Question

The active ingredient in aspirin is acetylsalicylic acid. A 2.51-g sample of acetylsalicylic acid required \(27.36 \mathrm{~mL}\) of \(0.5106 M \mathrm{NaOH}\) for complete reaction. Addition of \(13.68 \mathrm{~mL}\) of \(0.5106 M \mathrm{HCl}\) to the flask containing the aspirin and the sodium hydroxide produced a mixture with \(\mathrm{pH}=3.48\). Determine the molar mass of acetylsalicylic acid and its \(K_{\mathrm{a}}\) value. State any assumptions you must make to reach your answer.

Short answer

The molar mass of acetylsalicylic acid is approximately \(179.6 \frac{\text{g}}{\text{mol}}\), and its \(K_a\) value is approximately \(3.311 \times 10^{-4}\). The assumption made is that all the \([\text{H}^{+}]\) come from the dissociation of acetylsalicylic acid.

Step-by-step solution

Calculate the moles of NaOH and HCl

First, calculate the moles of NaOH that reacted with the given sample of acetylsalicylic acid by using the molarity and volume of NaOH: Moles of NaOH = Molarity of NaOH × Volume of NaOH (in L) Moles of NaOH = \(0.5106 \frac{\text{mol}}{\text{L}}\) × \(\frac{27.36 \text{ mL}}{1000} \text{ L}\) Moles of NaOH = \(0.01397 \text{ mol}\) Next, calculate the moles of HCl that were added to the mixture of aspirin and sodium hydroxide: Moles of HCl = Molarity of HCl × Volume of HCl (in L) Moles of HCl = \(0.5106 \frac{\text{mol}}{\text{L}}\) × \(\frac{13.68 \text{ mL}}{1000} \text{ L}\) Moles of HCl = \(0.006984 \text{ mol}\)

Determine the moles of acetylsalicylic acid

Since acetylsalicylic acid completely reacts with NaOH, the moles of acetylsalicylic acid will be equal to the moles of NaOH: Moles of acetylsalicylic acid = \(0.01397 \text{ mol}\)

Calculate the molar mass of acetylsalicylic acid

Now that we know the moles of acetylsalicylic acid, we can calculate its molar mass by using the mass given in the problem: Molar Mass of acetylsalicylic acid = \(\frac{\text{mass}}{\text{moles}}\) Molar Mass of acetylsalicylic acid = \(\frac{2.51 \text{ g}}{0.01397 \text{ mol}}\) Molar Mass of acetylsalicylic acid = \(179.6 \frac{\text{g}}{\text{mol}}\)

Find the concentration of hydrogen ions

We know the pH of the mixture after the addition of HCl is 3.48. To find the concentration of hydrogen ions \([\text{H}^{+}]\) in the solution, use the pH formula: pH = \(-\log [\text{H}^{+}]\) Rearrange this formula to find \([\text{H}^{+}]\): \([\text{H}^{+}]\) = \(10^{-\text{pH}}\) \([\text{H}^{+}]\) = \(10^{-3.48}\) \([\text{H}^{+}]\) = \(3.311 \times 10^{-4} \frac{\text{mol}}{\text{L}}\)

Calculate the \(K_a\) value for acetylsalicylic acid

Since we know the concentration of hydrogen ions, we can now find the \(K_a\) for acetylsalicylic acid. This can be done by understanding the following equation: Acetylsalicylic acid \(\leftrightarrow\) Acetylsalicylate\(^{-}\) + H\(^{+}\) Assuming that all the \([\text{H}^{+}]\) come from the dissociation of acetylsalicylic acid, we can form the equation for \(K_a\): \(K_a\) = \(\frac{[\text{Acetylsalicylate}^{-}] [\text{H}^{+}]}{[\text{Acetylsalicylic acid}]}\) Since the moles of acetylsalicylic acid (\(0.01397 \text{ mol}\)) and HCl (\(0.006984 \text{ mol}\)) are known, we can determine the moles of acetylsalicylate\(^{-}\) and acetylsalicylic acid after the reaction. Since one mole of HCl reacts with one mole of acetylsalicylate\(^{-}\): Moles of acetylsalicylate\(^{-}\) = \(0.01397 - 0.006984 = 0.006986\) Now, we convert the moles of acetylsalicylate\(^{-}\) and acetylsalicylic acid into concentrations using the total volume of the solution (27.36 mL NaOH + 13.68 mL HCl = 41.04 mL): \([\text{Acetylsalicylate}^{-}]\) = \(\frac{0.006986 \text{ mol}}{0.04104 \text{ L}}\) = \(0.1702 \frac{\text{mol}}{\text{L}}\) \([\text{Acetylsalicylic acid}]\) = \(\frac{0.006984 \text{ mol}}{0.04104 \text{ L}}\) = \(0.1701 \frac{\text{mol}}{\text{L}}\) Now, we can find the \(K_a\) value for acetylsalicylic acid: \(K_a\) = \(\frac{(0.1702) (3.311 \times 10^{-4})}{0.1701}\) \(K_a = 3.311 \times 10^{-4}\) So, the molar mass of acetylsalicylic acid is approximately \(179.6 \frac{\text{g}}{\text{mol}}\), and its \(K_a\) value is approximately \(3.311 \times 10^{-4}\). The assumption made is that all the \([\text{H}^{+}]\) come from the dissociation of acetylsalicylic acid.

Key concepts

Molar Mass Calculation

Determining the molar mass of a compound is essential in the field of chemistry because it relates the mass of the substance to the number of its particles, like atoms or molecules. In the case of acetylsalicylic acid, the molar mass calculation comes down to how many grams of the compound are present in one mole.

Here's how it's done: take the mass of your sample in grams and divide it by the number of moles of the compound you have. This number of moles is typically found through titration, as seen in our exercise where sodium hydroxide (NaOH) is used to neutralize the acid. The balanced equation for the reaction allows us to determine the stoichiometric relationship between the reactants, and from here the molar mass is a simple division away.

Acid Dissociation Constant (Ka)

The strength of an acid in aqueous solution is characterized by its acid dissociation constant, represented as Ka. The greater the value of Ka, the stronger the acid is. It provides insight into the degree to which an acid dissociates into its ions.

The calculation involves understanding the equilibrium concentration of the products and reactants. For acetylsalicylic acid, we assume that it dissociates into a hydrogen ion and the acetylsalicylate ion. By knowing the initial concentrations and the changes that occur due to the reaction (stoichiometry), we can plug these values into the Ka expression: Ka = \( \frac{[\text{C}^{-}][\text{H}^{+}]}{[\text{HA}]} \) where [C^{-}] is the concentration of the conjugate base, [H^{+}] is the concentration of hydrogen ions, and [HA] is the concentration of the un-dissociated acid.

pH Calculation

pH is a logarithmic measure used to indicate the acidity or basicity of a solution. It is defined as the negative logarithm of the hydrogen ion concentration. Calculating pH can be summarized by the equation pH = -log[H^+]. It's a straightforward process once you have the concentration of hydrogen ions in your solution.

To find the pH, as demonstrated in the exercise, you can take the exponent to which 10 must be raised (or '10 to the power of') the negative of the pH. This relationship shows why a pH of 7 (neutral) corresponds to a hydrogen ion concentration of 10^{-7} M (molarity).

Stoichiometry

Stoichiometry is the section of chemistry that involves using relationships from balanced chemical equations to calculate quantities of reactants and products. In any given chemical reaction, stoichiometry can tell us how much product we can expect from a certain amount of reactant, based on the molar ratios in the balanced equation.

In the context of this titration problem, stoichiometry allows us to relate the molar amounts of acetylsalicylic acid and NaOH in the reaction. These calculations are critical when determining the correct amount of titrant to add to the analyte in a titration, and for understanding how compounds react in fixed proportions to one another.