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Chapter 14: Problem 118

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

Expert verified
In a 0.25 M solution of ethylammonium chloride, the concentrations of all species present are: - Ethylammonium ion (C₂H₅NH₃⁺): approximately 0.25 M - Hydronium ion (H₃O⁺): \(3.73 \times 10^{-6}\) M - Ethylamine (C₂H₅NH₂): \(3.73 \times 10^{-6}\) M - Chloride ion (Cl⁻): 0.25 M

Step by step solution

01

Write the dissociation equation of ethylammonium chloride

First, let's write the dissociation equation for ethylammonium chloride: \[C_2H_5NH_3Cl \rightarrow C_2H_5NH_3^+ + Cl^-\]
02

Write the equilibrium expression for the reaction of ethylammonium ion with water

Next, we'll write the equilibrium expression for the reaction of ethylammonium ion (C₂H₅NH₃⁺) reacting with water to produce hydronium ion (H₃O⁺) and ethylamine (C₂H₅NH₂): \[C_2H_5NH_3^+ + H_2O \leftrightarrow H_3O^+ + C_2H_5NH_2\] The acid dissociation constant, \(K_a\), is given by: \[K_a = \frac{[H_3O^+][C_2H_5NH_2]}{[C_2H_5NH_3^+]}\]
03

Create an ICE table to represent the change in concentrations as a result of the reaction

We can create an ICE (Initial, Change, Equilibrium) table to help us determine the change in concentrations as a result of the reaction. It is important to note the initial concentration of the species and the change in the concentration due to the reaction. | |C₂H₅NH₃⁺|H₃O⁺|C₂H₅NH₂| |---------|-------|-----|------| |Initial |0.25 |0 |0 | |Change |-x |+x |+x | |Equilibrium|0.25-x |x |x | *Note: Since Cl⁻ ions do not participate in the reaction, their concentration remains constant at 0.25 M.
04

Substitute the equilibrium concentrations into the \(K_a\) expression and solve for x

We can now substitute the equilibrium concentrations from the ICE table into the \(K_a\) expression: \[K_a = \frac{x \times x}{0.25 - x}\] We can use the \(K_a\) value for ethylammonium ion (C₂H₅NH₃⁺), which is \(5.6 \times 10^{-11}\), to solve for x: \(5.6 \times 10^{-11} = \frac{x^2}{0.25 - x}\) Since the \(K_a\) value is very small, we can assume that x is also small and ignore it in the denominator, which will give us an approximation: \(5.6 \times 10^{-11} \approx \frac{x^2}{0.25}\) To solve for x, we can rearrange the equation and take the square root: x = \( \sqrt{5.6 \times 10^{-11} \times 0.25}\) x ≈ \(3.73 \times 10^{-6}\) M
05

Determine the concentrations of all species present in the solution

Now that we have the value for x, we can substitute it back into the respective equilibrium concentration values to determine the concentrations of all species present in the 0.25 M ethylammonium chloride solution: 1. Ethylammonium ion (C₂H₅NH₃⁺) concentration: \(0.25 - 3.73 \times 10^{-6}\) M ≈ 0.25 M 2. Hydronium ion (H₃O⁺) concentration: \(3.73 \times 10^{-6}\) M 3. Ethylamine (C₂H₅NH₂) concentration: \(3.73 \times 10^{-6}\) M 4. Chloride ion (Cl⁻) concentration: 0.25 M Thus, the concentrations of all species present in the 0.25 M solution of ethylammonium chloride are as follows: - C₂H₅NH₃⁺: approximately 0.25 M - H₃O⁺: \(3.73 \times 10^{-6}\) M - C₂H₅NH₂: \(3.73 \times 10^{-6}\) M - Cl⁻: 0.25 M

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Most popular questions from this chapter

Calculate the \(\mathrm{pH}\) of the following solutions. a. \(0.10 \mathrm{M} \mathrm{NaOH}\) b. \(1.0 \times 10^{-10} M \mathrm{NaOH}\) c. \(2.0 \mathrm{M} \mathrm{NaOH}\)

What mass of \(\mathrm{KOH}\) is necessary to prepare \(800.0 \mathrm{~mL}\) of a solution having a \(\mathrm{pH}=11.56\) ?

What mass of \(\mathrm{NaOH}(s)\) must be added to \(1.0 \mathrm{~L}\) of \(0.050 \mathrm{M}\) \(\mathrm{NH}_{3}\) to ensure that the percent ionization of \(\mathrm{NH}_{3}\) is no greater than \(0.0010 \%\) ? Assume no volume change on addition of \(\mathrm{NaOH}\)

Calculate the \(\mathrm{pH}\) of a \(2.0-M \mathrm{H}_{2} \mathrm{SO}_{4}\) solution.

The following illustration displays the relative number of species when an acid, HA, is added to water. a. Is HA a weak or strong acid? How can you tell? b. Using the relative numbers given in the illustration, determine the value for \(K_{\mathrm{a}}\) and the percent dissociation of the acid. Assume the initial acid concentration is \(0.20 \mathrm{M}\).
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