Question

Ksp for Ag2CrO4=5.4x10^-12. Calc molar solubility of Ag2CrO4 in 0.0050M K2CrO4

Answer 1

Ag₂CrO₄       2Ag⁺ + CrO₄^--

-x                      2x          x

Ksp = [Ag⁺]²      [CrO₄^--]

5.410^-12 = (2x)² x = 4x³

4x³ = 5.410^-12

x³ = (5.410^-12 )/4 = 1.3510^-12

x = (1.3510^-12 )^(1/3) = 1.105210^-4   M
Thus x is solubility of Ag₂CrO₄ .

Hence, molar solubility of Ag₂CrO₄ = 1.105210^-4   M (or moles/L)