Question

Indicate the equilibrium equation and constant expression for the autoprotolysis of liquid ammonia. If \(\mathrm{KNH}_{3}=10^{-22}\), how many molecules of ammonia are ionized in 1 mole of ammonia? Assume a density of \(0.771 \mathrm{~g} / \mathrm{ml}\) for ammonia.

Short answer

The autoprotolysis of liquid ammonia is represented by the equation: \( \mathrm{2NH}_{3} \rightleftharpoons \mathrm{NH}_{4}^{+} + \mathrm{NH}_{2}^{-} \) with the constant expression \( K = \frac{[\mathrm{NH}_{4}^{+}][\mathrm{NH}_{2}^{-}]}{[\mathrm{NH}_{3}]^2} \). We find the concentration of ionized species (NH₄⁺ and NH₂⁻) to be y = 10⁻¹¹ M. In 1 mole of ammonia, there are approximately \( 6.022 \times 10^{12} \) ionized ammonia molecules.

Step-by-step solution

Determine the autoprotolysis equation of liquid ammonia

: Autoprotolysis is a process in which two molecules of a solvent react with each other to produce one positive and negative ion. For liquid ammonia (NH₃), the autoprotolysis reaction is: \( \mathrm{2NH}_{3} \rightleftharpoons \mathrm{NH}_{4}^{+} + \mathrm{NH}_{2}^{-} \)

Write the constant expression for the autoprotolysis

: To write the constant expression (K) for the autoprotolysis of liquid ammonia, we use the equilibrium concentrations of the species: \( K = \frac{[\mathrm{NH}_{4}^{+}][\mathrm{NH}_{2}^{-}]}{[\mathrm{NH}_{3}]^2} \) Given \( KNH_{3} = 10^{-22} \), we can substitute this value into K: \( 10^{-22} = \frac{[\mathrm{NH}_{4}^{+}][\mathrm{NH}_{2}^{-}]}{[\mathrm{NH}_{3}]^2} \)

Set up the equilibrium table

: Next, let's set up the equilibrium table: Initial concentrations: - [NH₃] = x M - [NH₄⁺] = 0 M - [NH₂⁻] = 0 M Change in concentrations: - [NH₃] = -2y - [NH₄⁺] = +y - [NH₂⁻] = +y Equilibrium concentrations: - [NH₃] = x - 2y - [NH₄⁺] = y - [NH₂⁻] = y

Solve the equilibrium equation

: Now we need to solve the equilibrium equation using the equilibrium concentrations: \( 10^{-22} = \frac{y^2}{(x - 2y)^2} \) Since the ionization is very low, we can approximate x with 1 mole and ignore the 2y term: \( 10^{-22} = \frac{y^2}{(1)^2} \) Solving for y: \( y = \sqrt{10^{-22}} \) \( y = 10^{-11} \)

Calculate the number of ionized ammonia molecules

: We have found the equilibrium concentration of ionized ammonia species (NH₄⁺ and NH₂⁻) with y = 10⁻¹¹ M. To find the number of ionized ammonia molecules in 1 mole of ammonia, we use the following formula: number of ionized ammonia molecules = (concentration of ionized species) × (total number of molecules in 1 mole) Since there are 6.022 x 10²³ molecules in 1 mole (Avogadro's number): number of ionized ammonia molecules = (10⁻¹¹ M) × (6.022 x 10²³ molecules/mol) number of ionized ammonia molecules = \( 6.022 \times 10^{12} \) molecules Thus, there are approximately \( 6.022 \times 10^{12} \) ionized ammonia molecules in 1 mole of ammonia.

Key concepts

Equilibrium Equation

When discussing the autoprotolysis of ammonia, it's important to understand that this process involves the reaction between two molecules of ammonia to produce ions. This reaction reaches a point called equilibrium, where the rate of formation of products equals the rate of formation of reactants. In a system at equilibrium, concentrations of the reactants and products remain constant over time. The equilibrium equation for the autoprotolysis of ammonia is as follows:
\[ 2\text{NH}_3 \rightleftharpoons \text{NH}_4^+ + \text{NH}_2^- \]
This equation shows that two ammonia molecules react to form one ammonium ion \(\text{NH}_4^+\) and one amide ion \(\text{NH}_2^-\). Understanding this equation sets the foundation for analyzing the behavior of ammonia in this context, as it provides insight into how molecules interact at a molecular level.

Equilibrium Constant

The equilibrium constant, frequently represented as \(K\), is a crucial concept for understanding the extent of a chemical reaction at equilibrium. It quantifies the ratio of the concentration of products to reactants. For the autoprotolysis of ammonia, the equilibrium constant expression can be written as:
\[ K = \frac{[\text{NH}_4^+][\text{NH}_2^-]}{[\text{NH}_3]^2} \]
The given value \( K_{\text{NH}_3} = 10^{-22} \) reveals that this reaction produces a very small amount of ions at equilibrium. The smaller the value of \(K\), the less the reaction proceeds toward the formation of products. In this equation, the concentration of products \([\text{NH}_4^+]\) and \([\text{NH}_2^-]\) is much smaller compared to that of the reactant \([\text{NH}_3]\), indicating a low level of ionization.

Ionization

Ionization refers to the process by which a neutral molecule gains or loses electrons to form ions. For ammonia undergoing autoprotolysis, ionization involves two ammonia molecules reacting to form an ammonium ion and an amide ion.
In the equation \( 2\text{NH}_3 \rightleftharpoons \text{NH}_4^+ + \text{NH}_2^- \), the two product ions result from molecules interacting and exchanging hydrogen atoms. In this case, a very minimal quantity of ammonia is ionized, reflecting the weak autoprotolysis of ammonia. The small equilibrium constant \(10^{-22}\) supports the understanding that only a tiny fraction of ammonia molecules undergo ionization, which aligns with the calculation showing an equilibrium concentration of \(10^{-11}\) M for the ions formed. Therefore, ionization here has a minor impact on the overall amount of ammonia molecules.

Moles and Concentrations

In chemistry, moles and concentrations are fundamental in determining the extent of a reaction. The mole is a basic unit that measures the amount of substance, while concentration refers to the amount of solute in a specific volume of solution.
In this autoprotolysis reaction, we started with 1 mole of ammonia. At equilibrium, the concentration of ionized ammonia species \(\text{NH}_4^+\) and \(\text{NH}_2^-\) was found to be \(10^{-11}\) M. Using Avogadro's number, which is \(6.022 \times 10^{23}\) molecules per mole, we calculated the number of ionized molecules.
To find out how many ammonia molecules are ionized, we used the equation:
\[\text{Number of ionized molecules} = \text{Concentration of ionized species} \times \text{Avogadro's number} \]
Therefore, approximately \(6.022 \times 10^{12}\) ionized molecules are produced from 1 mole of ammonia, showing the limited extent of ionization in this reaction.