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Chapter 16: Problem 70

Write the chemical equation and the \(K_{b}\) expression for the reaction of each of the following bases with water: (a) propylamine, \(\mathrm{C}_{3} \mathrm{H}_{7} \mathrm{NH}_{2} ;\) (b) monohydrogen phosphate ion, \(\mathrm{HPO}_{4}^{2-} ;(\mathbf{c})\) benzoate ion, \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2}^{-}\)

Short Answer

Expert verified
(a) K_b = rac{[ ext{C}_3 ext{H}_7 ext{NH}_3^+][OH^-]}{[ ext{C}_3 ext{H}_7 ext{NH}_2]}; (b) K_b = rac{[H_2PO_4^-][OH^-]}{[HPO_4^{2-}]}; (c) K_b = rac{[C_6H_5CO_2H][OH^-]}{[C_6H_5CO_2^-]}.

Step by step solution

01

Identify the Base and Water Interaction

Each base reacts with water in a process of accepting a proton (H^+), resulting in the base's conjugate acid and a hydroxide ion (OH^-). This proton acceptance indicates that propylamine, monohydrogen phosphate ion, and benzoate ion are acting as Bronsted-Lowry bases.
02

Write the Propylamine Reaction with Water

Propylamine ( ext{C}_3 ext{H}_7 ext{NH}_2) reacts with water by accepting a proton from water to form propylammonium ion ( ext{C}_3 ext{H}_7 ext{NH}_3^+) and hydroxide ion (OH^-). The chemical equation is:  ext{C}_3 ext{H}_7 ext{NH}_2 + H_2O ightleftharpoons ext{C}_3 ext{H}_7 ext{NH}_3^+ + OH^-.
03

Write the Propylamine Base Ionization Constant (K_b)

The expression for K_b is derived from the equilibrium concentrations of products over reactants, excluding water because it's a liquid: K_b = rac{[ ext{C}_3 ext{H}_7 ext{NH}_3^+][OH^-]}{[ ext{C}_3 ext{H}_7 ext{NH}_2]}.
04

Write the Monohydrogen Phosphate Ion Reaction with Water

The monohydrogen phosphate ion (HPO_4^{2-}) can accept a proton from water forming dihydrogen phosphate ion (H_2PO_4^-) and hydroxide ion (OH^-). The balanced chemical equation is: HPO_4^{2-} + H_2O ightleftharpoons H_2PO_4^- + OH^-.
05

Write the Monohydrogen Phosphate Base Ionization Constant (K_b)

The K_b expression represents the following equilibrium: K_b = rac{[H_2PO_4^-][OH^-]}{[HPO_4^{2-}]}.
06

Write the Benzoate Ion Reaction with Water

The benzoate ion (C_6H_5CO_2^-) accepts a proton to form benzoic acid (C_6H_5CO_2H) and hydroxide ion (OH^-). The chemical equation is: C_6H_5CO_2^- + H_2O ightleftharpoons C_6H_5CO_2H + OH^-.
07

Write the Benzoate Base Ionization Constant (K_b)

The K_b expression for this reaction is calculated as follows: K_b = rac{[C_6H_5CO_2H][OH^-]}{[C_6H_5CO_2^-]}.

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Key Concepts

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

Chemical Equation
A chemical equation is a symbolic representation of a chemical reaction. In the context of base ionization, the equation shows how a base interacts with water to form a conjugate acid and hydroxide ions. For example, when propylamine (\( \text{C}_3\text{H}_7\text{NH}_2 \)) reacts with water, it accepts a proton from water. This forms a conjugate acid, propylammonium ion (\( \text{C}_3\text{H}_7\text{NH}_3^+ \)), and releases a hydroxide ion (\( \text{OH}^- \)). The transformation is succinctly shown as:
  • \( \text{C}_3\text{H}_7\text{NH}_2 + H_2O \rightleftharpoons \text{C}_3\text{H}_7\text{NH}_3^+ + OH^- \)

This representation helps in understanding the reactive nature of bases when dissolved in water.
Bronsted-Lowry Bases
Bronsted-Lowry theory is a fundamental principle in acid-base chemistry. According to this theory, a Bronsted-Lowry base is any species that can accept a proton (\( H^+ \)). This characteristic is what we observe in the base ionization reactions. When identifying Bronsted-Lowry bases, watch for the species that grabs a proton from water. For instance:
  • The monohydrogen phosphate ion (\( HPO_4^{2-} \)) acts as a base, taking a proton from water to become dihydrogen phosphate (\( H_2PO_4^- \)).
  • The benzoate ion (\( C_6H_5CO_2^- \)) similarly accepts a proton forming benzoic acid (\( C_6H_5CO_2H \)).

These reactions exemplify the essential role of bases in accepting protons and the dynamic nature of chemical reactions.
Equilibrium Expressions
In chemistry, equilibrium expressions quantify the balance between reactants and products in a reversible chemical reaction. For base ionization, this expression is known as the base ionization constant (\( K_b \)), which represents how well a base disassociates in water. Consider the ionization of propylamine:
  • The equilibrium expression for propylamine when balanced is given by:\[ K_b = \frac{[\text{C}_3\text{H}_7\text{NH}_3^+][OH^-]}{[\text{C}_3\text{H}_7\text{NH}_2]} \]
  • For monohydrogen phosphate:\[ K_b = \frac{[H_2PO_4^-][OH^-]}{[HPO_4^{2-}]} \]
  • And for benzoate:\[ K_b = \frac{[C_6H_5CO_2H][OH^-]}{[C_6H_5CO_2^-]} \]

These expressions allow chemists to calculate equilibrium concentrations, offering insights into the strength of a base in water.
Conjugate Acid
The concept of a conjugate acid is crucial when discussing Bronsted-Lowry bases. A conjugate acid forms when a base gains a proton. This means, within the base ionization reactions, the conjugate acid is the direct product from the base accepting a proton. For instance:
  • In the reaction involving propylamine, the conjugate acid formed is the propylammonium ion (\( \text{C}_3\text{H}_7\text{NH}_3^+ \)).
  • For the monohydrogen phosphate, the conjugate acid is dihydrogen phosphate (\( H_2PO_4^- \)).
  • The benzoate ion forms benzoic acid (\( C_6H_5CO_2H \)) as its conjugate acid.

Understanding conjugate acids is key to mastering acid-base reactions, providing clarity on the nature of acid-base interactions in various chemical processes.

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Most popular questions from this chapter

A solution is made by adding \(1.000 \mathrm{~g} \mathrm{Ca}(\mathrm{OH})_{2}(s), 100.0 \mathrm{~mL}\) of \(0.10 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4}\), and enough water to make a final volume of \(350.0 \mathrm{~mL}\). Assuming that all of the solid dissolves, what is the \(\mathrm{pH}\) of the final solution?

The acid-dissociation constant for chlorous acid \(\left(\mathrm{HClO}_{2}\right)\) is \(1.1 \times 10^{-2} .\) Calculate the concentrations of \(\mathrm{H}_{3} \mathrm{O}^{+}, \mathrm{ClO}_{2}^{-}\), and \(\mathrm{HClO}_{2}\) at equilibrium if the initial concentration of \(\mathrm{HClO}_{2}\) is \(0.0200 \mathrm{M}\)

Calculate the molar concentration of \(\mathrm{OH}^{-}\) in a \(0.050 \mathrm{M}\) solution of ethylamine \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{2} ; K_{b}=6.4 \times 10^{-4}\right) .\) Calculate the \(\mathrm{pH}\) of this solution.

At the boiling point of water \(\left(100^{\circ} \mathrm{C}\right), K_{w}=5.6 \times 10^{-13} .\) Calculate \(\left[\mathrm{H}^{+}\right]\) and \(\left[\mathrm{OH}^{-}\right]\) for a neutral solution at this temperature.

A particular sample of vinegar has a pH of 2.20 . If acetic acid is the only acid that vinegar contains \(\left(K_{a}=1.8 \times 10^{-5}\right)\), calculate the concentration of acetic acid in the vinegar.
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