Question

Base Solutions Write equations showing how each weak base ionizes water to form \(\mathrm{OH}^{-} .\) Also write the corresponding expression for \(K_{\mathrm{b}} .\) $$\begin{array}{l}{\text { a. } \mathrm{CO}_{3}^{2-}} \\ {\text { b. } \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}} \\ {\text { c. } \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{2}}\end{array}$$

Short answer

a. Ionization: \(\mathrm{CO}_3^{2-} + \mathrm{H}_2O \rightarrow \mathrm{HCO}_3^- + \mathrm{OH}^-\), \(K_{\mathrm{b}}\) Expression: \(K_{\mathrm{b}} = \frac{[\mathrm{HCO}_3^-][\mathrm{OH}^-]}{[\mathrm{CO}_3^{2-}]}\). b. Ionization: \(\mathrm{C}_6\mathrm{H}_5\mathrm{NH}_2 + \mathrm{H}_2O \rightarrow \mathrm{C}_6\mathrm{H}_5\mathrm{NH}_3^+ + \mathrm{OH}^-\), \(K_{\mathrm{b}}\) Expression: \(K_{\mathrm{b}} = \frac{[\mathrm{C}_6\mathrm{H}_5\mathrm{NH}_3^+][\mathrm{OH}^-]}{[\mathrm{C}_6\mathrm{H}_5\mathrm{NH}_2]}\). c. Ionization: \(\mathrm{C}_2\mathrm{H}_5\mathrm{NH}_2 + \mathrm{H}_2O \rightarrow \mathrm{C}_2\mathrm{H}_5\mathrm{NH}_3^+ + \mathrm{OH}^-\), \(K_{\mathrm{b}}\) Expression: \(K_{\mathrm{b}} = \frac{[\mathrm{C}_2\mathrm{H}_5\mathrm{NH}_3^+][\mathrm{OH}^-]}{[\mathrm{C}_2\mathrm{H}_5\mathrm{NH}_2]}\).

Step-by-step solution

Ionization of Carbonate Ion

To write the ionization equation of carbonate ion, \(\mathrm{CO}_3^{2-}\), in water, pair it with water \(\mathrm{H}_2O\) to produce hydroxide ions \(\mathrm{OH}^-\) and hydrogen carbonate ion \(\mathrm{HCO}_3^-\): \[\mathrm{CO}_3^{2-} + \mathrm{H}_2O \rightarrow \mathrm{HCO}_3^- + \mathrm{OH}^-\]

Expression for the Equilibrium Constant for Carbonate Ion

Now, write the expression for the base ionization constant \(K_{\mathrm{b}}\) for the carbonate ion: \[K_{\mathrm{b}} = \frac{[\mathrm{HCO}_3^-][\mathrm{OH}^-]}{[\mathrm{CO}_3^{2-}]}\]

Ionization of Aniline

For aniline \(\mathrm{C}_6\mathrm{H}_5\mathrm{NH}_2\), the ionization in water is written as: \[\mathrm{C}_6\mathrm{H}_5\mathrm{NH}_2 + \mathrm{H}_2O \rightarrow \mathrm{C}_6\mathrm{H}_5\mathrm{NH}_3^+ + \mathrm{OH}^-\]

Expression for the Equilibrium Constant for Aniline

Write the base ionization constant expression for aniline: \[K_{\mathrm{b}} = \frac{[\mathrm{C}_6\mathrm{H}_5\mathrm{NH}_3^+][\mathrm{OH}^-]}{[\mathrm{C}_6\mathrm{H}_5\mathrm{NH}_2]}\]

Ionization of Ethylamine

Ethylamine \(\mathrm{C}_2\mathrm{H}_5\mathrm{NH}_2\) ionizes in water to form: \[\mathrm{C}_2\mathrm{H}_5\mathrm{NH}_2 + \mathrm{H}_2O \rightarrow \mathrm{C}_2\mathrm{H}_5\mathrm{NH}_3^+ + \mathrm{OH}^-\]

Expression for the Equilibrium Constant for Ethylamine

The base ionization constant expression for ethylamine is: \[K_{\mathrm{b}} = \frac{[\mathrm{C}_2\mathrm{H}_5\mathrm{NH}_3^+][\mathrm{OH}^-]}{[\mathrm{C}_2\mathrm{H}_5\mathrm{NH}_2]}\]

Key concepts

Weak Base Ionization

Understanding the behavior of weak bases in aqueous solutions is key in the study of acid-base chemistry. Weak bases are substances that do not completely ionize in water, instead establishing an equilibrium between the base and its ionized form. When a weak base like aniline \(\mathrm{C}_6\mathrm{H}_5\mathrm{NH}_2\) is added to water, it reacts to a small extent with water molecules to produce hydroxide ions \(\mathrm{OH}^-\) and the conjugated acid of the base \(\mathrm{C}_6\mathrm{H}_5\mathrm{NH}_3^+\).

The extent of ionization is quantitatively described by the base ionization constant \(K_{\mathrm{b}}\). The value of \(K_{\mathrm{b}}\) indicates the strength of the base; the larger the value, the stronger the base. For example, the ionization of ethylamine, \(\mathrm{C}_2\mathrm{H}_5\mathrm{NH}_2\), occurs as follows:
\[\mathrm{C}_2\mathrm{H}_5\mathrm{NH}_2 + \mathrm{H}_2O \rightarrow \mathrm{C}_2\mathrm{H}_5\mathrm{NH}_3^+ + \mathrm{OH}^-\]
This equilibrium is vital for predicting the pH of the solution and understanding buffer actions in biological and chemical systems.

Hydroxide Ion Formation

Hydroxide ion formation is a critical aspect of base chemistry. When a base dissolves in water, it increases the concentration of hydroxide ions \(\mathrm{OH}^-\), thereby making the solution basic. The process in the context of the weak base is not complete, but rather an equilibrium between the reactants and products.

The general reaction often looks like this:
\[\mathrm{Base} + \mathrm{H}_2O \leftrightarrow \mathrm{Conjugate Acid} + \mathrm{OH}^-\]
For instance, when carbonate ion \(\mathrm{CO}_3^{2-}\) reacts with water, it forms bicarbonate \(\mathrm{HCO}_3^-\) and hydroxide ions:
\[\mathrm{CO}_3^{2-} + \mathrm{H}_2O \rightarrow \mathrm{HCO}_3^- + \mathrm{OH}^-\]
Hydroxide ions play a crucial role in determining the pH of a solution, which quantifies its acidity or basicity. The presence of hydroxide ions directly affects the pH, with more hydroxide ions resulting in a higher pH value, indicating a more basic solution.

Equilibrium Constant Expression

The concept of the equilibrium constant expression is central to understanding reactions that do not go to completion. For a weak base, the equilibrium constant for its reaction with water is expressed in terms of the base ionization constant \(K_{\mathrm{b}}\). It illustrates the ratio of the concentrations of the products to the reactants, with each raised to the power of their stoichiometric coefficients.

The general form for \(K_{\mathrm{b}}\) of a weak base is given by:
\[K_{\mathrm{b}} = \frac{[\mathrm{Conjugate Acid}][\mathrm{OH}^-]}{[\mathrm{Base}]}\]
Using ethylamine as an example, the base ionization constant expression is written as follows:
\[K_{\mathrm{b}} = \frac{[\mathrm{C}_2\mathrm{H}_5\mathrm{NH}_3^+][\mathrm{OH}^-]}{[\mathrm{C}_2\mathrm{H}_5\mathrm{NH}_2]}\]
Calculating the value of \(K_{\mathrm{b}}\) can help predict the position of equilibrium and the concentrations of species in solution at equilibrium. It also allows chemists to calculate the pOH and pH of the solution, which are measures of the basicity and acidity of the solution, respectively.