Problem 39 - Advanced Problems in Physical Chemistry for Competitive Examinations

Question

What is the aqueous ammonia concentration of a solution prepared by dissolving \(0.15\) mole of \(\mathrm{NH}_{4}^{+} \mathrm{CH}_{3} \mathrm{COO}^{-}\) in 1 L of water? Given: \(K_{\text {a }}\left(\mathrm{CH}_{3} \mathrm{COOH}\right)\) \(=1.8 \times 10^{-5} ; K_{\mathrm{b}}\left(\mathrm{NH}_{4} \mathrm{OH}\right)=1.8 \times 10^{-5}\) (a) \(8.3 \times 10^{-4} \mathrm{M}\) (b) \(0.15 \mathrm{M}\) (c) \(5.52 \times 10^{-3} \mathrm{M}\) (d) \(3.8 \times 10^{-4} \mathrm{M}\)

Step-by-Step Solution

Answer

The concentration of aqueous ammonia is (c) \(5.52 \times 10^{-3} \text{M}\)

Step 1 - Write Dissociation Reactions

Ammonium acetate (\( \text{NH}_4^+CH_3COO^- \)) completely dissociates in water into ammonium ions (\( \text{NH}_4^+ \)) and acetate ions (\( \text{CH}_3COO^- \)). Additionally, ammonium ions can partially dissociate in water, acting as an acid, and acetate ions can act as a base:1. For ammonium ion as acid: \(\text{NH}_4^+(aq) + H_2O(l) \leftrightarrow \text{NH}_3(aq) + H_3O^+(aq) \) 2. For acetate ion as base: \(\text{CH}_3COO^-(aq) + H_2O(l) \leftrightarrow \text{CH}_3COOH(aq) + OH^-(aq) \)

Step 2 - Apply the Equilibrium Constant

Determine the equilibrium concentration of \(\text{NH}_3\) using the equilibrium constant for ammonium hydroxide (\(K_b = 1.8 \times 10^{-5} \) for \(\text{NH}_4OH\)). Similarly, the equilibrium concentration for acetic acid from acetate ions will be the same as that for \(\text{NH}_3\) due to molar to molar reaction and equal \(\text{K}_a\) and \(\text{K}_b\) values. \(\text{NH}_4^+\) concentration remains approximately \(\text{0.15 M}\) as it's a weak acid and does not dissociate significantly.

Step 3 - Calculate the Concentration of Hydroxide Ions

The equilibrium expression for the dissociation of \(\text{NH}_4OH\) is: \([OH^-] = \sqrt{K_b [NH_4^+]} = \sqrt{(1.8 \times 10^{-5})(0.15)}\). Calculate the concentration of hydroxide ions to find the concentration of \(\text{NH}_3\).

Step 4 - Use the Calculated Hydroxide Ion Concentration

Since for every hydroxide ion produced, one ammonia (\( \text{NH}_3 \) molecule is produced, the concentration of aqueous ammonia will be the same as that of hydroxide ions.

Step 5 - Choose the Correct Answer

Find which option matches the calculated concentration of aqueous ammonia (\( \text{NH}_3 \) and select it as the correct answer.

Key Concepts

Acid-Base Equilibrium

Understanding acid-base equilibrium is essential when working with solutions like aqueous ammonia. The principle of acid-base equilibrium revolves around the reversible reactions of acids and bases in water and how they dissociate to form ions. When we consider a weak acid, such as ammonium (NH_4^+), it partially dissociates in water to form ammonia (NH_3) and hydronium ions (H_3O^+). Similarly, a weak base like acetate (CH_3COO^-) reacts with water to produce hydroxide ions (OH^-) and acetic acid (CH_3COOH).The balance between the forward and backward reactions is described by an equilibrium constant, which in the case of ammonium acting as an acid has a Kb value, and for acetic acid as a base has a Ka value. Both constants indicate the degree of ionization in a solution. In the exercise, these two ions stem from a common salt, ammonium acetate, which dissociates completely in water. The concentration of resulting NH_3 in the solution, therefore, becomes a question of calculating how much NH_4^+ ion dissociates in the water.

Molar Concentration Calculation

Molar concentration calculation is a fundamental concept in chemistry that explains how to determine the amount of substance present in a specific volume of solution. It is often expressed in moles per liter (M). In our case, determining the concentration of aqueous ammonia involves knowing the molar amount of ammonium acetate initially dissolved and the volume of water it is dissolved in.The initial formula for molar concentration is \[ C = \frac{n}{V} \]where C is the concentration, n is the number of moles, and V is the volume in liters. In the exercise, we start with a known number of moles of ammonium acetate and a volume of one liter. As ammonium acetate dissociates completely into ammonium and acetate ions, the molar concentration of these ions initially matches that of the salt. To find the concentration of aqueous ammonia, we must consider the dissociation equilibrium of ammonium ion, as only part of it will form ammonia.

Chemical Equilibrium Expression

The chemical equilibrium expression quantitatively represents the state of chemical equilibrium. It is given by the equilibrium constant (K), which is a dimensionless quantity calculated as the ratio of the product of the concentrations of the products raised to the power of their stoichiometric coefficients to the corresponding product for the reactants under the same conditions. For the dissociation of a weak base, such as ammonia in water, this expression can be written as:\[ K_{b} = \frac{[NH_{3}][OH^{-}]}{[NH_{4}^+]} \]The square brackets indicate the equilibrium molar concentrations of the species. Using the equilibrium constant, we can determine unknown concentrations in the equilibrium mixture. In our exercise, we used Kb to find the hydroxide ion concentration resulting from the equilibrium, knowing that the initial concentration of NH_4^+ ion remains roughly the same due to its weak acidity. Once we know the [OH^-], we can quickly infer the concentration of NH_3, since they are produced in a 1:1 ratio in the equilibrium reaction.