Problem 77 - Chemistry and Chemical Reactivity

Question

About this time, you may be wishing you had an aspirin. Aspirin is an organic acid (page 756 ) with a \(K_{\mathrm{a}}\) of 3.27 \(\times 10^{-4}\) for the reaction $$\mathrm{HC}_{9} \mathrm{H}_{7} \mathrm{O}_{4}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\ell) \rightleftarrows \mathrm{C}_{9} \mathrm{H}_{7} \mathrm{O}_{4}^{-}(\mathrm{aq})+\mathrm{H}_{3} \mathrm{O}^{+}(\mathrm{aq})$$ If you have two tablets, each containing \(0.325 \mathrm{g}\) of aspirin (mixed with a neutral "binder" to hold the tablet together), and you dissolve them in a glass of water to give 225 mL of solution, what is the \(\mathrm{pH}\) of the solution?

Step-by-Step Solution

Answer

The pH of the aspirin solution is approximately 2.64.

Step 1: Calculate Moles of Aspirin

First, find the molar mass of aspirin (C₉H₈O₄). Using periodic table values: C (12.01 g/mol), H (1.01 g/mol), O (16.00 g/mol), total molar mass of aspirin is 9(12.01) + 8(1.01) + 4(16.00) = 180.17 g/mol. Calculate moles of aspirin in two tablets with combined mass 0.650 g: \( n = \frac{0.650 \text{ g}}{180.17 \text{ g/mol}} \approx 0.00361 \text{ mol} \).

Step 2: Calculate Initial Concentration

Convert the volume of the solution from mL to L: 225 mL = 0.225 L. Calculate the initial concentration of aspirin: \( C = \frac{0.00361 \text{ mol}}{0.225 \text{ L}} \approx 0.01604 \text{ M} \).

Step 3: Use the Acid Dissociation Constant

Aspirin partially dissociates in water. Using the expression for the dissociation constant: \( K_{a} = \frac{[\text{C}_9\text{H}_7\text{O}_4^-][\text{H}_3\text{O}^+]}{[\text{HC}_9\text{H}_7\text{O}_4]} \). Initially, \([\text{C}_9\text{H}_7\text{O}_4^-]\) and \([\text{H}_3\text{O}^+]\) are both zero.

Step 4: Establish Equilibrium Expression

At equilibrium, let \( x \) be the amount of aspirin that dissociates. Thus, \([\text{C}_9\text{H}_7\text{O}_4^-] = x\), \([\text{H}_3\text{O}^+] = x\), and \([\text{HC}_9\text{H}_7\text{O}_4] = 0.01604 - x\). Substitute into the dissociation equation: \( K_{a} = \frac{x^2}{0.01604 - x} = 3.27 \times 10^{-4} \).

Step 5: Approximate and Solve for x

Assuming \( x \ll 0.01604 \), \( 0.01604 - x \approx 0.01604 \). Substitute to find \( x \): \( 3.27 \times 10^{-4} = \frac{x^2}{0.01604} \). Thus, \( x^2 = 3.27 \times 10^{-4} \times 0.01604 \) and solving gives \( x \approx 0.00228 \text{ M} \).

Step 6: Calculate pH

Since \( x = [\text{H}_3\text{O}^+] \), the concentration of hydronium ions is \( 0.00228 \text{ M} \). Calculate pH: \( \text{pH} = -\log(0.00228) \approx 2.64 \).

Key Concepts

Acid Dissociation Constant

In chemistry, understanding the acid dissociation constant, usually denoted by \( K_a \), is crucial when exploring chemical equilibrium in solutions. The \( K_a \) value is a quantitative measure of the strength of an acid in a solution, specifically its ability to donate protons. A reaction involving an acid, such as aspirin, interacting with water can be represented in equilibrium as:

• \( \text{HC}_9\text{H}_7\text{O}_4 + \text{H}_2\text{O} \rightleftharpoons \text{C}_9\text{H}_7\text{O}_4^- + \text{H}_3\text{O}^+ \)

The \( K_a \) value for aspirin is \( 3.27 \times 10^{-4} \). This relatively small value indicates that aspirin is a weak acid; it doesn't completely dissociate in water. During calculations, this dissociation constant helps us predict concentrations of products at equilibrium. It provides a basis for calculating other important values like the \( \text{pH} \) of the solution, by setting up equations that express the concentrations of ions formed during the dissociation process.

Molar Mass Calculation

Determining the molar mass of a compound, such as aspirin (\( \text{C}_9\text{H}_8\text{O}_4 \)), is the initial step in various calculations in chemistry. The molar mass is essentially the sum of the atomic masses of all atoms in a molecule, with the unit \( \text{g/mol} \). Let's calculate it for aspirin:

• Carbon: \( 9 \times 12.01 \text{ g/mol} = 108.09 \text{ g/mol} \)
• Hydrogen: \( 8 \times 1.01 \text{ g/mol} = 8.08 \text{ g/mol} \)
• Oxygen: \( 4 \times 16.00 \text{ g/mol} = 64.00 \text{ g/mol} \)
• Total molar mass of aspirin: \( 180.17 \text{ g/mol} \)

This molar mass allows us to convert between the mass of a substance and the number of moles. For example, if we have two aspirin tablets each weighing \( 0.325 \text{ g} \), we can convert this into moles using the molar mass:\[ n = \frac{0.650 \text{ g}}{180.17 \text{ g/mol}} \approx 0.00361 \text{ mol} \]Moles provide a "count" of how many chemical units are involved in a reaction, which is essential for further calculations like concentration and pH.

Solution Concentration

The concentration of a solution tells us how much of a solute is present in a given volume of solvent. It is often expressed in molarity (\( M \)), which is moles of solute per liter of solution. In this problem, we begin with 0.00361 moles of aspirin and dissolve it in \( 225 \text{ mL} \) of water, converting to liters gives \( 0.225 \text{ L} \).

The calculation for molarity is straightforward:

• \[ C = \frac{\text{moles of solute}}{\text{liters of solution}} \]
• \[ C = \frac{0.00361 \text{ mol}}{0.225 \text{ L}} \approx 0.01604 \text{ M} \]

This result indicates the initial concentration of aspirin in the solution before any dissociation occurs. Maintaining clear pathways from moles to concentration is key for subsequent steps in chemical equilibrium calculations, like finding the degree of dissociation.

pH Calculation

Calculating the \( \text{pH} \) of a solution is a common task in chemistry, as it provides insight into the acidity or basicity of the solution. The \( \text{pH} \) is calculated using the concentration of hydronium ions \( [\text{H}_3\text{O}^+] \), found after reaching equilibrium. In this case, aspirin dissociates partially, contributing to the hydronium concentration:

Given that all conditions are met, we assume equilibrium with formation of \( x \) moles of \( \text{H}_3\text{O}^+ \), calculated as:

• The dissociation equation: \( K_a = \frac{x^2}{0.01604 - x} = 3.27 \times 10^{-4} \).
• Assuming \( x \ll 0.01604 \), simplifies to \( x^2 = 3.27 \times 10^{-4} \times 0.01604 \).
• Solve for \( x \): \( x \approx 0.00228 \text{ M} \).

Finally, use this concentration to find the \( \text{pH} \):\[ \text{pH} = -\log(0.00228) \approx 2.64 \]The low pH confirms that the solution is acidic, consistent with the partial dissociation of aspirin producing hydronium ions.