Problem 110 - Chemistry: An Atoms-Focused Approach

Question

Solution \(A\) is \(0.0200 M\) in \(A g^{+}\) ions and \(P b^{2+}\) ions. You have access to two other solutions: (B) \(0.250 M\) NaCl and (C) \(0.250 M\) aBr. a. Which solution, B or C, would be the better one to add to solution \(A\) to separate \(A g^{+}\) ions from \(P b^{2+}\) by selective precipitation? b. Using the solution you selected in part a, is the separation of the two ions complete?

Step-by-Step Solution

Answer

Solution B (NaCl) would be a better choice for separating Ag+ ions from Pb2+ ions by selective precipitation. The separation of the two ions is not complete, as the remaining concentration of Ag+ ions is much higher than the equilibrium concentration in a saturated AgCl solution.

Step 1: Identify relevant solubility products

To determine which solution would be better for selective precipitation, we need to analyze the solubility products (Ksp) of the different possible combinations: 1. AgCl: Ksp = 1.8 x 10^-10 2. AgBr: Ksp = 5.0 x 10^-13 3. PbCl2: Ksp = 1.7 x 10^-5 4. PbBr2: Ksp = 6.3 x 10^-6

Step 2: Calculate the ion product (Q) for each combination

We now need to calculate the ion product Q for each combination of solution A with solutions B and C. Q is calculated using the same formula as Ksp, but with the actual concentrations given in the problem. a) Solution A + Solution B (with NaCl, forming AgCl and PbCl2) Q_AgCl = [Ag+][Cl-] = (0.0200 M)(0.250 M) = 5.0 x 10^-3 Q_PbCl2 = [Pb2+][Cl-]^2 = (0.0200 M)(0.250 M)^2 = 1.25 x 10^-3 b) Solution A + Solution C (with aBr, forming AgBr and PbBr2) Q_AgBr = [Ag+][Br-] = (0.0200 M)(0.250 M) = 5.0 x 10^-3 Q_PbBr2 = [Pb2+][Br-]^2 = (0.0200 M)(0.250 M)^2 = 1.25 x 10^-3

Step 3: Compare Q to Ksp for each combination

For a precipitate to form, Q must be greater than Ksp. We compare the calculated Q values to the corresponding Ksp values from Step 1 to determine which precipitates will form. a) Solution A + Solution B Q_AgCl > Ksp_AgCl (5.0 x 10^-3 > 1.8 x 10^-10) Q_PbCl2 < Ksp_PbCl2 (1.25 x 10^-3 < 1.7 x 10^-5) b) Solution A + Solution C Q_AgBr > Ksp_AgBr (5.0 x 10^-3 > 5.0 x 10^-13) Q_PbBr2 > Ksp_PbBr2 (1.25 x 10^-3 > 6.3 x 10^-6)

Step 4: Choose the better solution

From the results of Step 3, we can see that when adding Solution B (NaCl), only Ag+ ions will precipitate (as AgCl), while Pb2+ ions will remain in solution. When adding Solution C (aBr), both Ag+ and Pb2+ ions will precipitate. Therefore, Solution B (NaCl) would be a better choice for separating Ag+ ions from Pb2+ ions by selective precipitation. a. Answer: Solution B (NaCl)

Step 5: Determine if the separation is complete

Although Solution B selectively precipitates Ag+ ions as AgCl, we still need to evaluate if this separation is complete or if there are remaining Ag+ ions in the solution. To do this, we can compare the concentration of the remaining Ag+ ions to the solubility limit of AgCl: [Ag+] left = (Q_AgCl / [Cl-]) = (5.0 x 10^-3) / (0.250 M) = 2.0 x 10^-2 M Comparing the concentration of the remaining Ag+ ions to the equilibrium concentration of Ag+ in a saturated AgCl solution: [Ag+]sat = sqrt(Ksp_AgCl) = sqrt(1.8 x 10^-10) = 1.34 x 10^-5 M. Since the remaining concentration of Ag+ ions (2.0 x 10^-2 M) is much higher than the equilibrium concentration in a saturated AgCl solution (1.34 x 10^-5 M), the separation of the two ions is not complete. b. Answer: No, the separation of the two ions is not complete.

Key Concepts

Solubility Product (Ksp)

The solubility product constant, or Ksp, is a measure that helps us understand the solubility of different salts in water. It represents the level at which a solute dissolves in a solution to form its ions. Each ionic compound has its unique Ksp value.
These values are derived from the concentrations of the ions that make up the compound, particularly at equilibrium where the solution is saturated. The formula for calculating Ksp is typically given as a product of the concentrations of the ions, each raised to the power of its coefficient in the balanced chemical equation:

• For AgCl: \( K_{sp} = [Ag^+][Cl^-] \)
• For AgBr: \( K_{sp} = [Ag^+][Br^-] \)
• For PbCl2: \( K_{sp} = [Pb^{2+}][Cl^-]^2 \)
• For PbBr2: \( K_{sp} = [Pb^{2+}][Br^-]^2 \)

The smaller the Ksp value, the less soluble the compound is in water. This is crucial when planning selective precipitation since it helps predict which salts will form first.

Ion Product

The ion product, or Q, acts similarly to the Ksp but is used to describe the current state of the solution rather than its equilibrium state.
Q can be calculated using the actual concentrations of the ions in the solution you're studying. Just like Ksp, Q is expressed as the product of the ionic concentrations, keeping in mind their stoichiometry.
Here’s how it can guide us:

• If \( Q < K_{sp} \), the solution is unsaturated, and no precipitate will form.
• If \( Q = K_{sp} \), the solution is exactly saturated, and it's on the verge of forming a precipitate.
• If \( Q > K_{sp} \), the solution surpasses the solubility limit, leading to precipitation of the solute out of the solution.

In our scenario, calculating Q for combinations with NaCl (forming AgCl and PbCl2) and aBr (forming AgBr and PbBr2) allows us to determine which ions will precipitate first. When Q exceeds Ksp, precipitation is likely, guiding the selective precipitation process.

AgCl and AgBr Solubility

The solubility of AgCl and AgBr plays a significant role in selective precipitation processes.
Given that both silver chloride (AgCl) and silver bromide (AgBr) have very low solubility in water, they make excellent candidates for forming precipitates. However, their solubility differs, as indicated by their Ksp values:

• \( K_{sp} \) for AgCl is \( 1.8 \times 10^{-10} \)
• \( K_{sp} \) for AgBr is \( 5.0 \times 10^{-13} \)

AgBr is much less soluble than AgCl, which makes it more likely to precipitate out of solution compared to AgCl under the same conditions. However, the context of the surrounding ions and solutions affects these outcomes.
In our specific example, adding NaCl to solution A leads to the precipitation of AgCl because the ion product for AgCl exceeds its Ksp, a useful strategy for separating silver ions from other ions like Pb2+. Understanding these solubility differences is crucial for effectively deploying selective precipitation in chemical separations.