Answer :
For the first scenario with BaF2, we need to compare the ion product (Q) with the solubility product constant (Ksp) to determine if a precipitate will form. The ion product (Q) is calculated by multiplying the concentrations of the ions involved in the reaction.
a. For BaF2:
[Ba2+] = 0.040 M
[F-] = 0.0025 M
Q = [Ba2+][F-]
Q = 0.040 * 0.0025
Q = 1.0 x 10^-4
Since Q (1.0 x 10^-4) is greater than Ksp (1.0 x 10^-6), a precipitate of BaF2 will form in the solution.
For the second scenario with PbCl2:
[Pb2+] = 0.060 M
[Cl-] = 0.034 M
b. For PbCl2:
Q = [Pb2+][Cl-]
Q = 0.060 * 0.034
Q = 2.04 x 10^-3
Since Q (2.04 x 10^-3) is greater than Ksp (1.6 x 10^-5), a precipitate of PbCl2 will form in the solution.
a. For BaF2:
[Ba2+] = 0.040 M
[F-] = 0.0025 M
Q = [Ba2+][F-]
Q = 0.040 * 0.0025
Q = 1.0 x 10^-4
Since Q (1.0 x 10^-4) is greater than Ksp (1.0 x 10^-6), a precipitate of BaF2 will form in the solution.
For the second scenario with PbCl2:
[Pb2+] = 0.060 M
[Cl-] = 0.034 M
b. For PbCl2:
Q = [Pb2+][Cl-]
Q = 0.060 * 0.034
Q = 2.04 x 10^-3
Since Q (2.04 x 10^-3) is greater than Ksp (1.6 x 10^-5), a precipitate of PbCl2 will form in the solution.
