For PbCl₂ (K_sp = 2.4 × 10^-4), will a precipitate of PbCl₂ form when 0.10 L of 3.0 × 10^-2^MPb(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 to the solubility product constant KS1U1B1spS1U1B0. The equation for the dissolution of PbClS1U1B12S1U1B0 is:

PbClS1U1B12S1U1B0(s) ⇌ PbS1U1B12+ (aq) + 2ClS1U1B1- (aq)

The expression for Q is:

Q = [Pb2+] [Cl-]2

We need to calculate the molar concentrations of Pb2+ and Cl- ions after mixing the solutions.

The number of moles of Pb(NO3)2 in 0.10 L of 3.0 × 10S1U1P1-2S1S1P0S1U1P1M solution is:

moles of Pb(NO3)2 = M × V = (3.0 × 10S1U1P1-2S1S1P0S1U1P1) × (0.10 L) = 3.0 × 10S1U1P1-3S1S1P0 mol

Since each mole of Pb(NO3)2 produces one mole of Pb2+ ions, the molar concentration of Pb2+ in the solution is:

[Pb2+] = moles of Pb2+ / total volume of solution in liters = (3.0 × 10S1U1P1-3S1S1P0 mol) / (0.10 L + 0.4 L) = 6.0 × 10S1U1P1-4S1S1P0 M

Similarly, the number of moles of NaCl in 400 mL of 9.0 × 10S1U1P1-2S1S1P0S1U1P1M solution is:

moles of NaCl = M × V = (9.0 × 10S1U1P1-2S1S1P0S1U1P1) × (0.400 L) = 3.6 × 10S1U1P1-1S1S1P0 mol

Since each mole of NaCl produces one mole of Cl- ions, the molar concentration of Cl- in the solution is:

[Cl-] = moles of Cl- / total volume of solution in liters = (3.6 × 10S1U1P1-1S1S1P0 mol) / (0.10 L + 0.4 L) = 7.2 × 10S1U1P1-1S1S1P0 M

Substituting these values into the expression for Q:

Q = [Pb2+] [Cl-]2 = (6.0 × 10S1U1P1-4S1S1P0 M) × (7.2 × 10S1U1P1-1S1S1P0 M)2 = 2.9 × 10S1U1P1-13S1S1P0

Comparing Q to KS1U1B1spS1U1B0:

Q < KS1U1B1spS1U1B0

Therefore, no precipitate of PbClS1U1B12S1U1B0 will form, and the answer is B.

For PbCl2 (Ksp = 2

For PbCl₂ (K_sp = 2