Question

Consider a \(2.0-\mathrm{L}\) aqueous solution of \(4.17 \mathrm{M} \mathrm{NH}_{3}\), where \(21.0 \mathrm{~g}\) of \(\mathrm{NH}_{4} \mathrm{Cl}\) are dissolved. To this solution, \(4.8 \mathrm{~g}\) of \(\mathrm{CaCl}_{2}\) are added. (a) What is \(\left[\mathrm{OH}^{-}\right]\) before \(\mathrm{CaCl}_{2}\) is added? (b) Will a precipitate form? (c) What is \(\left[\mathrm{Ca}^{2+}\right]\) after equilibrium is established?

Short answer

What is the equilibrium concentration of calcium ions (Ca2+) after equilibrium is established? Answer: The equilibrium concentration of hydroxide ions (OH-) before adding CaCl2 is 1.85 x 10^-4 M. A precipitate will form after adding CaCl2. The equilibrium concentration of calcium ions (Ca2+) after equilibrium is established is 0.0161 M.

Step-by-step solution

Determine the concentration of OH- in the initial solution

To find the concentration of hydroxide ions (OH-) in our initial NH3 solution, we will use the equilibrium reaction of NH3 in water: NH3 (aq) + H2O (l) <=> NH4+ (aq) + OH- (aq) For this reaction, the equilibrium constant (Kb) of NH3 is 1.76 x 10^-5. Therefore, assuming the reaction can be written as 1:1, we get: Kb = ( [NH4+] * [OH-] ) / [NH3] We are given that the initial solution has 4.17 Molar NH3. With the help of an ICE (initial-change-equilibrium) table, we can set up the equilibrium concentrations of each species in the solution: Initial: [NH3] = 4.17 M, [NH4+] = 21.0 g / (53.49 g/mol) / 2.0 L = 0.196 M, [OH-] = 0 M Change: [NH3] = -x, [NH4+] = x, [OH-] = x Equilibrium: [NH3] = 4.17 - x, [NH4+] = 0.196 + x, [OH-] = x Now we can simply plug the equilibrium concentrations into the Kb expression: Kb = ( (0.196 + x) * x ) / (4.17 - x) We can approximate that x is much smaller than 4.17, so we get: ( (0.196 * x) / 4.17) =1.76 x 10^-5 x = 1.85 x 10^-4 M This is the equilibrium concentration of OH- in our solution before adding CaCl2.

Evaluate if a precipitate will form

After adding 4.8 g of CaCl2 into the solution, we need to check if Ca(OH)2 will form a precipitate or not. The solubility product constant (Ksp) for Ca(OH)2 is 5.5 × 10-6. We have the dissolution reaction: Ca(OH)2 (s) <=> Ca2+ (aq) + 2OH- (aq) Now, we can calculate the initial concentration of Ca2+ ions in the solution by dividing the amount of CaCl2 added by the molarity of the solution: Initial [Ca2+] = 4.8 g / (110.98 g/mol) / 2.0 L = 0.0216 M Next, we use the solubility product expression to write a simple formula to calculate the ion product (Qsp) of [Ca2+] and [OH-]: Qsp = [Ca2+] * [OH-]^2 Qsp = (0.0216) * (1.85 x 10^-4)^2 = 7.46 x 10^-9 As the ion product (Qsp) is larger than the solubility product (Ksp): 7.46 x 10^-9 > 5.5 × 10-6 Since the Qsp is greater than the Ksp, it means that the solution is supersaturated, and a precipitate will form.

Determine the equilibrium concentration of Ca2+

Since a precipitate forms, we need to find the equilibrium concentration of Ca2+ ions in the solution. We can use the Ksp expression for Ca(OH)2: Ksp = [Ca2+] * [OH-]^2 We already know the Ksp of Ca(OH)2 and the equilibrium concentration of OH- ions: 5.5 × 10-6 = [Ca2+] * (1.85 x 10^-4)^2 Now, we just need to solve for [Ca2+]: [Ca2+] = 5.5 × 10-6 / (1.85 x 10^-4)^2 = 0.0161 M Thus, the equilibrium concentration of Ca2+ ions in the solution after a precipitate has formed is 0.0161M.

Key concepts

Equilibrium Constant

The equilibrium constant, represented as K (Kc for concentrations, Kp for pressures), is a fundamental concept in chemical equilibrium. It provides a quantitative measure of the extent of a reaction at equilibrium. For a generic reaction where reactants A and B form products C and D, with lowercase letters representing stoichiometric coefficients, the equilibrium constant expression is written as:
\[\begin{equation} K = \frac{[\text{C}]^c\cdot[\text{D}]^d}{[\text{A}]^a\cdot[\text{B}]^b} \end{equation}\]
In these expressions, the concentrations of the products are raised to the power of their coefficients and divided by the concentrations of the reactants, also raised to their respective coefficients. Equilibrium favors the formation of products when K is large, and favors reactants when K is small.For the given exercise, we look at the weak base ammonia (\text{NH}_3) in water. Its equilibrium constant (\text{Kb}) reflects the base's ability to accept a proton, resulting in ammonium (\text{NH}_4^+) and hydroxide (\text{OH}^-) ions. An ICE table allows us to visualize changes from initial concentrations to equilibrium concentrations, ultimately leading us to calculate the equilibrium concentration of hydroxide ions.

Solubility Product

The solubility product constant (Ksp) is a special case of the equilibrium constant applied to the solubility of sparingly soluble salts. It quantifies the maximum amount of a compound that can dissolve in solution and is defined only for saturated solutions at a given temperature.
For the dissolution of a generic compound, such as \text{AB}_2, the reaction can be represented as:
\[\begin{equation} \text{AB}_2 (s) \rightleftharpoons \text{A}^{2+} (aq) + 2\text{B}^- (aq) \end{equation}\]
\[\begin{equation} \text{Ksp} = [\text{A}^{2+}][\text{B}^-]^2 \end{equation}\]
Understanding the \text{Ksp} is crucial when predicting the formation of a precipitate in a solution. If the ion product (Qsp), the product of the molar concentrations of the ions at any point in time, exceeds Ksp, the solution is supersaturated, and precipitation occurs to re-establish equilibrium. In the exercise provided, the solubility product concept is used to determine whether or not a precipitate of calcium hydroxide (\text{Ca(OH)}_2) will form when calcium chloride (\text{CaCl}_2) is added in excess to the solution.

Precipitation Reactions

Precipitation reactions involve the formation of a solid called a precipitate from a solution when two soluble salts react. The key to predicting a precipitation reaction lies in understanding the solubility rules and calculating the ion product (Qsp) to compare it with the solubility product constant (Ksp).
A reaction leading to the formation of an insoluble product is typically represented as:
\[\begin{equation} \text{aA}^{b+} (aq) + \text{bB}^{a-} (aq) \rightarrow \text{AbBb} (s) \end{equation}\]
For example, when solutions containing calcium ions (\text{Ca}^{2+}) and hydroxide ions (\text{OH}^-) are mixed, a precipitation reaction may occur if the product of ion concentrations exceeds the Ksp for calcium hydroxide.In the context of our exercise, the addition of \text{CaCl}_2 to the ammonium chloride (\text{NH}_4\text{Cl}) and ammonia (\text{NH}_3) solution results in an increase in the calcium ion concentration. By calculating the Qsp and comparing it with the Ksp of calcium hydroxide, we predict whether a precipitate will form. In our case, precipitation is indeed observed as Qsp > Ksp, signifying that the system must shift toward the formation of a solid to reach a new equilibrium.