjugate base in the system, this solution can act as a buffer solution.

pH of a buffer solution is given by the Henderson-Hasselbalch equation.

In this calculation, the H⁺ ion concentration from the dissociation of water has been neglected.

Q 05: Find the pH of the solution when 24.9 cm³ of NaOH is added to the solution.

Since the CH₃COOH and NaOH are equal in concentration and the stoichiometric ratio between NaOH and CH₃COOH is 1:1, it forms an equal number of CH₃COONa mols as the NaOH. Since there is a weak acid and its conjugate base in the system, this solution can act as a buffer solution.

pH of a buffer solution is given by the Henderson-Hasselbalch equation.

In this calculation, the H⁺ ion concentration from the dissociation of water has been neglected.

Q 06: Find the pH of the solution when 25 cm³ of NaOH is added to the solution.

Since the CH₃COOH and NaOH are equal in concentration, at this point all the CH₃COOH mol in the solution will completely react with the NaOH to form CH₃COONa and H₂O. Therefore, this is the equivalent point of the reaction. At this point, there is no excess acid or base. The pH of the solution is given by the amount of the conjugate base of the weak acid.

In this medium, the salt CH₃COONa will dissociate into CH₃COO^- ions and Na⁺ ions. These CH₃COO^- ions can act as a conjugate acid. In other words, CH₃COO^- reacts with water and results in OH^- ions in the water while regenerating the CH₃COOH.

In the second reaction, it shows that it liberates OH^- ions to the solution. Therefore, CH₃COO^- ions can act as a base. The dissociation constant of this conjugated base can be calculated as follows. Where the Ka of the CH₃COOH is 1.5 × 10^-5 mol dm^-3. And the dissociation constant of water is 1 × 10^-14 mol² dm^-6 at 25 °C.

OH^- ions in the solution can be calculated using the Kb value of the conjugate base. If the concentration of the OH^- ions in the solution is x,

At the equivalent point, the pH of the solution is 8.762. That means the solution is basic. In weak acid-strong base titrations, a basic equivalent point is obtained.

Q 07: Find the pH of the solution when 25.1 cm³ of NaOH is added to the solution

After the equivalent point, the weak acid has been completely reacted with the strong base. If 25.1 cm³ of NaOH is added to the solution, there is are excess of base in the solution. In this system, there are two types of bases that result in OH^- ions in the solution. One is NaOH, and the other one is the conjugate base of the CH₃COOH, which is CH₃COO^- ions.

However, the OH^- ions given by the CH₃COO^- ions are negligible due to the common ion effect. The OH^- ions given by the NaOH decrease the reaction between CH₃COO^- ions and the water according to the Le Chatelier principle. Therefore, the total OH^- ions in the system are approximately equal to the OH ions given by the dissociation of NaOH.

After the equivalent point, even a small addition of the base makes a significant change in pH. At the equivalent point, the pH was 8.76. After the addition of 0.1 cm³ of NaOH, the pH of the solution has dramatically changed to pH = 10.3.

Q 08: Find the pH of the solution when 30 cm³ of NaOH is added to the solution

At this point, too, the OH^- ions given by the CH₃COO^- ions are negligible due to the common ion effect. Therefore, the total OH^- ions are approximately equal to the OH^- ions given by the dissociation of the NaOH.

Q 09: Find the pH of the solution when 50 cm³ of NaOH is added to the solution

At this point, too, the OH^- ions given by the CH₃COO^- ions are negligible due to the common ion effect. Therefore, the total OH^- ions are approximately equal to the OH^- ions given by the dissociation of the NaOH.

Let’s tabulate the added NaOH volume and the obtained pH values in the above calculations.

Weak acid-strong base titration curve

Use the Interactive Titration Curve Simulator by Learnbin Lab to create pH titration curves