Problem 104 - 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 \(0.40 \mathrm{M} \mathrm{HCl}\). 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 the NH₃ gas into the HCl solution is approximately 8.95.

Step 1: Determine the moles of NH₃ gas using the Ideal Gas Law

First, use the Ideal Gas Law: \(PV = nRT\), where P is pressure, V is volume, n is the number of moles, R is the gas constant, and T is temperature in Kelvin. In this problem, we are given the volume, temperature, and pressure of the NH₃ gas: Volume (V): \(7.5~L\) Pressure (P): \(735~torr\) Temperature (T): \(22^\circ C = 295~K\) To use the Ideal Gas Law equation, pressure should be in atm. Convert torr to atm: \(735~torr \times \frac{1~atm}{760~torr} \approx 0.967~atm\) R (gas constant) = \(0.0821~\frac{L\cdot atm}{mol\cdot K}\) Now, we can solve for n (moles of \(NH_3\)): \(n = \frac{PV}{RT}\)

Step 2: Calculate the moles of NH₃ gas

Plug in the given values and solve for n: \(n = \frac{(0.967~atm)(7.5~L)}{(0.0821~\frac{L\cdot atm}{mol\cdot K})(295~K)}\) \(n \approx 0.299~mol\) There are approximately 0.299 moles of NH₃ gas that are bubbled into the HCl solution.

Step 3: Perform stoichiometric calculation for NH₃ reacting with HCl

The reaction between NH₃ and HCl is as follows: \(NH_3 + HCl \rightarrow NH_4Cl\) Since the reaction has a 1:1 stoichiometry, the moles of NH₄⁺ produced will be equal to the moles of NH₃. We also need to determine the moles of HCl in 0.5 L of 0.40 M HCl solution: Moles of HCl \(= M \times V = 0.40~M \times 0.50~L = 0.20~mol\) Since there are fewer moles of HCl than NH₃, HCl is the limiting reactant, and the reaction will consume all the HCl.

Step 4: Determine the concentration of NH₄⁺ and Cl⁻ ions

Upon complete reaction, the moles of NH₄⁺ ion produced are equal to the moles of HCl reacted (0.20 mol). Determine the concentration of NH₄⁺ in the 0.5 L solution: \[ NH_4^+~concentration = \frac{0.20~mol}{0.5~L} = 0.40~M \]

Step 5: Determine the pH of the final solution

Now, we can use the equilibrium expression for NH₃ to find the pH: \(K_b = \frac{[OH^-][NH_4^+]}{[NH_3]}\) , where \(K_b\) is the base dissociation constant. The \(K_b\) of NH₃ is \(1.8 \times 10^{-5}\). Since we have the value of \([NH_4^{+}] = 0.40 M\), we can find \([OH^{-}]\) and \([NH_3]\) in the final solution by plugging in the values and solving for \([OH^{-}]\). In the final solution, the remaining moles of NH₃ are \(0.299-0.20 = 0.099~mol\). Therefore, the concentration of NH₃ is: \([NH_3] = \frac{0.099~mol}{0.50~L} = 0.198~M\) Plugging the values in the equilibrium expression: \(1.8 \times 10^{-5} = \frac{[OH^-](0.40~M)}{(0.198~M)}\) Now, we can solve for \([OH^-]\): \([OH^-] = 8.91 \times 10^{-6}~M\)

Step 6: Calculate pH

Finally, we can use the relationship between \(pOH\) and \(pH\) to calculate the final pH: \(pOH = -\log{[OH^{-}]}\) \(pOH \approx 5.05\) Since \(pH + pOH = 14\), we can find the pH of the solution: \(pH = 14 - pOH = 14 - 5.05 \approx 8.95\) The pH of the resulting solution is approximately 8.95.

Key Concepts

Ideal Gas Law

Understanding the Ideal Gas Law is crucial to solving many problems in chemistry, especially when dealing with gases. This law combines several gas laws into one universal equation, which can be written as PV = nRT. Here, P stands for pressure, usually in atmospheres (atm), V is the volume in liters (L), n is the number of moles of the gas, R is a constant (0.0821 L*atm/mol*K), and T is the temperature in Kelvin (K).

To solve for any of the variables, you need to ensure that units are consistent; for instance, pressures should be converted to atm if they're given in a different unit like torr, and temperatures must be in Kelvin. In the example problem, converting the given pressure from torr to atm and using the Ideal Gas Law allowed us to find the moles of ammonia gas, which is the first step to figuring out the resulting pH of the solution after it reacts with hydrochloric acid.

Stoichiometry

Stoichiometry relates to the quantitative aspect of chemical reactions. By understanding the stoichiometry of a reaction, you can predict the amounts of reactants and products involved. It's based on the principle of conservation of mass, where the mass of reactants equals the mass of products in a chemical reaction.

For example, the reaction between ammonia (NH₃) and hydrochloric acid (HCl) to produce ammonium chloride (NH₄Cl) follows a 1:1 stoichiometry, meaning one mole of NH₃ reacts with one mole of HCl to produce one mole of NH₄Cl. Knowing this ratio allows us to determine which reactant is the limiting reactant (the one that determines the amount of product formed) and calculate the concentration of resulting ions in solution, both imperative for calculating the pH.

Acid-Base Chemistry

Acid-Base chemistry is a central topic in any discussion involving pH. Acids are substances that donate protons (H⁺ ions) in a reaction, while bases accept protons. The pH scale, ranging from 0 to 14, measures the acidity or basicity of a solution, with 7 being neutral.

In a reaction between an acid and a base, such as HCl (a strong acid) and NH₃ (a weak base), they form a salt -- in this case, ammonium chloride (NH₄Cl). This reaction can also influence the pH of the solution due to the formation of ammonium ions (NH₄⁺), which can slightly dissociate, releasing H⁺ ions back into the solution. During the process of calculating pH, it's important to account for these acid-base reactions and the strength of the weak base or weak acid involved.

Equilibrium Expressions

Equilibrium expressions are mathematical representations of the state where the rates of the forward and reverse reactions are equal, resulting in no net change in the concentrations of the reactants and products over time. The expression for the equilibrium constant (K) varies depending on whether we're dealing with a gas, acid, or base.

For the reaction involving ammonia and hydrochloric acid, we are concerned with the base dissociation constant (K_b) because NH₃ is a weak base. The expression for K_b is given as K_b = [OH⁻][NH₄⁺]/[NH₃], where the brackets denote the concentration of each species at equilibrium. By rearranging this expression and using the known concentrations, we can solve for the hydroxide ion concentration, which is then used to calculate the pH of the solution.