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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 (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

01

Write the hydrolysis reaction

First, we need to write the reaction for the hydrolysis of ethylammonium chloride in water. The reaction would be: \(C_2H_5NH_3Cl \rightleftharpoons C_2H_5NH_2+HCl\)
02

Write the equilibrium expression for the reaction

We can write the equilibrium expression (Ka) for the hydrolysis of ethylammonium chloride as: \(Ka = \frac{[C_2H_5NH_2][HCl]}{[C_2H_5NH_3Cl]}\)
03

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
04

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
05

Plug the concentrations into the equilibrium expression

We can plug in the expressions of concentrations in terms of x into the equilibrium expression Ka: \(Ka = \frac{x^2}{0.25-x}\)
06

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.
07

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.

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Most popular questions from this chapter

Rank the following 0.10\(M\) solutions in order of increasing \(\mathrm{pH.}\) a. HI, HF, NaF, NaI b. \(\mathrm{NH}_{4} \mathrm{Br}, \mathrm{HBr}, \mathrm{KBr}, \mathrm{NH}_{3}\) c. $C_{6} \mathrm{H}_{5} \mathrm{NH}_{3} \mathrm{NO}_{3}, \mathrm{NaNO}_{3}, \mathrm{NaOH}, \mathrm{HOC}_{6} \mathrm{H}_{5}, \mathrm{KOC}_{6} \mathrm{H}_{5}$ \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}, \mathrm{HNO}_{3}\)

Calculate the pH of the following solutions: a. 1.2\(M \mathrm{CaBr}_{2}\) b. 0.84$M \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{3} \mathrm{NO}_{3}\left(K_{\mathrm{b}} \text { for } \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}=3.8 \times 10^{-10}\right)$ c. 0.57$M \mathrm{KC}_{7} \mathrm{H}_{5} \mathrm{O}_{2}\left(K_{\mathrm{a}} \text { for } \mathrm{HC}_{7} \mathrm{H}_{5} \mathrm{O}_{2}=6.4 \times 10^{-5}\right)$

Hemoglobin (abbreviated Hb) is a protein that is responsible for the transport of oxygen in the blood of mammals. Each hemoglobin molecule contains four iron atoms that are the binding sites for \(\mathrm{O}_{2}\) molecules. The oxygen binding is pH- dependent. The relevant equilibrium reaction is $$ \mathrm{HbH}_{4}^{4+}(a q)+4 O_{2}(g) \rightleftharpoons \mathrm{Hb}\left(\mathrm{O}_{2}\right)_{4}(a q)+4 \mathrm{H}^{+}(a q) $$ Use Le Châtelier's principle to answer the following. a. What form of hemoglobin, HbH \(_{4}^{4+}\) or \(\mathrm{Hb}\left(\mathrm{O}_{2}\right)_{4},\) is favored in the lungs? What form is favored in the cells? b. When a person hyperventilates, the concentration of \(\mathrm{CO}_{2}\) in the blood is decreased. How does this affect the oxygen-binding equilibrium? How does breathing into a paper bag help to counteract this effect? (See Exercise \(146 .\) ) c. When a person has suffered a cardiac arrest, injection of a sodium bicarbonate solution is given. Why is this necessary? (Hint: CO, blood levels increase during cardiac arrest.)

a. The principal equilibrium in a solution of \(\mathrm{NaHCO}_{3}\) is $$ \mathrm{HCO}_{3}^{-}(a q)+\mathrm{HCO}_{3}^{-}(a q) \rightleftharpoons \mathrm{H}_{2} \mathrm{CO}_{3}(a q)+\mathrm{CO}_{3}^{2-}(a q) $$ Calculate the value of the equilibrium constant for this reaction. b. At equilibrium, what is the relationship between $\left[\mathrm{H}_{2} \mathrm{CO}_{3}\right]\( and \)\left[\mathrm{CO}_{3}^{2-}\right] ?$ c. Using the equilibrium $$ \mathrm{H}_{2} \mathrm{CO}_{3}(a q) \rightleftharpoons 2 \mathrm{H}^{+}(a q)+\mathrm{CO}_{3}^{2-}(a q) $$ derive an expression for the pH of the solution in terms of \(K_{\mathrm{a}_{1}}\) and \(K_{\mathrm{a}_{2}}\) using the result from part b. d. What is the pH of a solution of \(\mathrm{NaHCO}_{3} ?\)

Calculate the concentration of all species present and the pH of a \(0.020-M\) HF solution.
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