Question

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

Short answer

a. Ammonia ionizes as \(\mathrm{NH}_{3} + \mathrm{H}_{2}\mathrm{O} \rightleftharpoons \mathrm{NH}_{4}^{+} + \mathrm{OH}^{-}\) with \(K_{\mathrm{b}} = \frac{[\mathrm{NH}_{4}^{+}][\mathrm{OH}^{-}]}{[\mathrm{NH}_{3}]}\). b. Bicarbonate ionizes as \(\mathrm{HCO}_{3}^{-} + \mathrm{H}_{2}\mathrm{O} \rightleftharpoons \mathrm{H}_{2}\mathrm{CO}_{3} + \mathrm{OH}^{-}\) with \(K_{\mathrm{b}} = \frac{[\mathrm{H}_{2}\mathrm{CO}_{3}][\mathrm{OH}^{-}]}{[\mathrm{HCO}_{3}^{-}]}\). c. Methylamine ionizes as \(\mathrm{CH}_{3}\mathrm{NH}_{2} + \mathrm{H}_{2}\mathrm{O} \rightleftharpoons \mathrm{CH}_{3}\mathrm{NH}_{3}^{+} + \mathrm{OH}^{-}\) with \(K_{\mathrm{b}} = \frac{[\mathrm{CH}_{3}\mathrm{NH}_{3}^{+}][\mathrm{OH}^{-}]}{[\mathrm{CH}_{3}\mathrm{NH}_{2}]}\).

Step-by-step solution

Writing the Ionization Equations for the Bases

For each base, write an equation showing how it accepts a hydrogen ion from water, producing hydroxide ions and the conjugate acid. For ammonia, \(\mathrm{NH}_3 + \mathrm{H}_2\mathrm{O} \rightleftharpoons \mathrm{NH}_4^{+} + \mathrm{OH}^{-}\). For bicarbonate, \(\mathrm{HCO}_3^{-} + \mathrm{H}_2\mathrm{O} \rightleftharpoons \mathrm{H}_2\mathrm{CO}_3 + \mathrm{OH}^{-}\). And for methylamine, \(\mathrm{CH}_3\mathrm{NH}_2 + \mathrm{H}_2\mathrm{O} \rightleftharpoons \mathrm{CH}_3\mathrm{NH}_3^{+} + \mathrm{OH}^{-}\).

Writing the Kb Expression for Each Base

The expression for the base dissociation constant (\(K_{\mathrm{b}}\)) can be derived from the ionization equation. For \(K_{\mathrm{b}}\) of ammonia, the expression is \(K_{\mathrm{b}} = \frac{[\mathrm{NH}_4^{+}][\mathrm{OH}^{-}]}{[\mathrm{NH}_3]}\). For bicarbonate, it's \(K_{\mathrm{b}} = \frac{[\mathrm{H}_2\mathrm{CO}_3][\mathrm{OH}^{-}]}{[\mathrm{HCO}_3^{-}]}\). And for methylamine, the expression is \(K_{\mathrm{b}} = \frac{[\mathrm{CH}_3\mathrm{NH}_3^{+}][\mathrm{OH}^{-}]}{[\mathrm{CH}_3\mathrm{NH}_2]}\).

Key concepts

Base Ionization in Water

When a base dissolves in water, it undergoes a chemical reaction that increases the concentration of hydroxide ions \(\mathrm{OH}^{-}\) in the solution. This process is commonly referred to as base ionization in water. Weak bases, such as ammonia \(\mathrm{NH}_3\), bicarbonate \(\mathrm{HCO}_3^{-}\), and methylamine \(\mathrm{CH}_3\mathrm{NH}_2\), do not completely dissociate in water. Instead, they partially accept protons from water molecules, producing hydroxide ions \(\mathrm{OH}^{-}\) and a conjugate acid.

For instance, ammonia ionizes in water as follows:
\[\mathrm{NH}_3 + \mathrm{H}_2\mathrm{O} \rightleftharpoons \mathrm{NH}_4^{+} + \mathrm{OH}^{-}\]
Explaining this process clearly ensures students understand that during ionization, water acts as an acid by donating a proton to the weak base, converting it into its conjugate acid and forming hydroxide ions in the process.

Hydroxide Ion Formation

Hydroxide ions \(\mathrm{OH}^{-}\) are fundamental to the concept of base ionization. They are formed when a base accepts a proton from a water molecule, which then leaves behind the hydroxide ions. The presence of hydroxide ions is responsible for the basic nature of the solution.

Example of Hydroxide Ion Formation

Using the earlier example of ammonia in water, you can see that hydroxide ions are formed as:\[\mathrm{NH}_3 + \mathrm{H}_2\mathrm{O} \rightleftharpoons \mathrm{NH}_4^{+} + \mathrm{OH}^{-}\]
It is crucial for students to grasp that the increase in \(\mathrm{OH}^{-}\) concentration is directly linked to the solution's basicity—a higher concentration of hydroxide ions means that the solution is more basic.

Base Dissociation Constant (Kb)

The base dissociation constant, symbolized as \(K_{\mathrm{b}}\), quantifies the extent of ionization of a base in water. It's an equilibrium constant that provides insight into the base's strength: the higher the \(K_{\mathrm{b}}\), the stronger the base. For all bases, \(K_{\mathrm{b}}\) is calculated using the concentrations of the products (conjugate acid and \(\mathrm{OH}^{-}\)) divided by the concentration of the base.

Determining \(K_{\mathrm{b}}\) for Ammonia

For ammonia, the \(K_{\mathrm{b}}\) expression is formed from the ionization equation:\[K_{\mathrm{b}} = \frac{[\mathrm{NH}_4^{+}][\mathrm{OH}^{-}]}{[\mathrm{NH}_3]}\]
By including expressions for \(K_{\mathrm{b}}\) in the learning material, students can practice calculating the strength of different bases based on their respective \(K_{\mathrm{b}}\) values and therefore predict and compare their behaviors in aqueous solutions.