Problem 113 - Chemistry The Central Science In Si Units

Question

Aspirin has the structural formula CC(=O)Oc1ccccc1C(=O)O At body temperature \(\left(37^{\circ} \mathrm{C}\right), K_{a}\) for aspirin equals \(3 \times 10^{-5}\). If two aspirin tablets, each having a mass of \(325 \mathrm{mg}\), are dissolved in a full stomach whose volume is \(1 \mathrm{~L}\) and whose \(\mathrm{pH}\) is \(2,\) what percent of the aspirin is in the form of neutral molecules?

Step-by-Step Solution

Answer

Approximately 99.7% of aspirin is in the form of neutral molecules.

Step 1: Calculate the Molarity of Aspirin Solution

First, calculate the number of moles of aspirin in the dissolved tablets. Each tablet weighs 325 mg, so two tablets weigh 650 mg, which is 0.650 g. The molecular weight of aspirin (C9H8O4) is approximately 180 g/mol. \[\text{Moles of aspirin} = \frac{0.650 \text{ g}}{180 \text{ g/mol}} \approx 0.00361 \text{ mol}\]Since the aspirin is dissolved in 1 L, the molarity \( [HA] \) of the aspirin solution is:\[[HA] = \frac{0.00361 \text{ mol}}{1 \text{ L}} = 0.00361 \text{ M}\]

Step 2: Set up the Expression for the Equilibrium

Aspirin dissociates in water according to the equation:\[HA \rightleftharpoons H^+ + A^-\]The initial concentration of \(H^+\) is given by the pH of the solution:\[[H^+] = 10^{-2} \text{ M}\]Using the equilibrium constant expression for \(K_a\):\[K_a = \frac{[H^+][A^-]}{[HA]}\]

Step 3: Calculate \\(A^-\\) Concentration at Equilibrium

Given \(K_a = 3 \times 10^{-5}\) and the initial \([H^+] = 10^{-2} \text{ M}\):Assuming the concentration of \([A^-] = x\), the equilibrium concentrations are:\[[H^+] = 0.01 \text{ M (remains nearly unchanged because it's much higher than}\ x), \quad [HA] = 0.00361 - x\]Plug these into the \(K_a\) expression:\[3 \times 10^{-5} = \frac{0.01 \times x}{0.00361 - x}\]Assuming \(x\) is small compared to 0.00361:\[3 \times 10^{-5} \approx \frac{0.01 \times x}{0.00361}\]Solve for \(x\):\[x = \frac{3 \times 10^{-5} \times 0.00361}{0.01} = 1.083 \times 10^{-5} \text{ M}\]

Step 4: Calculate Percent of Neutral Aspirin Molecules

The concentration of neutral aspirin molecules \([HA]_0\) is given as initially 0.00361 M. At equilibrium, it will be:\[[HA]_{eq} = 0.00361 - 1.083 \times 10^{-5} \approx 0.00360 \text{ M}\]The percent of aspirin in the form of neutral molecules is:\[\frac{0.00360}{0.00361} \times 100 \approx 99.7\%\]

Key Concepts

Acid Dissociation Constant (Ka)

The Acid Dissociation Constant, known as \(K_a\), is a value that reflects the strength of an acid in solution. In simpler terms, it's an indicator of how well the acid can donate its hydrogen ions \(H^+\). For aspirin, the \(K_a\) value at body temperature \(37^{\circ} \mathrm{C}\) is given as \(3 \times 10^{-5}\). This relatively small number indicates that aspirin is a weak acid.
To understand aspirin's dissociation in water, consider the equilibrium:

• \(HA \rightleftharpoons H^+ + A^-\)

The equation shows that the acid \(HA\) (in this case, aspirin) partially dissociates into hydrogen ions \(H^+\) and its conjugate base \(A^-\).
At equilibrium, the ratio of concentrations of the products (\([H^+]\) and \([A^-]\)) to the reactant \([HA]\) is constant:

• \(K_a = \frac{[H^+][A^-]}{[HA]}\)

This expression helps determine how much of the aspirin remains undissociated in its neutral form, known as \([HA]\). A large \(K_a\) implies more dissociation, while a small \(K_a\) (our case) indicates limited dissociation.

pH and its Calculation

The term pH is a measure of the acidity of a solution. It specifically reflects the concentration of hydrogen ions \([H^+]\). pH is calculated using the formula:

• \(\text{pH} = -\log_{10}([H^+])\)

In your stomach, a pH of 2 means \([H^+] = 10^{-2} \text{ M}\). This is a highly acidic environment. As aspirin dissolves in the stomach with this pH, we're interested in how the amount of \([H^+]\) affects the dissociation equilibrium.
Because the pH value is low, the concentration of hydrogen ions is high, leading to the assumption that the initial \([H^+]\) stays nearly constant through the reaction. This aids us in simplifying the equilibrium calculations, as the equilibrium concentration of \(H^+\) does not drastically change from its initial concentration of \(10^{-2} \text{ M}\).
Understanding this concept allows calculation of how much aspirin exists in its ionized form \(A^-\) versus its neutral form \(HA\), aiding in determining the medical effectiveness of the aspirin.

Molarity and Concentrations

Molarity is a way to express the concentration of a solute in a given volume of solution. It is defined as the number of moles of solute per liter of solution.
In our scenario, two aspirin tablets with a total mass of 0.650 g dissolve in a 1 L solution. The molar mass of aspirin \(C_9H_8O_4\) is approximately 180 g/mol, allowing us to calculate moles of aspirin:

• \(\text{Moles of aspirin} = \frac{0.65 \text{ g}}{180 \text{ g/mol}} \approx 0.00361 \text{ mol}\)

Given that the solution's volume is 1 L, the molarity \([HA]\) is computed as:

• \([HA] = \frac{0.00361 \text{ mol}}{1 \text{ L}} = 0.00361 \text{ M}\)

Molarity is vital because it frames our understanding of how concentrated the aspirin is in the stomach solution. It sets the initial condition for the equilibrium and influences the calculations of \([HA]\) and \([A^-]\) to assess how much aspirin remains undissociated.