Question

One method of analyzing ammonium salts is to treat them with \(\mathrm{NaOH}\) and then heat the solution to remove the \(\mathrm{NH}_{3}\) gas formed. $$ \mathrm{NH}_{4}^{+}(a q)+\mathrm{OH}^{-}(a q) \longrightarrow \mathrm{NH}_{3}(g)+\mathrm{H}_{2} \mathrm{O}(l) $$ (a) Label the Brønsted-Lowry acid-base pairs. (b) If \(2.86 \mathrm{~L}\) of \(\mathrm{NH}_{3}\) at \(60^{\circ} \mathrm{C}(333 \mathrm{~K})\) and \(10^{5} \mathrm{~Pa}\) is produced by the reaction of \(\mathrm{NH}_{4} \mathrm{Cl}\), how many grams of \(\mathrm{NH}_{4} \mathrm{Cl}\) were in the original sample?

Short answer

Ammonium ion is the acid, hydroxide ion is the base. Initial sample had 5.51 g of NH4Cl.

Step-by-step solution

Identify the Acid-Base Pairs

Identify which components in the reaction act as Brønsted-Lowry acids and bases. In the given reaction:\[ \mathrm{NH}_4^+(aq) + \mathrm{OH}^-(aq) \rightarrow \mathrm{NH}_3(g) + \mathrm{H}_2O(l) \] - \(\mathrm{NH}_4^+(aq)\) donates a proton to become \(\mathrm{NH}_3(g)\), hence it is the acid.- \(\mathrm{OH}^-(aq)\) accepts a proton to become \(\mathrm{H}_2O(l)\), hence it is the base.- \(\mathrm{NH}_3(g)\) is the conjugate base and \(\mathrm{H}_2O(l)\) is the conjugate acid.

Calculate Moles of Ammonia

Use the ideal gas law to find the moles of \(\mathrm{NH}_3\) produced. The ideal gas law is given by:\[ PV = nRT \]Where:- \(P = 10^5 \mathrm{~Pa}\) (Pressure)- \(V = 2.86 \mathrm{~L} = 0.00286 \mathrm{~m}^3\) (Volume)- \(R = 8.314 \mathrm{~J}\mathrm{~mol}^{-1}\mathrm{~K}^{-1}\) (Ideal gas constant)- \(T = 333 \mathrm{~K}\) (Temperature)Re-arrange to solve for \(n\):\[ n = \frac{PV}{RT} = \frac{(10^5)(0.00286)}{8.314 \times 333} \approx 0.103 \mathrm{~mol} \]

Calculate Mass of Ammonium Chloride

Use the moles of \(\mathrm{NH}_3\) to find the moles of ammonium chloride \(\mathrm{NH}_4Cl\). Since the stoichiometry of the reaction between \(\mathrm{NH}_4^+\) and \(\mathrm{NH}_3\) is 1:1, the moles of \(\mathrm{NH}_{4}^{+}\) is also \(0.103\) mol.Find the mass of \(\mathrm{NH}_4Cl\) using its molar mass (53.49 g/mol):\[ \text{Mass of } \mathrm{NH}_4Cl = 0.103 \times 53.49 \approx 5.51 \mathrm{~g} \]

Key concepts

Understanding Acid-Base Reactions

In the Brønsted-Lowry theory, acids are proton donors, and bases are proton acceptors. This concept helps us understand acid-base reactions, like the one where ammonium ions react with hydroxide ions.

In the reaction:

• you see that { }(aq) donates a proton to become { } is acting as the acid.
• OH^-(aq), on the other hand, accepts a proton to become , making it the base.
• The resulting products, NH (g) and H2 (l), are the conjugate base and conjugate acid, respectively.

Understanding these pairs helps us see how acids and bases transform during a reaction.

By grasping this concept, you can predict and label the changes in many chemical reactions.

Applying the Ideal Gas Law

The ideal gas law is a powerful tool in chemistry. It's expressed as \( PV = nRT \), where:

• \(P\) is the pressure of the gas,
• \(V\) is the volume,
• \(n\) is the number of moles,
• \(R\) is the ideal gas constant (8.314 J/mol·K),
• \(T\) is the temperature in Kelvin.

In our example, we use this law to find the moles of \(NH_3\) produced when heated. By rearranging the formula to \( n = \frac{PV}{RT} \), we substitute in the given values:

• Pressure (\( P = 10^5 \) Pa),
• Volume (\( V = 0.00286 \) m³),
• Temperature (\( T = 333 \) K).

This gives us approximately \( 0.103 \) moles of \( NH_3 \).

Understanding this relationship allows you to calculate the amount of gas involved based on measurable properties.

Exploring Stoichiometry in Reactions

Stoichiometry is the study of the quantitative relationships in a chemical reaction. It answers questions about how much of each substance is involved or produced.

In our given reaction, the ratio of \( NH_4^+ \) ions to \( NH_3 \) molecules is 1:1. This means the moles of \( NH_3 \) calculated using the ideal gas law directly translate to moles of \( NH_4^+ \).

Using stoichiometry, you can determine the mass of reactants or products. We calculate the mass of \( NH_4Cl \):

• Since the molar mass of \( NH_4Cl \) is 53.49 g/mol and we have \( 0.103 \) moles,
• The mass is \( 0.103 \times 53.49 \approx 5.51 \) grams.

By applying stoichiometric principles, we ensure precise calculations in chemical processes.