Problem 101 - Chemistry An Atoms First Approach

Question

a. Calculate the molar solubility of AgBr in pure water. \(K_{\mathrm{sp}}\) for AgBr is \(5.0 \times 10^{-13}\). b. Calculate the molar solubility of \(\mathrm{AgBr}\) in \(3.0\) \(M\) \(\mathrm{NH}_{3} .\) The overall formation constant for \(\mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}^{+}\) is \(1.7 \times 10^{7}\) that is, $$\mathrm{Ag}^{+}(a q)+2 \mathrm{NH}_{3}(a q) \longrightarrow \mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}^{+}(a q) \quad K=1.7 \times 10^{7}$$ c. Compare the calculated solubilities from parts a and b. Explain any differences. d. What mass of AgBr will dissolve in \(250.0 \mathrm{mL}\) of \(3.0 M \mathrm{NH}_{3} ?\) e. What effect does adding \(\mathrm{HNO}_{3}\) have on the solubilities calculated in parts a and b?

Step-by-Step Solution

Answer

The molar solubility of AgBr in pure water is approximately 7.1x10⁻⁷ M, while its molar solubility in a 3.0 M ammonia solution is about 1.33x10⁻¹⁰ M, which is higher due to the formation of the complex ion Ag(NH₃)₂⁺. The mass of AgBr that will dissolve in 250.0 mL of 3.0 M NH₃ is approximately 6.25x10⁻¹⁰ g. Adding HNO₃ will not affect the solubility in pure water but will reduce the solubility of AgBr in ammonia solution as it reacts with NH₃ and subsequently decreases the concentration of the complex ion.

Step 1: Write the balanced dissociation reaction for AgBr in water.

AgBr (s) ⇌ Ag⁺ (aq) + Br⁻ (aq)

Step 2: Set up the ICE table

Since the Ksp expression is [Ag⁺][Br⁻], we will use the ICE table to find the equilibrium concentrations of Ag⁺ and Br⁻. Initial: [Ag⁺] = [Br⁻] = 0 Change: +s +s Equilibrium: [Ag⁺] = [Br⁻] = s

Step 3: Calculate the molar solubility, s, using Ksp.

Ksp = [Ag⁺][Br⁻] = (s)(s) = 5.0x10⁻¹³ (given) s² = 5.0x10⁻¹³ s (molar solubility) = √(5.0x10⁻¹³) ≈ 7.1x10⁻⁷ M b. Calculate the molar solubility of AgBr in 3.0 M NH₃.

Step 1: Write the balanced reaction for complex formation

Ag⁺ (aq) + 2NH₃ (aq) ⇌ Ag(NH₃)₂⁺ (aq), K = 1.7x10⁷ (given)

Step 2: Set up the ICE table

Since the K expression for complex formation is [Ag(NH₃)₂⁺]/([Ag⁺][NH₃]²), we will use the ICE table to find the equilibrium concentrations. Initial: [Ag⁺] - from part a = 7.1x10⁻⁷ M [NH₃] = 3.0 M (given) [Ag(NH₃)₂⁺] = 0 Change: [-7.1x10⁻⁷] -2x(+7.1x10⁻⁷) +7.1x10⁻⁷ Equilibrium: [0] [2.994] [7.1x10⁻⁷]

Step 3: Calculate the molar solubility of AgBr using K

K = [Ag(NH₃)₂⁺]/([Ag⁺][NH₃]²) = 1.7x10⁷ Solving for [Ag⁺], we obtain: [Ag⁺] = [Ag(NH₃)₂⁺]/(1.7x10⁷ * (NH₃)²) = 7.1x10⁻⁷/(1.7x10⁷ * 2.994²) ≈ 1.33x10⁻¹⁰ M c. Compare the calculated solubilities from parts a and b

Step 7: In part a, the molar solubility of AgBr in pure water was 7.1x10⁻⁷ M, while in part b, the molar solubility in 3.0 M ammonia solution was 1.33x10⁻¹⁰ M. The solubility of AgBr is higher in the ammonia solution due to the formation of a complex ion (Ag(NH₃)₂⁺) which reduces the concentration of free Ag⁺ ions, driving the reaction to the right and dissolving more of the AgBr solid. d. What mass of AgBr will dissolve in 250.0 mL of 3.0 M NH₃?

Step 1: Calculate the moles of AgBr that will dissolve

Step 8: Moles of AgBr = Molar solubility * Volume = (1.33x10⁻¹⁰ M)(0.250 L) ≈ 3.33x10⁻¹² mol

Step 2: Calculate the mass of AgBr using its molar mass

Step 9: Mass of AgBr = moles * molar mass = (3.33x10⁻¹² mol) * (187.77 g/mol; molar mass of AgBr) ≈ 6.25x10⁻¹⁰ g e. What effect does adding HNO₃ have on the solubilities calculated in parts a and b?

Adding HNO₃ (a strong acid) will increase the concentration of H⁺ ions in the solution. In part a (pure water), adding HNO₃ will not affect the AgBr solubility significantly as there is no equilibrium involving H⁺ ions in the dissociation of AgBr. In part b (ammonia solution), however, adding HNO₃ will react with NH₃ and form NH₄⁺ ions, thus reducing the concentration of NH₃ available to form the complex ion Ag(NH₃)₂⁺. This will shift the equilibrium of complex formation to the left, increasing the concentration of Ag⁺ ions and, consequently, reducing the solubility of AgBr.

Key Concepts

Ksp (Solubility Product Constant)

Understanding the Solubility Product Constant, or Ksp, is key to predicting the solubility of ionic compounds in water. Ksp is an equilibrium constant specific to the dissolution of solids. It represents the level at which a solute's ions are saturated in solution, beyond which any additional solute will precipitate.

Taking the example of AgBr dissolving in water, where AgBr (s) ⇌ Ag⁺ (aq) + Br⁻ (aq), we see that the Ksp equation becomes Ksp = [Ag⁺][Br⁻]. Here, the concentrations are those at equilibrium, where the rates of dissolution and precipitation are equal.

In practice, if you know the Ksp and the stoichiometry of the dissolution, you can calculate the molar solubility of the compound. For AgBr, with a given Ksp of 5.0 x 10⁻¹³, the molar solubility is the square root of the Ksp, assuming the dissolution produces equal molar amounts of Ag⁺ and Br⁻. The lower the Ksp value, the lower the solubility of the compound in water.

Complex Ion Formation

Moving beyond simple solubility, complex ion formation can significantly alter solubility patterns. Complex ions are species formed from a metal ion and one or more ligands, which are molecules or ions that can donate an electron pair to the metal ion.When a metal ion in solution, such as Ag⁺, encounters ligands like NH₃, they can form a complex ion, in this case Ag(NH₃)₂⁺. The formation of such complex ions is highly relevant because it can greatly increase the solubility of an original compound. A new equilibrium constant, called the formation constant (Kf), is used to represent the stability of the complex ion.In the presence of excess NH₃, the reaction Ag⁺ (aq) + 2NH₃ (aq) ⇌ Ag(NH₃)₂⁺ (aq) has its own equilibrium constant, Kf, which is significantly larger than Ksp. The effect is a larger dissolution of AgBr than would occur without NH₃, due to the formation of the stable Ag(NH₃)₂⁺ complex.

ICE Table Method

To quantify the dynamic between solute and solvent, or complex ion formation, chemists use an ICE table method—standing for Initial, Change, and Equilibrium. It’s a systematic way to organize the concentrations of reactants and products in an equilibrium reaction. For the solubility of AgBr in water, we begin with the concentrations of Ag⁺ and Br⁻ as zero (Initial). When AgBr starts dissolving (Change), both Ag⁺ and Br⁻ increase by s, where s is the solubility of AgBr. At Equilibrium, the concentrations are equal to s. Then, applying the Ksp expression allows us to solve for s.When considering complex ion formation, the ICE table adjusts to include the complex ion and the ligand. For AgBr in NH₃, the initial concentration of Ag⁺ is the solubility from the water-only scenario, and NH₃ has a known concentration. As NH₃ reacts with Ag⁺ (Change), we track the concentrations until reaching Equilibrium. The calculations here are more intricate, as they require accounting for the Kf of the complex ion. The ICE table method neatly handles these, allowing chemists to solve for the unknowns in multi-step equilibria.