Question

(a) If the molar solubility of Cd₃(AsO₄)₂ at 25 ^oC is 1.15e-07 mol/L, what is the K_sp at this temperature?

K_sp =  




(b) It is found that 6.47e-05 g of Ag₃AsO₄ dissolves per 100 mL of aqueous solution at 25 ^oC. Calculate the solubility-product constant for Ag₃AsO₄.

K_sp =  




(c) The K_sp of ScF₃ at 25 ^oC is 5.81e-24. What is the molar solubility of ScF₃?

solubility =  mol/L

Answer 1

a) you can write the dissociation reaction and the expression for the Ksp to understand the problem:

<----- the concentrations raised to its stoichiometric coefficient

From the molar solubility, you can calculate the molarity of each of the ions, using the ratios from the reaction:

Now you can plug in the concentrations of the ions into the Ksp expression:

b)

Similiar to the one above, but first you need to find the concentration:

You need the molecular weight of the compound---> MW=462.61

The concentration is, the moles divided by the volume:

<-----mass over the MW

<---moles over Volume( in liters)

And now you need to write the reaction and the Ksp expression:

From the concentration above and the molar ratios from the reaction calculate the concentrations of each ion:

And plugging in the concentrations into the Ksp expression:

c)

In this case you need to go backward:

first write the reaction and the Ksp expression:

From the reaction you can see that when the ScF3 dissociates into ions th F has three times the concentraition of Sc.

You can assume the concentration of Sc=X, and so F=3X

and using the Ksp expression:

and since the molar ration from the reaction 1 mol of ScF3 will dissociate into 1 mol of Sc, thus: