A) Use the molar solubility 1.08 x 10^5 M in pure water water to calculate Ksp for BaCrO4.
B) Use the molar solubility 1.55 x 10^5 M in pure water water to calculate Ksp for Ag2SO3
C) Use the molar solubility 2.22 x 10^8 M in pure water water to calculate Ksp for Pd(SCN)2.
1 Answer

A) Ksp = [Ba++][CrO4–]
For every x moles of BaCrO4 that dissolves in a liter of water, the concentration of Ba++ is x Molar, and the concentration of CrO4– is x Molar. Therefore,
Ksp = (x)(x) = x² = (1.08 x 10^5 M)² = 1.17 x 10^10 M².
B) Ksp = [Ag]²[SO3–]
For every x moles of Ag2SO3 that dissolves in a liter of water, the concentration of Ag+ is 2x Molar, and the concentration of SO3– is x Molar. Therefore,
Ksp = (2x)²(x) = 4x³ = (4)(1.55 x 10^5 M)³ = 1.49 x 10^14 M³.
C) Ksp = [Pd++][SCN]²
In the same manner as question B),
Ksp = (x)(2x)² = 4x³ = (4)(2.22 x 10^8 M)³ = 4.38 x 10^23 M³.