Problem 181 - Chemistry

Question

A certain acid, HA, has a vapor density of \(5.11 \mathrm{~g} / \mathrm{L}\) when in the gas phase at a temperature of \(25^{\circ} \mathrm{C}\) and a pressure of \(1.00 \mathrm{~atm}\). When \(1.50 \mathrm{~g}\) of this acid is dissolved in enough water to make \(100.0 \mathrm{~mL}\) of solution, the \(\mathrm{pH}\) is found to be \(1.80\). Calculate \(K_{\mathrm{a}}\) for the acid.

Step-by-Step Solution

Answer

The Ka of the acid HA is 2.39 x 10^-3.

Step 1: Find the molar mass of the acid

To find the molar mass of the acid, we will use the ideal gas law equation, PV = nRT. We will first rearrange the equation to find n (number of moles), and then we will use the vapor density to find the molar mass of the acid. Rearrange the ideal gas law equation: n = PV/RT Given: Vapor density = \(5.11 \frac{g}{L}\) Temperature (T) = \(25^{\circ}\) C = 298.15 K (converted to Kelvin) Pressure (P) = 1.00 atm R (gas constant) = 0.0821 \(\frac{L \cdot atm}{mol \cdot K}\) Volume (V) = 1 L (assuming 1 L of gas) Now we plug in the values into the equation: n = \(\frac{(1.00 \, atm)(1 \, L)}{(0.0821 \, \frac{L \cdot atm}{mol \cdot K})(298.15 \, K)}\) n = 0.04094 mol Now, we use the vapor density to find the molar mass of the acid: Molar mass = \(\frac{mass}{moles}\) Molar mass = \(\frac{5.11 \,\frac{g}{L}}{0.04094 \, \frac{mol}{L}}\) Molar mass = 124.7 g/mol

Step 2: Determine the concentration of the acid

Next, we will find the concentration of the acid, [HA], in the 100.0 mL solution. We are given the mass of HA dissolved in the solution (1.50 g) and its molar mass (124.7 g/mol) from Step 1. Concentration = \(\frac{moles}{volume}\) First, convert mass to moles of HA: moles = \(\frac{1.50 \, g}{124.7 \,\frac{g}{mol}}\) = 0.01203 mol Now, convert mL to L and find the concentration: [HA] = \(\frac{0.01203 \, mol}{0.100 \, L}\) = 0.1203 M

Step 3: Determine the H+ and A- ion concentrations

We are given the pH of the solution, which is 1.80. We will use this to find the concentration of H+ ions in the solution. [H+] = \(10^{-pH}\) [H+] = \(10^{-1.80}\) = 0.0158 M Since HA is a weak acid, it does not fully dissociate in water. However, we can safely assume that the concentration of [A-] ions is approximately equal to the concentration of [H+] ions because for each HA molecule that dissociates, one H+ and one A- ion are produced. [A-] = 0.0158 M Now that we know the initial concentration of HA and the concentrations of H+ and A- ions after dissociation, we can update [HA]: [HA] (updated) = 0.1203 M - 0.0158 M = 0.1045 M

Step 4: Calculate Ka

Finally, we can use the concentrations of HA, H+, and A- ions to calculate the Ka (acid dissociation constant) of the acid. Ka = \(\frac{[H^+][A^-]}{[HA]}\) Ka = \(\frac{(0.0158 \, M)(0.0158 \, M)}{0.1045 \, M}\) Ka = 2.39 x 10^-3 Therefore, the Ka of the acid HA is 2.39 x 10^-3.

Key Concepts

Ideal Gas Law

The Ideal Gas Law is a fundamental principle in chemistry that relates the properties of a gas through an equation. This law helps us calculate the amount of substance (molecules) in the gaseous state using the parameters of pressure, volume, temperature, and the universal gas constant. Here's the straightforward equation:
- **PV = nRT** - **P** is the pressure of the gas. - **V** is the volume the gas occupies. - **n** is the number of moles of the gas. - **R** is the universal gas constant, which is approximately 0.0821 L atm/mol K. - **T** is the temperature of the gas in Kelvin.
The formula can be rearranged to solve for various unknowns, such as n (number of moles), by dividing both sides by RT:
- **n = \( \frac{PV}{RT} \)**
In our context, understanding that we can determine the number of moles of the gas using the given pressure, volume, and temperature is key. The Ideal Gas Law allows us to convert information about a gas's macroscopic properties into the microscopic count of molecules, supporting various chemical calculations.

Vapor Density

Vapor density is an important concept when considering gases. It refers to the mass of one liter of gas relative to the mass of one liter of another substance, typically air or hydrogen. In our exercise, it helps us calculate the molar mass of the acid.
To find the molar mass using vapor density, the relationship is:
- **Molar Mass (g/mol) = \( \frac{\text{Vapor Density (g/L)}}{\text{Number of moles (mol/L)}} \)**
Vapor density is a useful parameter because it directly connects the mass and the moles of the gas, providing insights into the molecular weight of gaseous compounds. If you have the vapor density and know the number of moles from the Ideal Gas Law, you can directly calculate the molar mass, as demonstrated in the exercise.

Concentration Calculation

Calculating concentration is crucial in chemistry to understand how much solute is present in a given volume of solvent, specifically in our case where an acid is dissolved in water. Concentration is commonly expressed in molarity (M), which is moles of solute per liter of solution:
- **Concentration (M) = \( \frac{\text{moles of solute}}{\text{volume of solution in L}} \)**
In this problem, we calculate the concentration of the dissolved acid, HA, by first determining how many moles of HA are in 1.50 g using its molar mass. Then, by dividing these moles by the volume of the solution in liters, we find the acid's concentration.
This step is crucial as it sets the stage for further calculations, such as determining pH and calculating the acid dissociation constant, by providing a baseline concentration before any dissociation.

pH Calculation

The pH value is a measure of the acidity or basicity of a solution, detailing the concentration of hydrogen ions in the solution. The logarithmic scale used ranges from 0 (very acidic) to 14 (very basic), with 7 being neutral. For our weak acid dissociation:
- **pH = -log[H⁺]** - [H⁺] represents the concentration of hydrogen ions in molarity (M).
Given the pH, we can reverse the process to find the concentration of hydrogen ions:
- [H⁺] = 10^-pH
In the exercise, the acid's pH is given, allowing us to calculate [H⁺]. This is crucial because it links directly to how the acid dissociates in water, affecting the concentration of other ions such as [A-]. Understanding these calculations is vital to grasp the extent of acid dissociation, ultimately leading to the computation of the acid dissociation constant, Ka, which quantifies the strength of the acid.