Question

A sample of \(500 \mathrm{mg}\) of an unknown monoprotic acid was dissolved in \(50.0 \mathrm{~mL}\) of water and titrated with \(0.200 \mathrm{M}\) KOH. The acid required \(20.60 \mathrm{~mL}\) of base to reach the equivalence point. (a) What is the molar mass of the acid? (b) After \(10.30 \mathrm{~mL}\) of base had been added in the titration, the pH was found to be 4.20 . What is the \(\mathrm{p} K_{a}\) for the unknown acid?

Short answer

The molar mass of the unknown monoprotic acid is \(121.36 \mathrm{g/mol}\) and its \(\mathrm{p} K_{a}\) is 4.20.

Step-by-step solution

Calculate the moles of KOH required for titration to reach the equivalence point

The problem gives that 20.60 mL of 0.200 M KOH are required to reach the equivalence point. Using this information, we can now calculate the moles of KOH used in the titration: Moles of KOH = Volume of KOH (L) × Molarity of KOH = (20.60 mL × (1L / 1000 mL)) × 0.200 M = 0.00412 moles

Determine moles of the unknown acid

Since it is a monoprotic acid, at the equivalence point, the moles of the acid will be equal to the moles of the base. Therefore, the moles of the unknown acid are also 0.00412 moles.

Calculate the molar mass of the unknown acid

To find the molar mass of the unknown acid, we will use the given mass (500 mg) and the moles of the unknown acid calculated in step 2. Convert the mass to grams first. Mass of the unknown acid (g) = 500 mg × (1 g/1000 mg) = 0.5 g Molar mass of the unknown acid = mass of the unknown acid (g) / moles of the unknown acid Molar mass = 0.5 g / 0.00412 moles = 121.36 g/mol This is the molar mass of the unknown acid, which is the answer to part (a) of the problem.

Calculate the initial concentration of the unknown acid

We were given the initial volume of the solution as 50.0 mL. To find the initial concentration, we can use the moles of the unknown acid and the given volume. Initial concentration = moles of unknown acid / initial volume (L) Initial concentration = 0.00412 moles / (50.0 mL × (1 L / 1000 mL)) = 0.0824 M

Calculate the concentration of the unknown acid after adding 10.30 mL of 0.200 M KOH

Since 10.30 mL of 0.200 M KOH is exactly half of the volume required to reach the equivalence point, we are at the half-equivalence point of the titration. At this point, half of the unknown acid has reacted with KOH, and the remaining half is still in the initial form. Thus, the concentration of the unknown acid and its conjugate base is equal: Remaining moles of the unknown acid = 0.5 × 0.00412 moles = 0.00206 moles Concentration of the unknown acid and conjugate base = (0.00206 moles)/(50 mL + 10.30 mL) × (1 L / 1000 mL) = 0.0295 M

Calculate the pKa of the unknown acid using the pH after adding 10.30 mL of 0.200 M KOH

We were given that the pH at this point is 4.20. Since we are at the half-equivalence point, the pH is equal to the pKa: pH = pKa Thus, the pKa of the unknown acid is 4.20, which is the answer to part (b) of the problem.

Key concepts

Monoprotic Acid

A monoprotic acid is a simple yet fundamental concept in chemistry. It is an acid that donates only one proton (hydrogen ion) per molecule during a chemical reaction. This characteristic is crucial because it directly affects how the acid behaves during titration. Unlike polyprotic acids, which can donate more than one proton, monoprotic acids complete their reaction in a straightforward manner.

For instance, when a monoprotic acid reacts with a base like potassium hydroxide (KOH), one major thing happens:

• One mole of the acid reacts with one mole of the base.

This makes calculations easier, especially when determining how much base is needed to reach the equivalence point in a titration.

Common examples of monoprotic acids include hydrochloric acid (HCl) and acetic acid (CH₃COOH). Understanding monoprotic acids is essential for predicting the outcomes of reactions, especially in determining molar mass and pH changes.

Equivalence Point

In a titration, the equivalence point is a critical concept. It marks the stage in the process where the amount of acid equals the amount of base added, hence their reactants are completely reacted. At this point, the number of moles of hydrogen ions equals the number of moles of hydroxide ions, leading to a neutralization reaction.

This can be observed through the following changes:

• A noticeable change in the pH, often measured with an indicator or pH meter.
• A visual change, such as color change of an indicator in the solution.

Understanding the equivalence point is key because it allows chemists to calculate unknown concentrations or molar mass of the acid. In this problem, reaching the equivalence point meant that the same number of moles of KOH and the unknown acid had reacted.

This is why knowing the volume of KOH needed to reach the equivalence point (20.60 mL in this exercise) is crucial for further calculations.

pKa

The term "pKa" stands for the negative logarithm of the acid dissociation constant (Ka) of a solution. It is an important property, especially for weak acids, as it helps indicate the strength of the acid in a solution. A lower pKa value suggests a stronger acid, while a higher pKa value indicates a weaker one.

This ties into the titration process when:

• The pH at the half-equivalence point is equal to the pKa.

This relationship is helpful in identifying the pKa of an acid during titration, as we saw in the exercise. After adding half the required amount of KOH, the pH was 4.20, identifying the pKa of the unknown acid.

Learners need to understand pKa because it not only tells us about acid strength but also influences how the acid behaves in various chemical environments. It's a measure of how easily the acid releases its proton into the solution.