Problem 103 - Chemistry: The Central Science

Question

(a) A 0.1044-g sample of an unknown monoprotic acid requires \(22.10 \mathrm{~mL}\) of \(0.0500 \mathrm{M} \mathrm{NaOH}\) to reach the end point. What is the molecular weight of the unknown? (b) As the acid is titrated, the \(\mathrm{pH}\) of the solution after the addition of \(11.05 \mathrm{~mL}\) of the base is \(4.89\). What is the \(K_{a}\) for the acid? (c) Using Appendix D, suggest the identity of the acid. Do both the molecular weight and \(K_{a}\) value agree with your choice?

Step-by-Step Solution

Answer

The molecular weight of the unknown acid is approximately \(94.39 g/mol \) and the Ka value is approximately \(1.29 \times 10^{-5}\). Based on the molecular weight and Ka value, the identity of the acid is most likely benzoic acid, which has a molecular weight of 122.12 g/mol and a Ka of \(6.46 \times 10^{-5}\). Although there is some discrepancy between the values, benzoic acid most closely matches the calculated molecular weight and Ka value.

Step 1: Calculate moles of NaOH used in the titration

Use the volume and concentration of NaOH used in the titration to find the moles of NaOH, using the formula: moles = volume (in L) x concentration (in M). Moles of NaOH = \(22.10 \times 10^{-3} L \times 0.0500 M = 1.105 \times 10^{-3} mol\)

Step 2: Determine moles of the unknown acid

In a titration with a monoprotic acid and a base, the molar ratio of acid to base is 1:1. Therefore, the moles of the unknown acid are equal to the moles of NaOH. Moles of the unknown acid = \(1.105 \times 10^{-3} mol\)

Step 3: Calculate the molecular weight of the unknown acid

Use the moles and mass of the unknown acid to calculate the molecular weight, using the formula: molecular weight = mass (in g) / moles. Molecular weight of the unknown acid = \(0.1044 g / 1.105 \times 10^{-3} mol \approx 94.39 g/mol\)

Step 4: Determine the concentration of the acid at the half-titration point

At the half-titration point, half of the volume of the base is added, and the pH equals the pKa of the acid. To find the concentration of the unknown acid at the half-titration point, divide the moles of the acid by the total volume (in L), which includes the initial volume of the acid and the added volume of the base. Initial moles of acid = \(1.105 \times 10^{-3}\div2 = 5.525 \times 10^{-4} mol\) Total volume = \(0.01105 \times 2 = 0.02210 L\) (since half of the base has been added) Concentration of the unknown acid = \(5.525 \times 10^{-4} mol / 0.02210 L\approx 0.0250 M\)

Step 5: Calculate the Ka for the unknown acid

Using the pH value given at the half-titration point and the fact that pH = pKa at this point, we can find the Ka for the unknown acid. pH = 4.89 pKa = 4.89 Ka = \(10^{-pKa} = 10^{-4.89} \approx 1.29 \times 10^{-5}\)

Step 6: Identify the acid using molecular weight and Ka

By comparing the molecular weight and Ka value of the unknown acid with the values provided in Appendix D, we can suggest the identity of the acid. In this case, the molecular weight is approximately 94 g/mol, and the Ka is approximately \(1.29 \times 10^{-5}\). These values are close to the values for benzoic acid, which has a molecular weight of 122.12 g/mol and a Ka of \(6.46 \times 10^{-5}\). Although there is some discrepancy between the values, benzoic acid is the acid that most closely matches the calculated molecular weight and Ka value.

Key Concepts

Molecular Weight Calculation

The molecular weight (or molecular mass) of a compound is a critical piece of information that can be deduced from titration data. Molecular weight is calculated by dividing the mass of the compound (in grams) by the number of moles of the compound. Here's a simplified breakdown:

• Firstly, the number of moles of titrant (in this case, sodium hydroxide, NaOH) is determined from its volume and concentration. This titrant reacts with the unknown acid in a 1:1 molar ratio.
• Since the acid is monoprotic, meaning it donates one proton (H+) per molecule, the moles of the acid will be equal to the moles of NaOH used in the titration.
• With the weight of the acid sample and the calculated moles, the molecular weight is then found using the formula \[\begin{equation}Molecular Weight = \frac{Mass}{Moles}\end{equation}\]

When dealing with small amounts of a substance as in the exercise, it's crucial to convert volumes into liters and use precise measurements to ensure an accurate calculation of the molecular weight.

Acid Dissociation Constant (Ka)

Understanding the acid dissociation constant, Ka, is essential for characterizing the strength of an acid. Ka gives insight into the degree to which an acid can release protons into a solution. Here's a rundown of its importance:

• The lower the pKa, the stronger the acid, and conversely, the higher the pKa, the weaker the acid. pKa is the negative logarithm of the Ka value.
• In the titration of a monoprotic acid, the midpoint, where exactly half of the acid has been neutralized, is known as the half-equivalence point. At this juncture, the pH of the solution equals the pKa of the acid.
• From the pH at the half-equivalence point, we can convert to pKa and then to Ka using the relationship \[\begin{equation}Ka = 10^{-pKa}\end{equation}\]

This property helps in identifying the acid by associating a experimental pKa value with the known values of various acids.

Titration Curve Analysis

A titration curve graphically represents the change in pH of an acid solution as a base is added. The analysis of this curve provides a wealth of information:

• The starting pH tells us about the initial acidity of the solution. For a strong acid, this would be a low pH, whereas for a weak acid, it may start higher.
• The shape and inflection points of the curve indicate the moment the acid has been entirely neutralized by the base, known as the equivalence point. At this stage, the number of moles of hydrogen ions equals the number of moles of hydroxide ions added.
• The region around the equivalence point is crucial for determining the endpoint of the titration, where the indicator changes color. The steeper the curve, the more precise the endpoint.
• In our exercise, the pH at the half-equivalence point is crucial as it equals the pKa. For monoprotic acids, the titration curve will exhibit only one steep rise, marking one equivalence point.

Through careful analysis of the titration curve, one can understand the behavior and properties of the acid in question, necessary for identification and a deeper grasp of acid-base chemistry.