For PbCl₂ (K_sp = 2.4 * 10^-4), will a precipitate of PbCl₂ form when 0.10 L of 3.0 * 10^-2 M Pb(NO₃)₂ is added to 400 mL of 9.0 * 10^-2 M NaCl

Question

Quizplus · Accepted Answer

To determine whether a precipitate will form, we need to compare the reaction quotient Q with the solubility product constant KS1U1B1sp.S1U1B0. The balanced chemical equation for the dissolution of PbClS1U1B12S1U1B0 in water is:
PbClS1U1B12S1U1B0(s) ⇌ PbS1U1B12S1U1B0(aq) + ClS1U1B1–S1U1B1(aq)
The solubility product expression is:
KS1U1B1sp.S1U1B0 = [PbS1U1B12S1U1B0][ClS1U1B1–]S1U1B1
According to the balanced equation, 1 mol of PbClS1U1B12S1U1B0 produces 1 mol of PbS1U1B12S1U1B0. Therefore, the initial concentration of PbS1U1B12S1U1B0, [PbS1U1B12S1U1B0] in the solution is:
[PbS1U1B12S1U1B0] = (0.10 L)(3.0 F1F1F1S1F1F1F10 10S1U1P1-2S1S1P0 M) = 0.30 F1F1F1S1F1F1F10 10S1U1P1-2S1S1P0 M
NaCl is a strong electrolyte and completely dissociates into NaS1U1B11S1U1B0 and ClS1U1B1–S1U1B1 ions. Thus, the initial concentration of ClS1U1B1–S1U1B1 is:
[ClS1U1B1–]S1U1B1 = (0.40 L)(9.0 F1F1F1S1F1F1F10 10S1U1P1-2S1S1P0 M) = 3.6 F1F1F1S1F1F1F10 10S1U1P1-2S1S1P0 M
The reaction quotient is:
Q = [PbS1U1B12S1U1B0][ClS1U1B1–]S1U1B1 = (0.30 F1F1F1S1F1F1F10 10S1U1P1-2S1S1P0 M)(3.6 F1F1F1S1F1F1F10 10S1U1P1-2S1S1P0 M) = 1.1 F1F1F1S1F1F1F10 10S1U1P1-11S1S1P0
Comparing Q with KS1U1B1sp.S1U1B0, we find that:
Q < KS1U1B1sp.S1U1B0
Therefore, a precipitate of PbS1U1B12S1U1B0 will not form. The correct answer is B.

For PbCl2 (Ksp = 2

For PbCl₂ (K_sp = 2