Problem 112 - 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 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.

Step-by-Step Solution

Answer

The molecular weight of the unknown acid is \(94.39 \frac{\mathrm{g}}{\mathrm{mol}}\). The \(K_{a}\) value for the acid is \(1.29 \times 10^{-5}\). Comparing this \(K_{a}\) value with known acids suggests the unknown acid's identity.

Step 1: (a) Calculate moles of NaOH used

To find the molecular weight of the unknown acid, we first need to determine the moles of NaOH used in the titration. We can find this using the information given: the volume of NaOH solution used and its concentration. Moles of NaOH = volume × concentration Moles of NaOH = \(22.10 \mathrm{mL} \times 0.0500 \mathrm{M}\)

Step 2: (a) Convert mL to L

We must first convert the volume of NaOH solution from milliliters to liters. Volume of NaOH = \(22.10 \mathrm{mL} \times \frac{1 \mathrm{L}}{1000 \mathrm{mL}} = 0.0221 \mathrm{L}\)

Step 3: (a) Calculate moles of NaOH

Now, we can calculate the moles of NaOH used in the titration. Moles of NaOH = \(0.0221 \mathrm{L} \times 0.0500 \mathrm{M} = 0.001105 \mathrm{mol}\)

Step 4: (a) Calculate moles of unknown acid

Since the unknown acid is monoprotic, one mole of acid will react with one mole of NaOH. Therefore, the moles of the unknown acid are equal to the moles of NaOH used in titration. Moles of unknown acid = Moles of NaOH = \(0.001105 \mathrm{mol}\)

Step 5: (a) Calculate the molecular weight of the unknown acid

Now that we have the moles of the unknown acid, we can calculate its molecular weight using its mass. Molecular weight = \(\frac{\text{mass of the acid}}{\text{moles of the acid}}\) Molecular weight = \(\frac{0.1044 \mathrm{g}}{0.001105 \mathrm{mol}} = 94.39 \frac{\mathrm{g}}{\mathrm{mol}}\)

Step 6: (b) Calculate the concentration of remaining acid at half-titration

To find the \(K_{a}\) of the unknown acid, we can use the pH at half-titration. At half-titration, the moles of the acid and its conjugate base are equal. The volume of base at half-titration is given as \(11.05 \mathrm{mL}\) or \(0.01105 \mathrm{L}\). Moles of remaining acid = Initial moles of acid - moles of used NaOH Moles of remaining acid = \(0.001105 \mathrm{mol} - 0.0005525 \mathrm{mol} = 0.0005525 \mathrm{mol}\)

Step 7: (b) Calculate the concentrations of acid and conjugate base

To find the concentrations of the acid and conjugate base, we can first find the total volume of the solution at half-titration. Total volume = initial acid volume + base volume added For this problem, the initial volume of the acid solution is not given, so we cannot find the exact concentrations of the acid and conjugate base. However, since they have equal concentrations at half-titration and the pH depends on the ratio of concentrations, we can still find the unknown \(K_{a}\) using the given pH.

Step 8: (b) Calculate the \(K_{a}\) of the unknown acid

Using the pH at half-titration, we can calculate the \(-\log{[H^{+}]}\) relationship. \(pH =-\log{[H^{+}]}\) \(4.89 =-\log{[H^{+}]}\) First, find \([H^{+}]\): \([H^{+}] = 10^{-4.89}\) Next, apply the equation for \(K_{a}\): \(K_{a} = \frac{[H^{+}][A^{-}]}{[HA]}\) Since the concentrations of HA and A- are equal at half-titration: \(K_{a} = [H^{+}]\) \(K_{a} = 10^{-4.89} = 1.29 \times 10^{-5}\)

Step 9: (c) Identify the unknown acid

Compare the calculated \(K_{a}\) value with the given values of known acids in Appendix D. The unknown acid is most likely the one with a \(K_{a}\) value closest to the calculated value. After comparing, the identity of the unknown acid can be suggested.

Key Concepts

Molecular Weight Calculation

Understanding molecular weight is crucial for characterizing substances in chemistry. It represents the mass of one mole of a substance, often expressed in grams per mole (g/mol). In a titration experiment, where an unknown monoprotic acid is neutralized by a base like NaOH, determining the molecular weight of the acid involves a few steps.First, measure the amount of NaOH required to reach the end-point of the titration. Since NaOH is a strong base, we assume it fully dissociates and that one mole of NaOH will neutralize one mole of the monoprotic acid. By knowing the volume and molarity of NaOH solution used, we calculate the moles of NaOH and, hence, the moles of the unknown acid.To find the molecular weight of the acid, divide its mass by the moles calculated from the titration. This step is critical because it helps in identifying the substance by comparing its molecular weight to known values. For example, a 0.1044 g sample of an acid that requires 0.001105 mol of NaOH for neutralization has a molecular weight of 94.39 g/mol, computed simply by the division of mass by moles. This basic calculation opens doors to further analysis, such as identifying the compound or understanding its properties.

Acid Dissociation Constant (Ka)

The strength of an acid can be quantified by its acid dissociation constant, abbreviated as Ka. It reflects how well an acid donates protons when in solution. A higher Ka value denotes a stronger acid, while a lower Ka indicates a weaker acid.In a titration context, the Ka can be determined from the pH at the halfway point, where the amounts of the acid and its conjugate base are equal. Using the pH, one can calculate the concentration of hydrogen ions ([H+] concentration, since pH is the negative logarithm of this concentration. At half-titration, the concentration of [H+] is equal to the concentration of the acid's conjugate base because the solution contains equal amounts of the two.By inserting the measured [H+] value into the Ka expression, which under these conditions simplifies to Ka = [H+] due to the equal concentrations of the acid and its conjugate base, we can determine the acid's dissociation constant. For instance, if the pH at half-titration is 4.89, the Ka value is calculated to be 1.29 x 10^-5, which can help in identifying the acid when compared with known values.

pH Calculation

Calculating pH is an essential skill in chemistry that provides insight into the acidity or basicity of a solution. pH is determined by the concentration of hydrogen ions (H+) in the solution and is calculated as the negative logarithm of the [H+] concentration.During a titration, the pH of the solution will change as acid and base react. For example, after adding 11.05 mL of NaOH to an acid solution, if the pH is recorded as 4.89, we can find the [H+] by reversing the logarithmic relationship: [H+] = 10^-pH. With a pH of 4.89, the [H+] is found to be 10^-4.89 or 1.29 x 10^-5 M.This information not only helps in determining the strength of the acid being titrated — as it is closely related to the Ka value — but also in understanding buffer systems, calculating the equilibrium concentration of species in a solution, and more. Always remember, pH is not merely a number; it's a logarithmic scale that tells a lot about the chemical nature of the solution you're working with.

Stoichiometry of Acid-Base Reactions

Stoichiometry is the quantitative relationship between reactants and products in a chemical reaction. In acid-base reactions, stoichiometry helps us understand how acids and bases react in terms of mole ratios.For a monoprotic acid reacting with NaOH, the stoichiometry is straightforward: one mole of NaOH reacts with one mole of the acid. Since this is a 1:1 reaction, we can use the stoichiometry to calculate the unknown quantity of one reactant if the other is known. When conducting a titration, as in the exercise, you can calculate the amount of acid present initially by using the known amount of NaOH that has been added to reach the endpoint.Understanding stoichiometry is not just useful for titrations but also essential for predicting product quantities in reactions or for scaling up reactions from the laboratory to industrial scale. It forms the basis of many calculations in chemistry and requires a strong grasp of the mole concept and the balanced chemical equation for the reaction.

Molarity of Solution

Molarity, expressed as moles per liter (M), is a measure of the concentration of a solution. It tells us how much of a solute is present in a given volume of the solution. For instance, when preparing a solution of NaOH for a titration, knowing its molarity is vital to ensure precise and accurate results.To calculate molarity, divide the number of moles of solute by the volume of solution in liters. In the context of our titration example, the molarity of NaOH is given as 0.0500 M. We use this information to calculate how many moles of NaOH were used to reach the titration endpoint by multiplying the molarity by the volume used.Understanding molarity is essential for all chemical dilutions and reactions. This concept ensures that the calculations of reactants and products are accurate. Therefore, a solid understanding of how to calculate and apply the concept of molarity can be the key factor between a successful experiment and an erroneous one.