Question

What is \(\left[\mathrm{NH}_{4}^{+}\right]\) in a solution that contain \(0.02 \mathrm{M} \mathrm{NH}_{3}\left(K_{b}=1.8 \times 10^{-5}\right)\) and \(0.01 \mathrm{M}\) ' \(\mathrm{KOH}\) ? (a) \(9 \times 10^{-6}\) (b) \(1.8 \times 10^{-5}\) (c) \(3.6 \times 10^{-5}\) (d) None of these

Short answer

(c) \(3.6 \times 10^{-5}\)

Step-by-step solution

Write the Equilibrium Expression for Ammonia Reaction

The equilibrium expression for the reaction of ammonia with water is given by \[ \text{NH}_{3} + \text{H}_2O \rightleftharpoons \text{NH}_{4}^+ + \text{OH}^- \] This will help determine the concentration of ammonium ion, \(\text{NH}_{4}^+\).

Determine the Concentration of OH- Due to KOH

KOH is a strong base and dissociates completely in water, so the concentration of \(\text{OH}^-\) ions in the solution is equal to the initial concentration of KOH, which is \(0.01\,\text{M}\).

Write the Expression for Kb and add OH- From KOH

The equilibrium constant expression for the reaction of ammonia with water (Kb) is \[ K_{b} = \frac{\left[\text{NH}_{4}^+\right]\left[\text{OH}^-\right]}{\left[\text{NH}_{3}\right]} \] Add the \(\text{OH}^-\) concentration from KOH to the equilibrium concentration of OH- from the ammonia reaction.

Calculate [OH-] at Equilibrium and [NH4+] Using Kb

Assuming that the contribution of \(\text{OH}^-\) from the ionization of \(\text{NH}_{3}\) is small compared to the amount from KOH, then the total \(\text{OH}^-\) concentration is approximately \(0.01\,\text{M}\).Plug into the Kb expression and solve for \(\left[\text{NH}_{4}^+\right]\):\[ 1.8 \times 10^{-5} = \frac{\left[\text{NH}_{4}^+\right](0.01)}{0.02} \]Which gives \(\left[\text{NH}_{4}^+\right] = \frac{1.8 \times 10^{-5} \times 0.02}{0.01} = 3.6 \times 10^{-5}\).

Verify the Answer Against the Provided Options

After calculating the concentration of ammonium ions, compare the value to the given options. The calculated value of \(3.6 \times 10^{-5}\) matches option (c).

Key concepts

Chemical Equilibrium

Chemical equilibrium is a dynamic state where the rates of the forward and reverse reactions are equal, resulting in no net change in the concentrations of reactants and products over time. Despite being a state of 'balance', the reactions continue to occur at the molecular level but do so at the exact same rate, leading to a constant concentration of the respective molecules involved.

In the context of the exercise, ammonia (\text{NH}\(_3\)) in aqueous solution reaches equilibrium with its reaction to form ammonium ions (\text{NH}\(_4^+\)) and hydroxide ions (\text{OH}\(^-\)). Understanding equilibrium is crucial for solving problems in physical chemistry as it helps predict the quantities of substances present in a reaction mixture at any given time.

Acid-Base Equilibria

Acid-base equilibria involve the transfer of protons (H\(^+\)) between acid and base species. The balance between the products and reactants in acid-base reactions is governed by their respective equilibrium constants, Ka for acids and Kb for bases.

The ammonia solution problem requires knowledge of how bases such as \text{NH}\(_3\) (a weak base) react with water, a process which is partially reversible, hence an equilibrium state is established. The presence of KOH, a strong base that dissociates completely, affects the equilibrium by increasing the concentration of \text{OH}\(^-\) ions in the solution. Calculating the ammonium ion concentration (\text{NH}\(_4^+\)) in the presence of another base (KOH) necessitates an understanding that strong bases like KOH will shift the position of the equilibrium.

Equilibrium Constant Expressions

Equilibrium constant expressions quantitatively describe the position of equilibrium in a reaction. For the reaction of a weak base with water, the equilibrium constant (Kb) is written as the concentration of the products over the concentration of the unreacted base.

The equation is given by: \[\begin{equation}K_b = \frac{[\text{NH}_4^+][\text{OH}^-]}{[\text{NH}_3]}\end{equation}\]Having a higher concentration of \text{OH}\(^-\) due to KOH affects the number of \text{NH}\(_4^+\) ions produced at equilibrium. By utilizing the Kb expression, the equilibrium concentration of \text{NH}\(_4^+\) can be calculated, accounting for the fact that KOH contributes a significant amount of \text{OH}\(^-\) ions to the system, which is not to be ignored when performing such calculations.