Question

Write the equilibrium constant expression for each of the following weak bases: (a) \(\mathrm{NH}_{2} \mathrm{OH}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftarrows \mathrm{NH}_{3} \mathrm{OH}^{+}(a q)+\mathrm{OH}^{-}(a q)\) (b) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftarrows\) $$ \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{3}^{+}(a q)+\mathrm{OH}^{-}(a q) $$ (c) \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{NH}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(l) \rightleftarrows\) $$ \left(\mathrm{CH}_{3}\right)_{2} \mathrm{NH}_{2}^{+}(a q)+\mathrm{OH}^{-}(a q) $$

Short answer

(a) \( K_b = \frac{[\mathrm{NH}_{3} \mathrm{OH}^{+}][\mathrm{OH}^{-}]}{[\mathrm{NH}_{2} \mathrm{OH}]} \), (b) \( K_b = \frac{[\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{3}^{+}][\mathrm{OH}^{-}]}{[\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}]} \), (c) \( K_b = \frac{[\left( \mathrm{CH}_{3} \right)_{2} \mathrm{NH}_{2}^{+}][\mathrm{OH}^{-}]}{[\left( \mathrm{CH}_{3} \right)_{2} \mathrm{NH}]} \).

Step-by-step solution

Identifying the Reaction Components

In each given reaction, identify the weak base and the products formed from its reaction with water. For reaction (a), the weak base is \( \mathrm{NH}_{2} \mathrm{OH} \). For reaction (b), it is \( \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2} \). For reaction (c), it is \( \left( \mathrm{CH}_{3} \right)_{2} \mathrm{NH} \).

Write the General Equilibrium Constant Expression

For a generic reaction of a base \( \mathrm{B} \) with water, the products include \( \mathrm{BH}^{+} \) and \( \mathrm{OH}^{-} \). The equilibrium constant \( K_b \) is expressed as: \[ K_b = \frac{[\mathrm{BH}^{+}][\mathrm{OH}^{-}]}{[\mathrm{B}]} \] Water is not included in the expression as it is a pure liquid.

Apply the Equilibrium Expression to Reaction (a)

For \( \mathrm{NH}_{2} \mathrm{OH} \): determine the products to be \( \mathrm{NH}_{3} \mathrm{OH}^{+} \) and \( \mathrm{OH}^{-} \). Thus, the expression is: \[ K_b = \frac{[\mathrm{NH}_{3} \mathrm{OH}^{+}][\mathrm{OH}^{-}]}{[\mathrm{NH}_{2} \mathrm{OH}]} \]

Apply the Equilibrium Expression to Reaction (b)

For \( \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2} \): determine the products to be \( \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{3}^{+} \) and \( \mathrm{OH}^{-} \). Thus, the expression is: \[ K_b = \frac{[\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{3}^{+}][\mathrm{OH}^{-}]}{[\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}]} \]

Apply the Equilibrium Expression to Reaction (c)

For \( \left( \mathrm{CH}_{3} \right)_{2} \mathrm{NH} \): determine the products to be \( \left( \mathrm{CH}_{3} \right)_{2} \mathrm{NH}_{2}^{+} \) and \( \mathrm{OH}^{-} \). Thus, the expression is: \[ K_b = \frac{[\left( \mathrm{CH}_{3} \right)_{2} \mathrm{NH}_{2}^{+}][\mathrm{OH}^{-}]}{[\left( \mathrm{CH}_{3} \right)_{2} \mathrm{NH}]} \]

Combine Expressions

Write down all individual equilibrium constant expressions from the previous steps: - (a) \( K_b = \frac{[\mathrm{NH}_{3} \mathrm{OH}^{+}][\mathrm{OH}^{-}]}{[\mathrm{NH}_{2} \mathrm{OH}]} \)- (b) \( K_b = \frac{[\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{3}^{+}][\mathrm{OH}^{-}]}{[\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}]} \)- (c) \( K_b = \frac{[\left( \mathrm{CH}_{3} \right)_{2} \mathrm{NH}_{2}^{+}][\mathrm{OH}^{-}]}{[\left( \mathrm{CH}_{3} \right)_{2} \mathrm{NH}]} \)

Key concepts

Understanding Weak Bases

In chemistry, weak bases are substances that do not completely dissociate in water. They partially react with water to form hydroxide ions (\(\mathrm{OH}^-\)) and the corresponding conjugate acid. Unlike strong bases, weak bases do not fully ionize, which means only a fraction of the base molecules react with water. This characteristic makes them interesting for studying equilibrium dynamics.

For instance, in the reaction of hydroxylamine (\(\mathrm{NH}_{2} \mathrm{OH}\)) and water:

• Hydroxylamine acts as a weak base.
• It partially converts into \(\mathrm{NH}_{3} \mathrm{OH}^{+}\) and \(\mathrm{OH}^-\).

The equilibrium constant expression (\(K_b\)) for each weak base helps us understand how readily the base generates hydroxide ions in solution. The bigger the \(K_b\), the stronger the base, but since weak bases have small \(K_b\) values, they do not produce a lot of \(\mathrm{OH}^-\) ions in their solutions.

Chemical Equilibrium in Weak Base Reactions

When dealing with chemical reactions of weak bases in water, it is important to understand the concept of chemical equilibrium. Equilibrium refers to a state where the rate of the forward reaction equals the rate of the backward reaction.

In a weak base reaction:

• The forward reaction involves the base combining with water to form hydroxide ions and the conjugate acid.
• In the backward reaction, these products reconvert to the original base and water.

This balance is dynamic, meaning that even though the concentrations of reactants and products remain constant, the molecules still continuously react with each other.

The equilibrium constant (\(K_b\)) gives insight into this balance and the extent to which a weak base dissociates in water. It is calculated using the concentrations of the products and the reactant at equilibrium:\[K_b = \frac{[\mathrm{Conjugate\ Acid}][\mathrm{OH}^-]}{[\mathrm{Weak\ Base}]}\]This consistent equation helps predict how changes in concentration or conditions affect the equilibrium position of acid-base reactions.

Understanding Acid-Base Reactions

Acid-base reactions are fundamental chemical reactions that involve the transfer of protons (H+) between reactants. Within this context, dealing with weak bases involves their interaction with water, showcasing unique features of acid-base chemistry.

In such a reaction, a proton is transferred from the water molecule to the base:

• The water acts as an acid, donating a proton.
• The weak base accepts the proton, forming its conjugate acid.
• This process increases the concentration of \(\mathrm{OH}^-\) ions, making the solution basic.

An example, such as the interaction of \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}\) with water, highlights the equilibrium set up between reactants and products, as they form \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{3}^{+} \)+ and \(\mathrm{OH}^-\).These concepts are key because understanding them allows prediction of factors such as pH changes, reaction direction when stress is applied, or how buffers work.