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Chapter 16: Problem 51

Ethanolamine, \(\mathrm{HOC}_{2} \mathrm{H}_{4} \mathrm{NH}_{2}\), is a viscous liquid with an ammonialike odor; it is used to remove hydrogen sulfide from natural gas. A \(0.15 M\) aqueous solution of ethanolamine has a pH of \(11.34\). What is \(K_{b}\) for ethanolamine?

Short Answer

Expert verified
The \( K_b \) for ethanolamine is approximately \( 3.20 \times 10^{-5} \).

Step by step solution

01

Write the Chemical Equation

First, we need to write the equilibrium reaction for ethanolamine acting as a base: \( \text{HOCH}_2\text{CH}_2\text{NH}_2 + \text{H}_2\text{O} \rightleftharpoons \text{HOCH}_2\text{CH}_2\text{NH}_3^+ + \text{OH}^- \).
02

Calculate the pOH

Given a pH of 11.34, we can calculate the pOH using the formula: \( \text{pOH} = 14 - \text{pH} \). Thus, \( \text{pOH} = 14 - 11.34 = 2.66 \).
03

Find the Concentration of OH-

Convert the pOH to \([\text{OH}^-]\) concentration using: \([\text{OH}^-] = 10^{-\text{pOH}}\). Therefore, \([\text{OH}^-] = 10^{-2.66} \approx 2.19 \times 10^{-3} \text{ M} \).
04

Set up the Kb Expression

Using the formula \( K_b = \frac{[\text{HOCH}_2\text{CH}_2\text{NH}_3^+][\text{OH}^-]}{[\text{HOCH}_2\text{CH}_2\text{NH}_2]} \). At equilibrium, \([\text{HOCH}_2\text{CH}_2\text{NH}_3^+] = [\text{OH}^-] \) because the base produces equal amounts of these ions, and the initial concentration of ethanolamine is given as 0.15 M.
05

Substitute Values and Solve for Kb

Substitute the known concentrations into the expression: \( K_b = \frac{(2.19 \times 10^{-3})(2.19 \times 10^{-3})}{0.15} \approx 3.20 \times 10^{-5} \).

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Key Concepts

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

Ethanolamine
Ethanolamine, known chemically as \( ext{HOCH}_2 ext{CH}_2 ext{NH}_2\), is an organic compound often utilized for removing hydrogen sulfide from natural gas due to its properties as a weak base. It's a colorless, viscous liquid emitting an ammonia-like scent. To understand its chemical behavior in aqueous solutions, it's essential to recognize its role as a base. As a base, ethanolamine can accept protons from water molecules. This reaction raises the solution's pH, providing it with alkaline characteristics.
  • In water, ethanolamine deprotonates, leading to a formation of ethanolammonium ions \( ext{HOCH}_2 ext{CH}_2 ext{NH}_3^+\) alongside hydroxide ions \( ext{OH}^-\).
  • The presence of these hydroxide ions increases the solution's pH, demonstrating ethanolamine's basic properties.
Understanding how ethanolamine behaves in solution aids in identifying its basicity level, which is quantified by its base ionization constant \(K_b\). This constant helps predict how strongly it tends to pull protons from water, converting to ethanolammonium ions.
Kb calculation
The calculation of the base ionization constant, \(K_b\), for ethanolamine in water is critical to understanding its strength as a base. Here, \(K_b\) quantifies the tendency of ethanolamine molecules to deprotonate and shift the equilibrium of the reaction towards the formation of ethanolammonium and hydroxide ions.

When determining \(K_b\), you should follow these steps:
  • Begin with writing the equilibrium reaction: \( ext{HOCH}_2 ext{CH}_2 ext{NH}_2 + ext{H}_2 ext{O} \rightleftharpoons ext{HOCH}_2 ext{CH}_2 ext{NH}_3^+ + ext{OH}^-\).
  • Use the initial concentration of ethanolamine and determine the equilibrium concentration of \([ ext{OH}^-]\).
  • Set up the \(K_b\) expression: \( K_b = \frac{[ ext{HOCH}_2 ext{CH}_2 ext{NH}_3^+][ ext{OH}^-]}{[ ext{HOCH}_2 ext{CH}_2 ext{NH}_2]}\), and substitute the known values.
  • Calculate \(K_b\) from these concentrations. For example, in a \(0.15 \, ext{M}\) solution with \([ ext{OH}^-] = 2.19 \times 10^{-3} \, ext{M}\), the base ionization constant is approximately \(3.20 \times 10^{-5}\).
This constant informs us about ethanolamine's potency as a base, with higher \(K_b\) pointing to stronger basicity.
pH and pOH relationship
The relationship between pH and pOH is a fundamental concept in understanding the acidity or basicity of solutions. The sum total of pH and pOH in any aqueous solution is always 14, at 25掳C. This relationship allows us to derive one measure if the other is known.

In the context of the ethanolamine solution:
  • Given that the solution has a pH of 11.34, we can calculate the pOH using: \( pOH = 14 - pH \).
  • Thus, the pOH is \(14 - 11.34 = 2.66\).
  • This pOH gives a direct insight into the hydroxide ion concentration, \([ ext{OH}^-]\), showing the solution's basic nature.
  • Using the relationship \([ ext{OH}^-] = 10^{- ext{pOH}}\), for a pOH of 2.66, the concentration is approximately \(2.19 \times 10^{-3} \, ext{M}\).
The simplicity of the pH and pOH relationship makes it a powerful tool in evaluating acid-base reactions, especially when assessing solutions like that of ethanolamine. Understanding this relationship is crucial for tackling problems involving equilibrium and ionization in solutions.

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Most popular questions from this chapter

Write the equation for the acid ionization of the \(\mathrm{Zn}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}^{2+}\) ion.

Sodium benzoate, \(\mathrm{NaC}_{7} \mathrm{H}_{5} \mathrm{O}_{2}\), is used as a preservative in foods. Consider a 50.0-mL sample of \(0.250 \mathrm{M} \mathrm{NaC}_{7} \mathrm{H}_{5} \mathrm{O}_{2}\) being titrated by \(0.200 M\) HBr. Calculate the \(\mathrm{pH}\) of the solution: a. when no \(\mathrm{HBr}\) has been added; \(\mathrm{b}\). after the addition of \(50.0 \mathrm{~mL}\) of the HBr solution; \(\mathrm{c}\). at the equivalence point; \(\mathrm{d}\). after the addition of \(75.00 \mathrm{~mL}\) of the \(\mathrm{HBr}\) solution. The \(K_{b}\) value for \(\mathrm{NaC}_{7} \mathrm{H}_{5} \mathrm{O}_{2}\) is \(1.6 \times 10^{-10}\)

a. When \(0.10\) mol of the ionic solid NaX, where \(X\) is an unknown anion, is dissolved in enough water to make \(1.0 \mathrm{~L}\) of solution, the \(\mathrm{pH}\) of the solution is \(9.12 .\) When \(0.10 \mathrm{~mol}\) of the ionic solid \(\mathrm{ACl}\), where \(\mathrm{A}\) is an unknown cation, is dissolved in enough water to make \(1.0 \mathrm{~L}\) of solution, the \(\mathrm{pH}\) of the solution is \(7.00 .\) What would be the \(\mathrm{pH}\) of \(1.0 \mathrm{~L}\) of solution that contained \(0.10\) mol of AX? Be sure to document how you arrived at your answer. b. In the AX solution prepared above, is there any \(\mathrm{OH}^{-}\) present? If so, compare the \(\left[\mathrm{OH}^{-}\right]\) in the solution to the \(\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]\) c. From the information presented in part a, calculate \(K_{b}\) for the \(\mathrm{X}^{-}(a q)\) anion and \(K_{a}\) for the conjugate acid of \(\mathrm{X}^{-}(a q)\). d. To \(1.0 \mathrm{~L}\) of solution that contains \(0.10 \mathrm{~mol}\) of \(\mathrm{AX}\), you add \(0.025\) mol of \(\mathrm{HCl}\). How will the \(\mathrm{pH}\) of this solution compare to that of the solution that contained only NaX? Use chemical reactions as part of your explanation; you do not need to solve for a numerical answer. e. Another \(1.0 \mathrm{~L}\) sample of solution is prepared by mixing \(0.10\) mol of \(\mathrm{AX}\) and \(0.10 \mathrm{~mol}\) of \(\mathrm{HCl}\). The \(\mathrm{pH}\) of the resulting solution is found to be \(3.12\). Explain why the \(\mathrm{pH}\) of this solution is \(3.12\). f. Finally, consider a different \(1.0\) -L sample of solution that contains \(0.10\) mol of \(A X\) and \(0.1\) mol of \(\mathrm{NaOH}\). The \(\mathrm{pH}\) of this solution is found to be \(13.00\). Explain why the \(\mathrm{pH}\) of this solution is \(13.00\). g. Some students mistakenly think that a solution that contains \(0.10 \mathrm{~mol}\) of \(\mathrm{AX}\) and \(0.10 \mathrm{~mol}\) of \(\mathrm{HCl}\) should have a \(\mathrm{pH}\) of \(1.00\). Can you come up with a reason why students have this misconception? Write an approach that you would use to help these students understand what they are doing wrong.

A chemist prepares dilute solutions of equal molar concentrations of \(\mathrm{NH}_{3}, \mathrm{NH}_{4} \mathrm{Br}, \mathrm{NaF}\), and \(\mathrm{NaCl}\). Rank these solutions from highest \(\mathrm{pH}\) to lowest \(\mathrm{pH}\).

What is the \(\mathrm{pH}\) of a solution in which \(40 \mathrm{~mL}\) of \(0.10 \mathrm{M}\) \(\mathrm{NaOH}\) is added to \(25 \mathrm{~mL}\) of \(0.10 \mathrm{M} \mathrm{HCl}\) ?
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