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Chapter 15: Problem 128

A solution contains \(0.675 \mathrm{~g}\) of ethylamine, \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{2}\), per \(100.0 \mathrm{~mL}\) of solution. Electrical conductivity measurements at \(20^{\circ} \mathrm{C}\) show that \(0.98 \%\) of the ethylamine has reacted with water. Write the equation for this reaction. Calculate the \(\mathrm{pH}\) of the solution.

Short Answer

Expert verified
The pH of the solution is approximately 11.17.

Step by step solution

01

Write the Reaction Equation

The reaction of ethylamine (\(\mathrm{C}_{2}\mathrm{H}_{5}\mathrm{NH}_{2}\)) with water can be represented as follows:\[\mathrm{C}_2\mathrm{H}_5\mathrm{NH}_2 (aq) + \mathrm{H}_2\mathrm{O} (l) \rightleftharpoons \mathrm{C}_2\mathrm{H}_5\mathrm{NH}_3^+ (aq) + \mathrm{OH}^- (aq)\] This is a typical base ionization reaction where ethylamine accepts a proton from water, forming the ethylammonium ion \(\mathrm{C}_2\mathrm{H}_5\mathrm{NH}_3^+\) and hydroxide ion \(\mathrm{OH}^-\).
02

Calculate Molarity of Ethylamine

First, find the molar mass of ethylamine \(\mathrm{C}_2\mathrm{H}_5\mathrm{NH}_2\). The atomic masses are approximately: \(\mathrm{C} = 12.01\), \(\mathrm{H} = 1.008\), \(\mathrm{N} = 14.01\). Thus, the molar mass is:\[2(12.01) + 5(1.008) + 14.01 + 2(1.008) = 45.08 \text{ g/mol}\]Using this, calculate the molarity: \[\text{Molarity} = \frac{0.675 \text{ g}}{45.08 \text{ g/mol}} \times \frac{1}{0.1 \text{ L}} = 0.1495 \text{ M}\]
03

Determine the Reacted Concentration

Given that \(0.98\%\) of the ethylamine has reacted, the concentration of reacted ethylamine is:\[0.1495 \times \frac{0.98}{100} = 0.0014651 \text{ M}\]This represents the concentration of \(\mathrm{C}_2\mathrm{H}_5\mathrm{NH}_3^+\) and \(\mathrm{OH}^-\) ions in the solution.
04

Find Hydroxide Ion Concentration

Since \(0.0014651 \text{ M}\) corresponds to \(\mathrm{OH}^-\), we can calculate the \(\mathrm{pOH}\):\[\mathrm{pOH} = -\log(0.0014651) = 2.83\]
05

Calculate the pH

Utilize the relationship between \(\mathrm{pH}\) and \(\mathrm{pOH}\):\[\mathrm{pH} + \mathrm{pOH} = 14\]\[\mathrm{pH} = 14 - 2.83 = 11.17\]
06

Conclusion: Reporting the Result

The \(\mathrm{pH}\) of the ethylamine solution, based on the degree of ionization and the given percentage reacted, is approximately \(11.17\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Ethylamine
Ethylamine, with the chemical formula \(\mathrm{C}_2\mathrm{H}_5\mathrm{NH}_2\), is a simple organic compound belonging to the class of amines. These are derivatives of ammonia, where one or more hydrogen atoms are replaced by alkyl groups.
  • Ethylamine is a weak base, which means it partially ionizes in water.
  • It has a notable fishy odor and is used in a range of applications from organic synthesis to pharmaceuticals.
When ethylamine dissolves in water, it tends to accept a proton (\(\mathrm{H}^+\)) from water molecules. This process results in the formation of ethylammonium ion \(\mathrm{C}_2\mathrm{H}_5\mathrm{NH}_3^+\) and hydroxide ion \(\mathrm{OH}^-\). Thus, an important aspect of studying ethylamine is understanding its behavior in water and its tendency to increase the pH of solutions due to hydroxide ion production.
Molarity Calculation
Molarity is a measure of the concentration of a solute in a solution. It represents the number of moles of a solute per liter of solution. Calculating molarity is a fundamental skill in chemistry as it helps to quantify substances.Here is how we calculate the molarity of ethylamine in the given problem:
  • Firstly, find the molar mass of ethylamine, which is calculated by adding the atomic masses of each constituent atom: \(\text{C}_2\text{H}_5\text{NH}_2\).
  • Ethylamine has: \[2(12.01) + 5(1.008) + 14.01 + 2(1.008) = 45.08 \text{ g/mol}\]
  • Next, determine the number of moles of ethylamine present in the solution: \(\frac{0.675 \text{ g}}{45.08 \text{ g/mol}}\).
  • Calculate the molarity: divide the moles by the volume of solution in liters. So, \(\text{Molarity} = 0.1495 \text{ M}\).
This calculation shows how concentrated the ethylamine is in the solution, providing a basis for further exploration of its reactivity and its impact on the solution's properties.
pH Calculation
The pH of a solution is a measure of its acidity or basicity. It is calculated from the concentration of hydrogen ions \(\mathrm{H}^+\) in the solution. For basic solutions like those involving ethylamine, the calculation also involves the related concept of \(\mathrm{pOH}\).To find the \(\mathrm{pH}\), we first need to determine the \(\mathrm{pOH}\) because ethylamine significantly affects the concentration of \(\mathrm{OH}^-\) ions:
  • First, calculate the concentration of \(\mathrm{OH}^-\) in the solution due to the ionization of ethylamine.
  • With \(0.98\%\) of ethylamine reacting, calculate the reacted molarity as \(0.1495 \times \frac{0.98}{100} = 0.0014651 \text{ M}\).
  • Find \( \mathrm{pOH} \) using \(-\log(0.0014651) = 2.83\).
  • To find \( \mathrm{pH} \), apply the formula: \(\mathrm{pH} + \mathrm{pOH} = 14\), giving \(\mathrm{pH} = 11.17\).
Thus, the solution is basic, indicated by a pH greater than 7. The pH calculation is crucial for determining how the solution may interact with other substances or systems.
Hydroxide Ion Concentration
Hydroxide ions \(\mathrm{OH}^-\) play a pivotal role in defining the basic nature of aqueous solutions. Their concentration directly influences both \(\mathrm{pH}\) and \(\mathrm{pOH}\).In this particular reaction with ethylamine:
  • The hydroxide ions are produced as ethylamine accepts a proton from water, forming \(\mathrm{OH}^-\).
  • Using the ionization percentage of ethylamine, the concentration of \(\mathrm{OH}^-\) is deduced to be \(0.0014651 \text{ M}\).
  • This concentration helps determine the \(\mathrm{pOH}\) as \(2.83\), which is part of the crucial step in calculating \(\mathrm{pH}\).
Understanding \(\mathrm{OH}^-\) concentrations is essential for predicting the behavior of basic solutions. It opens doors to exploring more advanced concepts like buffer solutions and titration curves 鈥� essential components of a chemistry toolkit.

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Most popular questions from this chapter

In the following reaction of tetrafluoroboric acid, \(\mathrm{HBF}_{4}\), with the acetate ion, \(\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{O}_{2}^{-},\) the formation of tetrafluoroborate ion, \(\mathrm{BF}_{4}^{-}\), and acetic acid, \(\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\) is favored. $$ \mathrm{HBF}_{4}+\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{O}_{2}^{-} \longrightarrow \mathrm{BF}_{4}^{-}+\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2} $$ Which is the weaker base, \(\mathrm{BF}_{4}^{-}\) or acetate ion?

A shampoo solution at \(25^{\circ} \mathrm{C}\) has a hydroxide-ion concentration of \(1.5 \times 10^{-9} M .\) Is the solution acidic, neutral, or basic?

Natural gas frequently contains hydrogen sulfide, \(\mathrm{H}_{2} \mathrm{~S}\). \(\mathrm{H}_{2} \mathrm{~S}\) is removed from natural gas by passing it through aqueous ethanolamine, \(\mathrm{HOCH}_{2} \mathrm{CH}_{2} \mathrm{NH}_{2}\) (an ammonia derivative), which reacts with the hydrogen sulfide. Write the equation for the reaction. Identify each reactant as either a Lewis acid or a Lewis base. Explain how you arrived at your answer.

In the following reaction of trichloroacetic acid, \(\mathrm{HC}_{2} \mathrm{Cl}_{3} \mathrm{O}_{2},\) with formate ion, \(\mathrm{CHO}_{2}^{-},\) the formation of trichloroacetate ion, \(\mathrm{C}_{2} \mathrm{Cl}_{3} \mathrm{O}_{2}^{-},\) and formic acid, \(\mathrm{HCHO}_{2}\) is favored. $$ \mathrm{HC}_{2} \mathrm{Cl}_{3} \mathrm{O}_{2}+\mathrm{CHO}_{2}^{-} \longrightarrow \mathrm{C}_{2} \mathrm{Cl}_{3} \mathrm{O}_{2}^{-}+\mathrm{HCHO}_{2} $$ Which is the stronger acid, trichloroacetic acid or formic acid? Exnlain.

You want to make up \(3.00 \mathrm{~L}\) of aqueous hydrochloric acid, \(\mathrm{HCl}(a q),\) that has a \(\mathrm{pH}\) of \(2.00 .\) How many grams of concentrated hydrochloric acid will you need? Concentrated hydrochloric acid contains 37.2 mass percent of \(\mathrm{HCl}\).
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