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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) Propylamine: \(\mathrm{C}_{3} \mathrm{H}_{7} \mathrm{NH}_{2} + \mathrm{H}_{2}\mathrm{O} \rightleftharpoons \mathrm{C}_{3} \mathrm{H}_{7} \mathrm{NH}_{3}^+ + \mathrm{OH}^-\) \(K_b = [\mathrm{C}_{3} \mathrm{H}_{7} \mathrm{NH}_{3}^+][\mathrm{OH}^-] / [\mathrm{C}_{3} \mathrm{H}_{7}\mathrm{NH}_{2}]\) (b) Monohydrogen phosphate ion: \(\mathrm{HPO}_{4}^{2-} + \mathrm{H}_{2}\mathrm{O} \rightleftharpoons \mathrm{H}_{2}\mathrm{PO}_{4}^- + \mathrm{OH}^-\) \(K_b = [\mathrm{H}_{2}\mathrm{PO}_{4}^-][\mathrm{OH}^-] / [\mathrm{HPO}_{4}^{2-}]\) (c) Benzoate ion: \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2}^- + \mathrm{H}_{2}\mathrm{O} \rightleftharpoons \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOH} + \mathrm{OH}^-\) \(K_b = [\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOH}][\mathrm{OH}^-] / [\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2}^-]\)

Step by step solution

01

Write the reaction

Propylamine reacts with water as follows: \(\mathrm{C}_{3} \mathrm{H}_{7} \mathrm{NH}_{2} + \mathrm{H}_{2}\mathrm{O} \rightleftharpoons \mathrm{C}_{3} \mathrm{H}_{7} \mathrm{NH}_{3}^+ + \mathrm{OH}^-\)
02

Write the equilibrium expression

\[ [\mathrm{C}_{3} \mathrm{H}_{7} \mathrm{NH}_{3}^+][\mathrm{OH}^-] / [\mathrm{C}_{3} \mathrm{H}_{7}\mathrm{NH}_{2}][\mathrm{H}_{2}\mathrm{O}] \]
03

Write the \(K_b\) expression

Since the concentration of water is considered constant, we can write: \[ K_b = [\mathrm{C}_{3} \mathrm{H}_{7} \mathrm{NH}_{3}^+][\mathrm{OH}^-] / [\mathrm{C}_{3} \mathrm{H}_{7}\mathrm{NH}_{2}] \] (b) Monohydrogen phosphate ion, \(\mathrm{HPO}_{4}^{2-}\)
04

Write the reaction

Monohydrogen phosphate ion reacts with water as follows: \(\mathrm{HPO}_{4}^{2-} + \mathrm{H}_{2}\mathrm{O} \rightleftharpoons \mathrm{H}_{2}\mathrm{PO}_{4}^- + \mathrm{OH}^-\)
05

Write the equilibrium expression

\[ [\mathrm{H}_{2}\mathrm{PO}_{4}^-][\mathrm{OH}^-] / [\mathrm{HPO}_{4}^{2-}][\mathrm{H}_{2}\mathrm{O}] \]
06

Write the \(K_b\) expression

Since the concentration of water is considered constant, we can write: \[ K_b = [\mathrm{H}_{2}\mathrm{PO}_{4}^-][\mathrm{OH}^-] / [\mathrm{HPO}_{4}^{2-}] \] (c) Benzoate ion, \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2}^{-}\)
07

Write the reaction

Benzoate ion reacts with water as follows: \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2}^- + \mathrm{H}_{2}\mathrm{O} \rightleftharpoons \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOH} + \mathrm{OH}^-\)
08

Write the equilibrium expression

\[ [\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOH}][\mathrm{OH}^-] / [\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2}^-][\mathrm{H}_{2}\mathrm{O}] \]
09

Write the \(K_b\) expression

Since the concentration of water is considered constant, we can write: \[ K_b = [\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOH}][\mathrm{OH}^-] / [\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2}^-] \]

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Key Concepts

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

Chemical Equation
Understanding a chemical equation is essential in the study of chemistry. It depicts a chemical reaction where reactants transform into products. A chemical equation is balanced when the number of atoms for each element is the same on both the reactant and product sides.

For a base reacting with water, the general formula is Base + H2O ⇌ Conjugate Acid + OH−. In the case of propylamine (C3H7NH2), the chemical equation with water would be:
C3H7NH2 + H2O ⇌ C3H7NH3+ + OH−.

This shows how propylamine accepts a proton from water, becoming its conjugate acid, propylammonium ion (C3H7NH3+), and producing hydroxide ions.
Kb Expression
The equilibrium constant for a base, designated as Kb, quantifies a base's strength in water. It is calculated from the concentrations of the products divided by the concentration of the reactants. The stronger the base, the higher the Kb value.

For a base like propylamine, the Kb expression is formulated without water since its concentration is constant and does not affect the equilibrium in a dilute solution. Thus, the Kb for propylamine becomes:
Kb = [C3H7NH3+][OH−] / [C3H7NH2]

Remember, a comprehensible Kb expression is key to understanding how the base interacts with water and forms its conjugate acid and hydroxide ions.
Base Reaction with Water
Bases react with water in a process known as hydrolysis. This involves the base accepting a hydrogen ion (H+) from water, forming the conjugate acid of the base and hydroxide ions (OH−).

An example is the benzoate ion (C6H5CO2−), which reacts with water to form benzoic acid (C6H5COOH) and hydroxide ions:
C6H5CO2− + H2O ⇌ C6H5COOH + OH−

This interaction shows the base's ability to raise the pH of a solution by producing hydroxide ions.
Equilibrium Expression
An equilibrium expression represents the ratio of product concentrations to reactant concentrations at equilibrium. It is related to the nature of reversible reactions, where the forward and reverse reactions occur at equal rates.

For a base like monohydrogen phosphate ion (HPO42−), the equilibrium expression is:
[H2PO4−][OH−] / [HPO42−]

Importance of Water Concentration

In dilute aqueous solutions, water's concentration remains effectively constant, thus it is typically omitted from the expression. Ultimately, understanding this concept is vital for predicting the position of equilibrium and calculating the pH of the solution.

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Most popular questions from this chapter

(a) Give the conjugate base of the following Bronsted-Lowry acids: (i) \(\mathrm{HIO}_{3},(\mathbf{i} \mathbf{i}) \mathrm{NH}_{4}^{+} .(\mathbf{b})\) Give the conjugate acid of the following Bronsted-Lowry bases: (i) \(\mathrm{O}^{2-},(\mathbf{i} \mathbf{i}) \mathrm{H}_{2} \mathrm{PO}_{4}^{-}\)

Addition of the indicator methyl orange to an unknown solution leads to a yellow color. The addition of bromthymol blue to the same solution also leads to a yellow color. (a) Is the solution acidic, neutral, or basic? (b) What is the range (in whole numbers) of possible pH values for the solution? (c) Is there another indicator you could use to narrow the range of possible pH values for the solution?

Deuterium oxide (\(\mathrm{D}_{2} \mathrm{O},\) where \(\mathrm{D}\) is deuterium, the hydrogen-2 isotope) has an ion-product constant, \(K_{w}\) , of \(8.9 \times 10^{-16}\) at \(20^{\circ} \mathrm{C}\). Calculate \(\left[\mathrm{D}^{+}\right]\) and \(\left[\mathrm{OD}^{-}\right]\) for pure (neutral) \(\mathrm{D}_{2} \mathrm{O}\) at this temperature.

Consider two solutions, solution \(\mathrm{A}\) and solution \(\mathrm{B} .\left[\mathrm{H}^{+}\right]\) in solution \(\mathrm{A}\) is 250 times greater than that in solution B. What is the difference in the pH values of the two solutions?

The average \(\mathrm{pH}\) of normal arterial blood is \(7.40 .\) At normal body temperature \(\left(37^{\circ} \mathrm{C}\right), K_{w}=2.4 \times 10^{-14} .\) Calculate \(\left[\mathrm{H}^{+}\right],\left[\mathrm{OH}^{-}\right],\) and \(\mathrm{pOH}\) for blood at this temperature.
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