For PbCl₂,K_sp = 1.7 × 10−⁵.What will occur if 250 mL of 0.12 M Pb(NO₃)₂ is mixed with 250 mL of 0.070 M NaCl?

Question

Quizplus · Accepted Answer

First we need to write the balanced chemical equation for the reaction between Pb(NO3)2 and NaCl:
Pb(NO3)2 (aq) + 2NaCl (aq) → PbCl2 (s) + 2NaNO3 (aq)
The solubility product expression for PbCl2 is:
Ksp = [Pb2+][Cl-]2
To calculate Q, we need to plug in the concentrations of Pb2+ and Cl- ions in the mixture:
[Pb2+] = 0.12 M × 250 mL / 500 mL = 0.06 M
[Cl-] = 0.070 M × 250 mL / 500 mL = 0.035 M
Q = [Pb2+][Cl-]2 = (0.06 M)(0.035 M)2 = 7.35 × 10^-5
Comparing Q to Ksp, we get:
Q > Ksp, which means the ion product exceeds the Ksp, and a precipitate (PbCl2) will form. Therefore, the answer is A.

For PbCl2,Ksp = 1

For PbCl₂,K_sp = 1