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From each of the following ion concentrations in a solution (BaF2 and PbCl2), predict whether a precipitate will form in the solution. a. a {[Ba2 ] = 0.040 M, [F-] = 0.0025 M. Ksp 1.0 x 10-6 b. b [Pb2 ] = 0.060 M, [Cl-]= 0.034 M. Ksp 1.6 × 10–5



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.

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