Problem 112 - Chemistry The Central Science

Question

A sample of 7.5 \(\mathrm{L}\) of \(\mathrm{NH}_{3}\) gas at \(22^{\circ} \mathrm{C}\) and 735 torr is bubbled into a 0.50 -L solution of 0.40 \(\mathrm{M}\) HCl. Assuming that all the \(\mathrm{NH}_{3}\) dissolves and that the volume of the solution remains \(0.50 \mathrm{L},\) calculate the \(\mathrm{pH}\) of the resulting solution.

Step-by-Step Solution

Answer

The pH of the resulting solution after bubbling 7.5 L of NH3 gas into a 0.50-L solution of 0.40 M HCl is 8.69.

Step 1: Convert gas volume and pressure to moles

The ideal gas law states that: PV = nRT Where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature in Kelvin. We have the following initial values: - Volume of NH3 gas, V = 7.5 L - Pressure of NH3 gas, P = 735 torr - Temperature, T = 22°C = 295 K - Ideal gas constant, R = 62.364 L torr / K mol (for this exercise) We can solve for the moles of NH3 gas (n) using these values: n = PV / RT

Step 2: Calculate moles of NH3 and HCl

Substitute the known values into the equation: n = (735 torr * 7.5 L) / (62.364 L torr / K mol * 295 K) n = 5.625 mol Now, we can calculate the initial moles of HCl in the solution: Moles of HCl = (concentration) * (volume of solution) Moles of HCl = 0.4 M * 0.5 L = 0.2 mol

Step 3: Write the reaction and determine the limiting reactant

The reaction between NH3 and HCl is given by: NH3 + HCl → NH4+ + Cl- The moles of NH3 and HCl are used to determine the limiting reactant. Since we have 5.625 mol of NH3 and only 0.2 mol of HCl, HCl is the limiting reactant, and all of it will react with NH3.

Step 4: Calculate final concentrations of NH3, HCl, and NH4+ ions

The moles of NH3 that will react with HCl: 5.625 mol - 0.2 mol = 5.425 mol The final concentrations in the 0.50 L solution are: - NH3: 5.425 mol / 0.5 L = 10.85 M - HCl: 0 mol (all reacted) - NH4+: 0.2 mol / 0.5 L = 0.4 M (formed from the reaction)

Step 5: Determine the pH of the solution

Since NH3 is a weak base, we'll have to use the Kb value for the base ionization constant. For NH3, Kb = 1.8 * 10^(-5). The ionization of NH3 in water can be written as: NH3 + H2O ⇌ NH4+ + OH- The equilibrium expression for this reaction is: Kb = ([NH4+][OH-]) / [NH3] We can substitute the known values and solve for [OH-]: 1.8 * 10^(-5) = (0.4 M * [OH-]) / 10.85 M Solve for [OH-]: [OH-] = 4.91 * 10^(-6) M Now, we can calculate the pOH of the solution: pOH = -log10([OH-]) = 5.31 Finally, we can calculate the pH using the relationship pH + pOH = 14: pH = 14 - pOH = 14 - 5.31 = 8.69 The pH of the resulting solution is 8.69.

Key Concepts

Understanding the Ideal Gas Law

The Ideal Gas Law is a fundamental equation in chemistry that relates the pressure, volume, and temperature of a gas to the number of moles of the gas present. Its formula is expressed as \( PV = nRT \), where \( P \) denotes the pressure of the gas, \( V \) represents the volume, \( n \) is the number of moles, \( R \) is the ideal gas constant, and \( T \) is the temperature measured in Kelvin. To use this law effectively, convert all measurements into the correct units. In our exercise, pressure in torr and volume in liters were used along with the specific ideal gas constant that accommodates these units. Solving for \( n \) allows us to find the amount of gas present, which is crucial for further calculations in the problem.

Limiting Reactant Analysis

In chemistry, when chemicals react, the limiting reactant is the substance that is completely consumed first and therefore determines the amount of product that can be formed. This is an important concept when dealing with chemical reactions because it affects the theoretical yield of the reaction. To determine the limiting reactant, compare the mole ratio of the reactants with the stoichiometry of the balanced equation. In the exercise, \( HCl \) is the limiting reactant because it has fewer moles compared to \( NH_3 \) given the 1:1 reaction ratio of \( NH_3 \) to \( HCl \). Identifying the limiting reactant helps in calculating the exact amounts of products formed, as seen in the calculation of the concentration of \( NH_4^+ \) ions.

pH Calculation Explained

The pH scale is used to specify the acidity or basicity of an aqueous solution. It is calculated as \( pH = -\log_{10}[H^+] \), where \( [H^+] \) represents the concentration of hydrogen ions in the solution. For basic solutions, we typically compute the concentration of hydroxide ions \( [OH^-] \) and then find the pOH, which is \( pOH = -\log_{10}[OH^-] \). The pH of the solution can then be found using the relationship \( pH + pOH = 14 \), a consequence of the water dissociation constant at 25°C. In our exercise, after determining the concentration of \( OH^- \) from the ionization of the weak base, we calculated pOH and subsequently found the pH of the resulting solution.

Weak Base Ionization

Weak bases only partially ionize in water, establishing an equilibrium between the un-ionized base and the ions produced. The ionization can be represented as \( B + H_2O \leftrightarrows BH^+ + OH^- \), with \( B \) being the weak base. The equilibrium constant for this reaction, \( K_b \), is called the base ionization constant. Lower values of \( K_b \) indicate weaker bases. To calculate the pH of a solution containing a weak base, we must set up and solve the equilibrium expression. This involves knowing the concentration of the base and the value of \( K_b \) to find the concentration of \( OH^- \) ions. With \( [OH^-] \) determined, we proceed to calculate pOH, and from there, the pH, as shown in our example with ammonia \( (NH_3) \) as the weak base undergoing ionization.