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.

Short answer

The pH of the resulting solution after dissolving \(NH_3\) gas in an HCl solution is approximately 11.26.

Step-by-step solution

1. Calculate moles of \(NH_3\)

To determine the moles of \(NH_3\), we'll use the Ideal Gas Law, which states: \(PV=nRT\), where P is pressure, V is volume, n is the number of moles, R is the ideal gas constant, and T is the temperature in Kelvin. First, convert the given values to the appropriate units: - Volume (\(V\)): \(7.5 L\) - Temperature (\(T\)): \(22^{\circ}C + 273.15 = 295.15K\) - Pressure (\(P\)): \(\frac{735 torr}{760} = 0.967atm\) Next, we'll use the value for R in atm, which is \(0.0821 \frac{L \cdot atm}{mol \cdot K}\): \(0.967atm \cdot 7.5L = n \cdot 0.0821 \frac{L \cdot atm}{mol \cdot K} \cdot 295.15K\) Now, we solve for n: \(n = \frac{0.967atm \cdot 7.5L}{0.0821 \frac{L \cdot atm}{mol \cdot K} \cdot 295.15K} = 0.297 mol\)

2. Calculate moles of HCl

We are given the concentration and volume of the HCl solution: Concentration of HCl: \(0.40M\) Volume of HCl solution: \(0.50L\) Now we can find the moles of HCl using the formula: Moles of HCl = Concentration × Volume Moles of HCl = \(0.40M \times 0.50L = 0.2 mol\)

3. Determine the reaction and resulting moles

\(^NH_3\) and HCl react in a 1:1 molar ratio: \(NH_3 + HCl \rightarrow NH_4Cl\) Since we have more moles of \(NH_3\) than HCl, we will have some leftover \(NH_3\) after the reaction. To find out how much, subtract the moles of HCl from the moles of \(NH_3\): Remaining moles of \(NH_3 = 0.297 \text{mol} - 0.2 \text{mol} = 0.097 \text{mol}\)

4. Calculate the concentration of the resulting solution

We are told that the volume of the solution remains constant at \(0.50L\). To find the concentration of the leftover \(NH_3\), we use the formula: Concentration = Moles ÷ Volume Concentration of \(NH_3 = \frac{0.097 \text{mol}}{0.50L} = 0.194M\)

5. Calculate the pH of the resulting solution

Since \(NH_3\) is a weak base, it will react with the water in the solution, forming \(OH^-\). To find the concentration of \(OH^-\), we can use the formula with the base ionization constant (\(K_b\)) of \(NH_3\), which is approximately \(1.8 \times 10^{-5}\): \(K_b = [NH_4^+][OH^-] / [NH_3]\) Assuming that the change in concentrations due to the reaction is negligible, we have: \(1.8 \times 10^{-5} = [x][x] / [0.194]\) Solving for x, we get: \([OH^-] = x = 1.83 \times 10^{-3}\) Now, we'll use the relationship between \(OH^-\) and \(H^+\) concentrations to determine the \(H^+\) concentration: \(K_w = [H^+][OH^-]\), where \(K_w = 1 \times 10^{-14}\) \([H^+] = \frac{1 \times 10^{-14}}{1.83 \times 10^{-3}} = 5.46 \times 10^{-12}\) Finally, we can calculate the pH using the formula: pH = \(-\log[H^+]\) pH = \(-\log(5.46 \times 10^{-12}) = 11.26\) The pH of the resulting solution is approximately 11.26.

Key concepts

Ideal Gas Law

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

• \(P\) is the pressure of the gas, measured in atmospheres (atm).
• \(V\) is the volume of the gas in liters (L).
• \(n\) stands for the number of moles of the gas.
• \(R\) is the ideal gas constant, approximately \(0.0821 \frac{L \, atm}{mol \, K}\).
• \(T\) is the temperature in Kelvin (K).

To calculate the moles of a gas, you must ensure all units are appropriately converted. For example, temperature should always be in Kelvin, which can be obtained by adding 273.15 to the Celsius temperature. Pressure may also need to be converted to atm if initially given in torr or another unit. Understanding this law is crucial, as it allows us to relate different physical properties of gases in chemical reactions.

Moles and Concentration

Understanding moles and concentration is essential in chemistry to quantify substances in reactions. A "mole" is a unit that represents \(6.022 \times 10^{23}\) entities, usually atoms or molecules, and helps in correlating the amount of a substance with its mass and volume. Concentration, often expressed in molarity (M), is a measure of how much solute is present in a given volume of solvent. The formula is:

• Molarity \(= \frac{\text{moles of solute}}{\text{volume of solution in Liters}}\)

In an acid-base reaction, understanding the concentration helps determine reactant quantities and analyze their interactions. For instance, knowing the molarity of hydrochloric acid (HCl) allows us to calculate the moles of HCl present in a solution by multiplying the molarity by the solution volume. This information is vital for understanding reactions and calculating subsequent equilibrium concentrations.

Acid-Base Reaction

Acid-Base reactions involve the transfer of protons (H\(^+\)) from an acid to a base. The reaction between ammonia (NH\(_3\)) and hydrochloric acid (HCl) demonstrates a classic acid-base interaction:

• NH\(_3\) (base) accepts a proton from HCl (acid) to form NH\(_4^+\) and Cl\(^-\).
• The net reaction is \(\text{NH}_3 + \text{HCl} \rightarrow \text{NH}_4\text{Cl}\).

This reaction occurs in a 1:1 molar ratio, meaning one mole of NH\(_3\) reacts with one mole of HCl to form one mole of NH\(_4\)Cl. The completion of the reaction depends on the limiting reactant, which is the substance completely consumed in the reaction. Understanding the stoichiometry of the reaction helps in predicting the amounts of reactants and products, which is crucial when determining the remaining reactant concentrations in the solution.

Weak Base Equilibria

Weak bases like ammonia (NH\(_3\)) do not completely ionize in solution, meaning only a small fraction of the base forms hydroxide ions (OH\(^-\)) in equilibrium. This weak ionization is governed by its base ionization constant, \(K_b\), which for ammonia is approximately \(1.8 \times 10^{-5}\).For the weak base ammonia, the reaction in water is:

• NH\(_3\) + H\(_2\text{O}\) \(\rightleftharpoons\) NH\(_4^+\) + OH\(^-\)

The equilibrium expression is given by \(K_b = \frac{[\text{NH}_4^+][\text{OH}^-]}{[\text{NH}_3]}\). Knowing \(K_b\) allows us to calculate the concentration of OH\(^-\) ions in the solution, which then helps determine the pH. In a typical setup, after calculating \([OH^-]\), use the relationship \(K_w = [H^+][OH^-]\) (where \(K_w\) is the water ionization constant, \(1 \times 10^{-14}\)) to find \([H^+]\). The pH can then be found using the formula \(\text{pH} = -\log[H^+]\). Understanding this principle is crucial for calculating the pH of solutions containing weak bases.