Question

At \(28^{\circ}\) and 0.982 atm, gaseous compound HA has a density of \(1.16 \mathrm{~g} / \mathrm{L}\). A quantity of \(2.03 \mathrm{~g}\) of this compound is dissolved in water and diluted to exactly \(1 \mathrm{~L}\). If the \(\mathrm{pH}\) of the solution at \(25^{\circ} \mathrm{C}\) is 5.22 (due to ionization of \(\mathrm{HA}\) ), calculate the \(K_{\mathrm{a}}\) of the acid.

Short answer

The \(K_a\) of the acid is approximately \(5.22 \times 10^{-10}\).

Step-by-step solution

Convert Conditions to STP

Since the density is given at non-STP conditions (28°C and 0.982 atm), we need to first adjust using the Ideal Gas Law to find the molar mass of HA at STP. We know that at STP (0°C and 1 atm), the molar volume is 22.4 L/mol. However, given the different temperature and pressure, we use the corrected form of the gas law: \[ PV = nRT \] Rearranging for molar mass (M):\[ M = \frac{dRT}{P} \] where \(R = 0.0821\) L atm/mol K, \(d = 1.16\) g/L, \(T = 28 + 273 = 301\) K, and \(P = 0.982\) atm.

Calculate Molar Mass of HA

Using the formula from Step 1:\[ M = \frac{1.16 \times 0.0821 \times 301}{0.982} \approx 29.175 \text{ g/mol} \] This is the molar mass of the gaseous compound HA.

Calculate Molarity of HA Solution

We have 2.03 g of HA dissolved in 1 L of water, so to find the molarity, use:\[ \text{Molarity (M)} = \frac{\text{mass of solute (g)}}{\text{molar mass (g/mol)} \times \text{volume (L)}} \]\[ M = \frac{2.03}{29.175} \approx 0.0696 \text{ mol/L} \]

Determine Ion Concentrations from pH

The pH is given as 5.22. To find the concentration of \(\text{H}^+\):\[ \text{pH} = -\log[\text{H}^+] \]\[ [\text{H}^+] = 10^{-5.22} \approx 6.03 \times 10^{-6} \text{ mol/L} \]

Set Up Expression for Ka

The dissociation of HA is given by:\[ \text{HA} \rightleftharpoons \text{H}^+ + \text{A}^- \]Let \(x\) represent the concentration of ionized HA, which equals \([\text{H}^+]\). Then initially, \([ ext{HA}] = 0.0696\) and after equilibrium, it becomes \(0.0696 - x\). So, the \(K_a\) expression is:\[ K_a = \frac{[\text{H}^+][\text{A}^-]}{[\text{HA}]} = \frac{(6.03 \times 10^{-6})^2}{0.0696 - 6.03 \times 10^{-6}} \]

Calculate Ka Value

Now, compute \(K_a\):\[ K_a \approx \frac{(6.03 \times 10^{-6})^2}{0.0696} \approx 5.22 \times 10^{-10} \] Thus, the acid dissociation constant \(K_a\) is approximately \(5.22 \times 10^{-10}\).

Key concepts

Ideal Gas Law

The Ideal Gas Law is a fundamental principle that relates the pressure, volume, temperature, and number of moles of a gas in a single equation. The formula is expressed as \( PV = nRT \), where:

• \( P \) is the pressure of the gas in atmospheres (atm),
• \( V \) is the volume in liters (L),
• \( n \) is the number of moles,
• \( R \) is the ideal gas constant (0.0821 L atm/mol K), and
• \( T \) is the temperature in Kelvin (K).

In the context of our exercise, the Ideal Gas Law helps to adjust the conditions from non-standard temperature and pressure to calculate the molar mass of HA. By rearranging the equation to \( M = \frac{dRT}{P} \), we use the given density, temperature, and pressure to find the molar mass. This step is crucial because it allows us to link the observed properties of the gas to the chemical information required for further calculations.

Molar Mass

Molar mass is the mass of one mole of a substance, and it plays an essential role in converting between mass and moles. It is measured in grams per mole (g/mol) and is derived from the resulting atomic masses of the elements in a compound. In the problem above, the molar mass of the gaseous compound HA was calculated using the density and the information given by the Ideal Gas Law. This value is crucial for determining the number of moles of HA when dissolved in solution. By calculating the molar mass as approximately 29.175 g/mol, we pave the way for subsequent steps such as calculating molarity. Understanding molar mass is foundational for stoichiometry and allows chemists to quantify how substances will react in various conditions.

Ionization

Ionization is the process whereby a neutral molecule is converted into ions, usually when dissolved in a solvent such as water. For acids like HA, ionization is crucial as it leads to the formation of hydronium ions \( \text{H}^+ \). This process is what makes acids acidic and is quantified by the degree of ionization, which can be observed via pH measurements.In the exercise, the ionization of HA in solution is responsible for the observed pH of 5.22. The dissociation equation \( \text{HA} \rightleftharpoons \text{H}^+ + \text{A}^- \) shows the equilibrium set up between the reactants and products. For weak acids, this equilibrium is not complete, meaning not all HA molecules dissociate. Therefore, calculating the concentration of dissociated HA helps in determining the strength of the acid, as represented by the acid dissociation constant \( K_a \).

pH Calculation

pH is a measure of the acidity or basicity of a solution, calculated as the negative logarithm of the hydronium ion concentration: \( \text{pH} = -\log[\text{H}^+] \). A pH less than 7 indicates an acidic solution, while a pH greater than 7 indicates a basic solution.In the given problem, the pH is provided as 5.22, allowing us to back-calculate the concentration of \( \text{H}^+ \) ions using the equation \( [\text{H}^+] = 10^{-5.22} \). This value is crucial for setting up the expression for the acid dissociation constant \( K_a \). By understanding pH and its calculation, students can deduce the acidity strength of the solution and how it correlates with the concentration of dissociated ions in equilibrium. This helps in evaluating the behavior of weak acids like HA in aqueous solutions.