Question

Write a balanced net ionic equation and the corresponding equilibrium equation for the reaction of the following weak bases with water: (a) Dimethylamine, \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{NH}\) (b) Aniline, \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}\) (c) Cyanide ion, CN

Short answer

(a) Dimethylamine forms \((\text{CH}_3)_2\text{NH}_2^+ + \text{OH}^-\); (b) Aniline forms \(\text{C}_6\text{H}_5\text{NH}_3^+ + \text{OH}^-\); (c) CN⁻ forms \(\text{HCN} + \text{OH}^-\).

Step-by-step solution

Recognize the Weak Base

Identify the weak base in each case:(a) Dimethylamine: \((\text{CH}_3)_2\text{NH}\)(b) Aniline: \(\text{C}_6\text{H}_5\text{NH}_2\)(c) Cyanide ion: \(\text{CN}^-\).

Write the Reaction with Water

For each weak base, set up the reaction with water, where the base accepts a proton. Water acts as an acid donating an H⁺ ion.(a) \((\text{CH}_3)_2\text{NH} + \text{H}_2\text{O} \rightleftharpoons (\text{CH}_3)_2\text{NH}_2^+ + \text{OH}^-\)(b) \(\text{C}_6\text{H}_5\text{NH}_2 + \text{H}_2\text{O} \rightleftharpoons \text{C}_6\text{H}_5\text{NH}_3^+ + \text{OH}^-\)(c) \(\text{CN}^- + \text{H}_2\text{O} \rightleftharpoons \text{HCN} + \text{OH}^-\).

Write the Net Ionic Equations

Since these reactions involve weak bases, the major players in the net ionic equation are the weak base and the produced ions:(a) \((\text{CH}_3)_2\text{NH} + \text{H}_2\text{O} \rightleftharpoons (\text{CH}_3)_2\text{NH}_2^+ + \text{OH}^-\)(b) \(\text{C}_6\text{H}_5\text{NH}_2 + \text{H}_2\text{O} \rightleftharpoons \text{C}_6\text{H}_5\text{NH}_3^+ + \text{OH}^-\)(c) \(\text{CN}^- + \text{H}_2\text{O} \rightleftharpoons \text{HCN} + \text{OH}^-\).

Write the Equilibrium Constants

For each of these weak base reactions, write the equilibrium expression in terms of concentrations:(a) \(K_b = \frac{[(\text{CH}_3)_2\text{NH}_2^+][\text{OH}^-]}{[(\text{CH}_3)_2\text{NH}]}\)(b) \(K_b = \frac{[\text{C}_6\text{H}_5\text{NH}_3^+][\text{OH}^-]}{[\text{C}_6\text{H}_5\text{NH}_2]}\)(c) \(K_b = \frac{[\text{HCN}][\text{OH}^-]}{[\text{CN}^-]}\).

Key concepts

Weak Bases

In order to understand weak bases, it's important to differentiate them from strong bases. Weak bases do not completely dissociate in aqueous solutions. Instead, they partially react with water molecules. This means that not all molecules of the base will accept protons from water; only a small fraction does. A classic example of this is ammonia (NH₃), which forms ammonium ions (NH₄⁺) and hydroxide ions (OH⁻) when dissolved in water, but not completely.
Some examples of weak bases include:

• Dimethylamine, \(( ext{CH}_3)_2 ext{NH}\)
• Aniline, \( ext{C}_6 ext{H}_5 ext{NH}_2\)
• Cyanide ion, \(\text{CN}^-\)

These bases tend to react with water in reversible reactions, which are central to understanding their behavior. The importance lies in the fact that in solutions, equilibrium is established where both the base and its conjugate acid form are present in certain ratios.

Equilibrium Expressions

When discussing reactions involving weak bases, an essential concept is chemical equilibrium. In simple terms, equilibrium is the state where the concentrations of all reactants and products remain constant over time.For weak bases, when they react with water, this interaction reaches a point where the forward reaction rate equals the reverse reaction rate.To express this equilibrium mathematically, we use the equilibrium constant (K_b) for the base.
This is a measure of the base's strength, which is defined by:\[K_b = \frac{{[\text{Conjugate Acid}][\text{OH}^-]}}{{[\text{Base}]}}\]For instance, the equilibrium expression for dimethylamine acting as a weak base is:\[K_b = \frac{{[(\text{CH}_3)_2\text{NH}_2^+][\text{OH}^-]}}{{[(\text{CH}_3)_2\text{NH}]}}\]This is a major tool in calculating the concentrations of the species involved in these reactions at equilibrium. For students, understanding how to set up equilibrium expressions is vital, as it provides a way to predict reaction extents and concentrations.

Proton Transfer Reactions

Proton transfer reactions are key to understanding acid-base chemistry. In such reactions, a proton (H^+) is transferred from an acid to a base. Within the context of weak bases, a water molecule acts as the proton donor or acid.
When a weak base interacts with water:

• The base accepts a proton from the water molecule.
• The water becomes a hydroxide ion (OH^-), contributing to the basicity of the solution.

For example, in the proton transfer reaction involving cyanide ion:\[\text{CN}^- + \text{H}_2\text{O} \rightleftharpoons \text{HCN} + \text{OH}^-\]The cyanide ion (\text{CN}^-) acts as a base, accepting a proton from the water, turning into hydrogen cyanide (\text{HCN}). This fundamental interaction underscores the dynamic nature of proton transfers and highlights the importance of considering both the reacting and produced species in acid-base chemistry.