Suppose you wish to make 0.5 L of a solution containing 0.0075 mole of NaCl and 0.075 Pb(NO3)2. Ksp=1.2×10^-5 (a)Will the precipitate form. (b) Calculate concentration of ions after reaction of PbCl2.

To determine whether a precipitate will form when mixing solutions of NaCl and Pb(NO3)2, we need to analyze the solubility product constant (Ksp) of lead(II) chloride (PbCl2) and the concentrations of the ions involved in the reaction. Let's break this down step by step.

Understanding the Reaction

When NaCl and Pb(NO3)2 are mixed, they dissociate in solution:

• NaCl → Na⁺ + Cl⁻
• Pb(NO3)2 → Pb²⁺ + 2NO3⁻

The relevant reaction for precipitate formation is:

Pb²⁺ + 2Cl⁻ ⇌ PbCl2 (s)

Calculating Ion Concentrations

First, we need to find the concentrations of the ions in the final solution. Since we are preparing 0.5 L of solution, we can calculate the concentration of each ion based on the moles provided.

Concentration of NaCl

For NaCl, we have:

• Moles of NaCl = 0.0075 moles
• Volume of solution = 0.5 L

The concentration of Cl⁻ ions is:

[Cl⁻] = Moles of Cl⁻ / Volume = 0.0075 moles / 0.5 L = 0.015 M

Concentration of Pb(NO3)2

For Pb(NO3)2, we have:

• Moles of Pb(NO3)2 = 0.075 moles
• Volume of solution = 0.5 L

The concentration of Pb²⁺ ions is:

[Pb²⁺] = Moles of Pb²⁺ / Volume = 0.075 moles / 0.5 L = 0.15 M

Determining Precipitate Formation

To see if a precipitate will form, we need to calculate the ion product (Q) and compare it to the Ksp of PbCl2.

The ion product Q is calculated as follows:

Q = [Pb²⁺][Cl⁻]²

Substituting the concentrations we found:

Q = (0.15)(0.015)² = (0.15)(0.000225) = 0.00003375

Comparing Q to Ksp

Now we compare Q to Ksp:

• Ksp of PbCl2 = 1.2 × 10⁻⁵
• Q = 0.00003375

Since Q (0.00003375) is greater than Ksp (1.2 × 10⁻⁵), this indicates that the solution is supersaturated, and a precipitate of PbCl2 will indeed form.

Calculating Ion Concentrations After Precipitation

Once PbCl2 precipitates, we need to find the new concentrations of the ions in solution. The reaction consumes Pb²⁺ and Cl⁻ ions according to the stoichiometry:

1 Pb²⁺ + 2 Cl⁻ → PbCl2 (s)

Let x be the amount of PbCl2 that precipitates. Since we know that the reaction consumes 1 mole of Pb²⁺ for every 2 moles of Cl⁻, we can express the changes in concentrations:

• [Pb²⁺] after precipitation = 0.15 - x
• [Cl⁻] after precipitation = 0.015 - 2x

To find x, we can set Q equal to Ksp at equilibrium:

Ksp = (0.15 - x)(0.015 - 2x)²

However, since the amount of PbCl2 that precipitates will be small compared to the initial concentrations, we can approximate:

Ksp ≈ (0.15)(0.015)²

Solving this will give us the final concentrations of Pb²⁺ and Cl⁻ after precipitation occurs.

In summary, a precipitate of PbCl2 will form, and the concentrations of the ions will adjust based on the stoichiometry of the reaction. This approach allows us to understand the dynamics of solubility and precipitation in ionic solutions.