Question

Calculate the concentrations of all species present in a \(0.25-M\) solution of ethylammonium chloride \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{3} \mathrm{Cl}\right)\)

Short answer

In a 0.25 M solution of ethylammonium chloride (C2H5NH3Cl), the hydrolysis reaction can be represented as \(C_2H_5NH_3Cl \rightleftharpoons C_2H_5NH_2+HCl\). The equilibrium expression (Ka) for this reaction is \(Ka = \frac{[C_2H_5NH_2][HCl]}{[C_2H_5NH_3Cl]}\). Using the given Ka value of 2.3 × 10^(-10), we can solve for x (the change in the concentration of the species) using the expressions [C2H5NH3Cl] = 0.25 - x, [C2H5NH2] = x, and [HCl] = x. Finally, the equilibrium concentrations of C2H5NH3Cl, C2H5NH2, and HCl can be determined using the calculated value of x.

Step-by-step solution

Write the hydrolysis reaction

First, we need to write the reaction for the hydrolysis of ethylammonium chloride in water. The reaction would be: \(C_2H_5NH_3Cl \rightleftharpoons C_2H_5NH_2+HCl\)

Write the equilibrium expression for the reaction

We can write the equilibrium expression (Ka) for the hydrolysis of ethylammonium chloride as: \(Ka = \frac{[C_2H_5NH_2][HCl]}{[C_2H_5NH_3Cl]}\)

Find the initial concentrations of all species

The given concentration of ethylammonium chloride is 0.25 M. Initially, we assume that the hydrolysis process has not started yet. Therefore, the initial concentrations of all species can be written as: [C2H5NH3Cl] = 0.25 M [C2H5NH2] = 0 M [HCl] = 0 M

Write the changes in concentrations during the reaction

As the reaction proceeds, some of the C2H5NH3Cl will be converted into C2H5NH2 and HCl. We can represent these changes in concentration using the variable x: [C2H5NH3Cl] = 0.25 - x [C2H5NH2] = x [HCl] = x

Plug the concentrations into the equilibrium expression

We can plug in the expressions of concentrations in terms of x into the equilibrium expression Ka: \(Ka = \frac{x^2}{0.25-x}\)

Use the Ka value to solve for x

The value of the Ka for ethylammonium ion (C2H5NH3+) is 2.3 × 10^(-10). We can use this value to solve for x: \(2.3 × 10^(-10) = \frac{x^2}{0.25 - x}\) Solve for x using the quadratic formula or by making an approximation (assuming x << 0.25 M) and simplifying the expression.

Determine the equilibrium concentrations of all species

Once we have found the value of x, we can use it to find the equilibrium concentrations of all species: [C2H5NH3Cl] = 0.25 - x [C2H5NH2] = x [HCl] = x These values represent the concentrations of each species in the 0.25 M solution of ethylammonium chloride at equilibrium.

Key concepts

hydrolysis reaction

The process of hydrolysis involves a chemical reaction with water that breaks down compounds into simpler forms. In the case of ethylammonium chloride ( C_2H_5NH_3^+Cl^- ), hydrolysis results in the formation of ethylamine ( C_2H_5NH_2 ) and hydrochloric acid (HCl). This occurs when the ethylammonium ( C_2H_5NH_3^+ ) ion reacts with water.

Hydrolysis is important because it demonstrates how chemical compounds can transform in an aqueous environment. By understanding hydrolysis, students can predict the resulting species in a solution and how these contribute to equilibrium.

• For our chemical equation, the hydrolysis can be represented as:
  [C_2H_5NH_3Cl ightleftharpoons C_2H_5NH_2 + HCl]
• This means the chloride ion is not directly participating in the hydrolysis, and we focus on the behavior of the ethylammonium ion in water.

equilibrium expression

The equilibrium expression is a mathematical way to describe the relationship between the concentrations of reactants and products at equilibrium. For the hydrolysis of ethylammonium chloride, we derive an equilibrium expression based on the balanced chemical equation.
The general form for an equilibrium expression is:
Ka is defined as

• Ka = \(\frac{[\text{products}]}{[\text{reactants}]}\)
• For this exercise, it becomes: Ka = \(\frac{[C_2H_5NH_2][HCl]}{[C_2H_5NH_3Cl]}\)

Typically, equilibrium constants express the extent to which a reaction proceeds to form products. A large Ka implies a reaction favoring product formation, while a small Ka indicates reactants dominate.

In our solution, the reported value of Ka for ethylammonium ion (C_2H_5NH_3^+) is very small (2.3 × 10^(-10)), indicating minimal product formation.

equilibrium concentrations

Equilibrium concentrations refer to the amounts of each species present in a reaction mixture when the rate of the forward reaction equals the rate of the backward reaction. With the given initial concentrations—such as the case of ethylammonium chloride—we start with 0.25 M of this compound, and nothing of the products ( C_2H_5NH_2 and HCl).

During hydrolysis:

• Ethylammonium chloride’s concentration decreases by an amount x, transforming into equivalent amounts of products.
• The concentrations at equilibrium are represented as:
  • [C_2H_5NH_3Cl] = 0.25 - x
  • [C_2H_5NH_2] = x
  • [HCl] = x

To solve for x, we use the equilibrium constant Ka, substituting the expressions for equilibrium concentrations into the equilibrium expression. Sometimes, for very small Ka values, we may approximate to simplify calculations, assuming x is negligible compared to initial concentrations.