Question

EXPERIMENT 7: DETERMINATION OF KS FOR A SLIGHTLY SOLUBLE SALT QUESTIONS Average Ksp = 3.17x10-12 1. Based on the value of the Key that you calculated for AgeCrO4, how many grams of Ag2 CrO 4 will dissolve per Liter of solution? 2. In the second part of the experiment, why was the [AgNO3l larger than the (K,CrO4l? What would have happened if they had been the same?

Answer 1

Let us assume the molar solubility of Ag₂CrO₄ is x mol/lit

As in water Ag₂SO₄ dissociates as, Ag₂CrO₄(aq)= 2Ag⁺(aq)+ CrO₄^2-(aq)

So, the solubility product will be, K_sp= (2x)²(x) = 4x³

As per the given solubility product is K_sp= 3.17X10^-12

So, 4x³=3.17X10^-12

or, x³= 3.17X10^-12/4

or, x³= 0.7925X10^-12

or, x= 0.925X10^-6

So in 1 lit water the amount of Ag₂SO₄ dissolved is 0.925X10^-6 moles

The molecular weight of Ag₂CrO₄ is 331.73 g/mol

So, 1 mole Ag₂CrO₄ is 331.73 g

Hence 0.925X10^-6 moles Ag₂​​​​​​​CrO₄ is (331.73X0.925X10^-6) g = 0.0307 g

So in 1 lit water the amount of Ag₂SO₄ dissolved is 0.0307 g.