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Methylamine, \(\mathrm{CH}_{3} \mathrm{NH}_{2},\) is a weak base. \(\mathrm{CH}_{3} \mathrm{NH}_{2}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\ell) \rightleftharpoons \mathrm{CH}_{3} \mathrm{NH}_{3}^{+}(\mathrm{aq})+\mathrm{OH}^{-}(\mathrm{aq})\) If the \(p H\) of a \(0.065 \mathrm{M}\) solution of the amine is \(11.70,\) what is the value of \(K_{\mathrm{b}} ?\)

Short Answer

Expert verified
The value of \(K_b\) is approximately \(4.18 \times 10^{-4}.\)

Step by step solution

01

Find the Concentration of OH^-

The given pH of the solution is 11.70. Use the relationship between pH and pOH to find pOH: \[pOH = 14 - pH = 14 - 11.70 = 2.30.\]
02

Calculate [OH^-] from pOH

Use the formula to find [OH^-] from pOH: \[ [OH^-] = 10^{-pOH} = 10^{-2.30} \] Calculating gives \[ [OH^-] = 5.01 \times 10^{-3} \; \text{M}.\]
03

Assume Initial Conditions for Ice Chart

For methylamine's dissociation, the initial concentration of \([\mathrm{CH}_3\mathrm{NH}_2]\) is 0.065 M, and the initial concentrations of \([\mathrm{CH}_3\mathrm{NH}_3^+ ]\) and \([\mathrm{OH}^- ]\) are both 0 M. At equilibrium, the concentrations change by \(-x\) for \([\mathrm{CH}_3\mathrm{NH}_2 ]\), and \(+x\) for both products.
04

Write Equilibrium Concentrations

The equilibrium concentrations are as follows:- \([\mathrm{CH}_3\mathrm{NH}_2 ] = 0.065 - x\)- \([\mathrm{CH}_3\mathrm{NH}_3^+ ] = x\)- \([\mathrm{OH}^- ] = x\)Since \([\mathrm{OH}^-] = 5.01 \times 10^{-3} \; \text{M}\), assume \(x = [\mathrm{OH}^-]\).
05

Write Kb Expression

The expression for the base dissociation constant \(K_b\) is:\[ K_b = \frac{{[\mathrm{CH}_3\mathrm{NH}_3^+][\mathrm{OH}^-]}}{[\mathrm{CH}_3\mathrm{NH}_2]} \] Using the equilibrium concentrations, this becomes:\[ K_b = \frac{(5.01 \times 10^{-3})(5.01 \times 10^{-3})}{0.065 - 5.01 \times 10^{-3}}.\]
06

Solve for Kb

Calculate \(K_b\) using the expression from Step 5:\[ K_b = \frac{(5.01 \times 10^{-3})^2}{0.065 - 5.01 \times 10^{-3}} \]\[ K_b \approx \frac{25.10 \times 10^{-6}}{0.05999} \approx 4.18 \times 10^{-4}.\]

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Key Concepts

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

Methylamine
Methylamine is an organic compound with the formula \( \mathrm{CH}_3\mathrm{NH}_2 \). It is classified as a weak base. Understanding its behavior in water is essential for acid-base equilibrium studies. Methylamine accepts protons, meaning it's a prototypical Bronsted-Lowry base. When it dissolves in water, methylamine partially dissociates to form its conjugate acid, \( \mathrm{CH}_3\mathrm{NH}_3^+ \), and hydroxide ions, \( \mathrm{OH}^- \). This partial dissociation is a key feature of weak bases. In solutions like the one described in the exercise, where the initial concentration is 0.065 M, the behavior of methylamine can be analyzed using equilibrium concepts to determine its dissociation constant, \( K_b \). Understanding methylamine's equilibrium helps in predicting the solution's pH and efficiency in accepting protons during reactions.
pH and pOH
The pH and pOH scales are critical for measuring the acidity and basicity of solutions. They are related by the equation: \[ pH + pOH = 14 \] at 25°C. In the original exercise, the pH is given as 11.70. This high pH value indicates that the solution is basic, as values above 7 are characteristic of bases. From the pH, we can find the pOH by subtracting the pH from 14, resulting in a pOH of 2.30. This calculation highlights the inverse relationship: as pH increases, indicating more basicity, pOH decreases.
Understanding pH and pOH helps predict the behavior of solutions like methylamine. In basic solutions, fewer hydrogen ions are present, meaning more hydroxide ions exist, impacting shifts in chemical equilibrium.
Base Dissociation Constant
The base dissociation constant, \( K_b \), is a measure of a base's strength in water. It provides insight into how well a base, such as methylamine, dissociates into its ions. The expression for \( K_b \) is formulated from the equilibrium concentrations of the ions, represented as:
\[ K_b = \frac{{[\mathrm{CH}_3\mathrm{NH}_3^+][\mathrm{OH}^-]}}{[\mathrm{CH}_3\mathrm{NH}_2]} \]
In this equation, \([\mathrm{CH}_3\mathrm{NH}_3^+]\) and \([\mathrm{OH}^-]\) denote the concentrations of the products, while \([\mathrm{CH}_3\mathrm{NH}_2]\) is the concentration of the undissociated base at equilibrium. By substituting these equilibrium values, derived from the ICE chart, into the expression, we can calculate \( K_b \). The resulting value provides a quantitative measure of methylamine's basicity, helping predict its behavior in various chemical scenarios.
ICE Chart Method
The ICE chart method is a systematic approach for solving chemical equilibrium problems. It stands for Initial, Change, and Equilibrium, accounting for concentration changes as a reaction reaches equilibrium.
  • Initial: Start by identifying initial concentrations. In our methylamine example, 0.065 M for methylamine, and 0 M for products.
  • Change: Define the changes in concentration as reaction progress. A typical notation is \( -x \) for reactants and \( +x \) for products.
  • Equilibrium: Combine initial concentrations and changes to find final equilibrium concentrations.
    Using these values to express the \( K_b \) formula, we solve for the equilibrium constant value.
    ICE charts are a powerful tool enabling clear visualization of shifts in equilibrium, crucial for accurately determining unknown quantities like \( K_b \) from experimental data.

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Most popular questions from this chapter

Prove that \(K_{\mathrm{a} 3} \times K_{\mathrm{b} 1}=K_{\mathrm{w}}\) for phosphoric acid, \(\mathrm{H}_{3} \mathrm{PO}_{4},\) by adding the chemical equilibrium expressions that correspond to the third ionization step of the acid in water with the first of the three successive steps of the reaction of phosphate ion, \(\mathrm{PO}_{4}^{3-},\) with water.

For each of the following cases, decide whether the \(\mathrm{pH}\) is less than \(7,\) equal to \(7,\) or greater than 7 (a) 25 mL of \(0.45 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4}\) is mixed with \(25 \mathrm{mL}\) of 0.90 M NaOH. (b) 15 mL of \(0.050 \mathrm{M}\) formic acid, \(\mathrm{HCO}_{2} \mathrm{H}\), is mixed with \(15 \mathrm{mL}\) of \(0.050 \mathrm{M} \mathrm{NaOH}\). (c) \(25 \mathrm{mL}\) of \(0.15 \mathrm{M} \mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\) (oxalic acid) is mixed with \(25 \mathrm{mL}\) of \(0.30 \mathrm{M} \mathrm{NaOH}\). (Both \(\mathrm{H}^{+}\) ions of oxalic acid are removed with NaOH.)

Explain why benzenesulfonic acid is a Bronsted acid.

Trimethylamine, \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{N},\) is a common reagent. It interacts readily with diborane gas, \(\mathrm{B}_{2} \mathrm{H}_{6}\). The latter dissociates to \(\mathrm{BH}_{3}\), and this forms a complex with the amine, \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{N} \rightarrow \mathrm{BH}_{3} .\) Is the \(\mathrm{BH}_{3}\) fragment a Lewis acid or a Lewis base?

You have three solutions labeled \(A, B\), and \(C\). You know only that each contains a different cation \(-\mathrm{Na}^{+}, \mathrm{NH}_{4}^{+},\) or \(\mathrm{H}_{3} \mathrm{O}^{+} .\) Each has an anion that does not contribute to the solution \(\mathrm{pH}\) (eg. \(\mathrm{Cl}^{-}\) ). You also have two other solutions, Y and \(\mathrm{Z}\) each containing a different anion, \(\mathrm{Cl}^{-}\) or \(\mathrm{OH}^{-}\) with a cation that does not influence solution \(\mathrm{pH}\) \(\left(\mathrm{e} \cdot \mathrm{g} \cdot, \mathrm{K}^{+}\right) .\) If equal amounts of \(\mathrm{B}\) and \(\mathrm{Y}\) are mixed, the result is an acidic solution. Mixing A and Z gives a neutral solution, whereas \(\mathrm{B}\) and Z give a basic solution. Identify the five unknown solutions. (Adapted from D. H. Barouch: Voyages in Conceptual Chemistry, Boston, Jones and Bartlett, \(1997 .)\)

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