Please help with the following problems. Provide step by step calculations for each.
5.[a] The molar solubility of barium fluoride, BaF2, is 7.5 x 10-3 M. Calculate the solubility product
[b]Calculate the molar solubility of lead fluoride, PbF2 in water, Ksp = 2.7 x 10-8
6.Calculate the molar solubility of bismuth sulfide, Bi2S3 (Ksp = 1.6 x 10-72) in (a) pure water, and
(b) 0.05 M cadmium sulfide, CdS
See below for the solution to your chemistry questions.
5.[a] The molar solubility of barium fluoride, BaF2, is 7.5 x 10-3 M. Calculate the solubility product constant
BaF2 ---> Ba2+ + 2F-
1 1 2
[Ba2+] = [BaF2] = 7.5 * 10-3 M
[F-] = 2[BaF2] = 1.5 * 10-2 M
So the solubility product ...
This posting helps with general chemical problems. The solution explains how to use molar solubility product to calculate the molar solubility of chemicals, including barium fluoride, lead fluoride, and bismuth sulfide. Step by step calculation are provided for each problem.