Problem 62 - Chemistry The Central Science In Si Units

Question

The active ingredient in aspirin is acetylsalicylic acid \(\left(\mathrm{HC}_{9} \mathrm{H}_{7} \mathrm{O}_{4}\right)\), a monoprotic acid with \(K_{a}=3.3 \times 10^{-4}\) at \(25^{\circ} \mathrm{C}\). What is the \(\mathrm{pH}\) of a solution obtained by dissolving one regular aspirin tablet, containing \(100 \mathrm{mg}\) of acetylsalicylic acid, in \(200 \mathrm{~mL}\) of water?

Step-by-Step Solution

Answer

The pH of the solution is approximately 3.02.

Step 1: Convert mass to moles

First, we need to determine the number of moles of the acetylsalicylic acid (HC₉H₇O₄). The molar mass of HC₉H₇O₄ is about 180.16 g/mol. Convert the given mass of aspirin to moles using the formula: \[ n = \frac{m}{M} \] where \( n \) is the number of moles, \( m = 0.1 \) g (100 mg converted to grams), and \( M = 180.16 \) g/mol. \[ n = \frac{0.1}{180.16} \approx 5.55 \times 10^{-4} \text{ moles} \]

Step 2: Calculate the concentration of the solution

To find the concentration of HC₉H₇O₄ in the solution, divide the number of moles by the volume of the solution in liters. The volume is 200 mL, which is 0.2 L. The concentration \( C \) is therefore: \[ C = \frac{n}{V} = \frac{5.55 \times 10^{-4}}{0.2} = 2.775 \times 10^{-3} \text{ M} \]

Step 3: Write the ionization equation and expression for Ka

Acetylsalicylic acid (HC₉H₇O₄) ionizes in water according to the equation: \[ \mathrm{HC}_{9}\mathrm{H}_{7}\mathrm{O}_{4} \rightleftharpoons \mathrm{H}^+ + \mathrm{C}_{9}\mathrm{H}_{7}\mathrm{O}_{4}^- \] The expression for the acid dissociation constant \( K_a \) is: \[ K_a = \frac{[\mathrm{H^+}][\mathrm{C}_{9}\mathrm{H}_{7}\mathrm{O}_{4}^-]}{[\mathrm{HC}_{9}\mathrm{H}_{7}\mathrm{O}_{4}]} \]

Step 4: Establish the equilibrium expression

Assuming \( x \) is the change in concentration of \( \mathrm{H^+} \) ion at equilibrium, we have: \([\mathrm{H^+}] = x\), \( [\mathrm{C}_{9}\mathrm{H}_{7}\mathrm{O}_{4}^-] = x \), and \([\mathrm{HC}_{9}\mathrm{H}_{7}\mathrm{O}_{4}] = 2.775 \times 10^{-3} - x\). Since \( K_a = 3.3 \times 10^{-4} \), we can write: \[ 3.3 \times 10^{-4} = \frac{x^2}{2.775 \times 10^{-3} - x} \]

Step 5: Simplify the equation assuming \( x \) is small

Given that \( K_a \) is small, we assume \( x \) is much smaller than 2.775 \times 10^{-3}. This simplifies the denominator: \[ 3.3 \times 10^{-4} = \frac{x^2}{2.775 \times 10^{-3}} \] Solving for \( x \): \[ x^2 = 3.3 \times 10^{-4} \times 2.775 \times 10^{-3} \] \[ x^2 = 9.1575 \times 10^{-7} \] \[ x = \sqrt{9.1575 \times 10^{-7}} \approx 9.57 \times 10^{-4} \text{ M} \]

Step 6: Calculate the pH

The \( \mathrm{pH} \) of the solution is given by \( \mathrm{pH} = -\log[\mathrm{H^+}] \). Substitute the value of \( x \) from Step 5, which represents \([\mathrm{H^+}]\), to find the \( \mathrm{pH} \): \[ \mathrm{pH} = -\log(9.57 \times 10^{-4}) \approx 3.02 \]

Key Concepts

Acetylsalicylic Acid

Acetylsalicylic acid, often recognized as the active ingredient in aspirin, is commonly used for its pain-relieving and anti-inflammatory properties. Chemically, it is represented by the formula \(\mathrm{HC}_{9}\mathrm{H}_{7}\mathrm{O}_{4}\). This monoprotic acid has a single ionizable hydrogen atom, meaning it can donate one proton \(\mathrm{H^+}\) to a solution.
At room temperature, 25°C, its acid dissociation constant \(K_a\) is valued at \(3.3 \times 10^{-4}\). This constant measures how easily the acid donates its proton in water. More simply, \(K_a\) tells us how "strong" or "weak" the acid is, with a smaller \(K_a\) indicating a weaker acid that doesn't dissociate completely.
Understanding the chemistry of acetylsalicylic acid is key to calculating solutions like the one in this exercise. It helps us to determine how the acid interacts in aqueous environments and affects the pH.

pH Calculation

Calculating the pH of a solution involves understanding its acidic or basic nature. pH is a scale used to specify the acidity or basicity of an aqueous solution.
In the context of acetylsalicylic acid, pH calculations help us know how acidic the resulting solution will be once it is dissolved in water. The calculation entails several steps:

• First, write the ionization equation to see how the acid dissociates in water: \(\mathrm{HC}_{9}\mathrm{H}_{7}\mathrm{O}_{4} \rightleftharpoons \mathrm{H^+} + \mathrm{C}_{9}\mathrm{H}_{7}\mathrm{O}_{4}^-\).
• The concentration of hydrogen ions \(\mathrm{H^+}\) in the solution is directly used to determine pH since \(\mathrm{pH} = -\log[\mathrm{H^+}]\).
• Next, use the given \(K_a\) value, the initial acid concentration, and assume that the change \(x\) due to dissociation is small enough that it does not significantly alter the initial concentration.

By these steps, the pH of the solution was found to be approximately 3.02, indicating an acidic environment.

Molar Mass Conversion

Molar mass conversion is essential in chemistry to transition between mass and moles, the latter being central in any calculation involving chemical reactions. Molar mass is the weight of one mole of a substance and is expressed in grams per mole (g/mol).
For acetylsalicylic acid, its molar mass is approximately 180.16 g/mol. To find out how many moles are present in a given mass of this compound, you use the equation: \[ n = \frac{m}{M} \] where \(m\) is the mass in grams and \(M\) is the molar mass.
For the problem at hand, converting 100 mg of acetylsalicylic acid (which is 0.1 g) to moles helps us progress with concentration and subsequent pH calculations. By using the formula, we determine that the solution contains approximately \(5.55 \times 10^{-4}\) moles of the acid.
This step is foundational because these moles are then used to find concentration, enabling us to eventually calculate the pH. Understanding molar mass conversion ensures accuracy in chemical calculations, which is critical for understanding solution properties.