Solubility?

Question

What will be the solubility of bismuth iodide in g/100g water in pure water at 25°C? What will its solubility be in g/100g water in a 6.00x10^-6 M potassium iodide solution? What will its solubility be in g/100g water in a 3.50x10^-5 M bismuth nitrate solution?

I have the answers so I need to see the steps to get there.

Answer 1

I can not calculate since you did not give the solubility product of solubility of BiI3. I found 8.1 10^-19

you write BiI3---> Bi+++ +3I

8.110^-19 = [Bi][I+]^3 if c is the concentration of BiI3 , the concentration of Bi+++ is the same that of I- =3c

so 8.1 10^-19= c*27c^3= 27c^4

you find c^4 = 0.310^-19 c= 1.3*10^-5

Mw BiI3= 207+3*127=588

m= 1.310^-5*588*0.1 =0.77mg

i can not go further

Answer 2

whilst a answer is asserted to be saturated, it skill that the answer is protecting the utmost quantity of solute it could carry. as a fashion to reply to your question: particular, it fairly is the type you come across solubility in g/L. The Ksp is termed the solubility product consistent. that's a ratio of the multiplication of the respective concentrations of the products shaped from dissolving the solute to the concentration of the solute. If the solute is a organic good, liquid, or gas (meaning its concentration will not be able to get replaced), then its concentration will take transport of as a million (or team spirit) interior the ratio. NiCl2 (s) --> Ni(2+) (aq) + 2 Cl(-) (aq) as a result, Ksp = [Ni(2+)] x [Cl(-a million)]^2. Moles NiCl2 = 2.a million g/129.596 g/mol = 0.0.5 mol Moles Ni(2+) = 0.0.5 mol NiCl2 x (a million mol Ni(2+)/a million mol NiCl2) = 0.0.5 mol Moles Cl(-) = 0.0.5 mol NiCl2 x (2 mol Cl(-)/a million mol NiCl2) = 0.030 mol as a result, at equilibrium, [Ni(2+)] = 0.0.5 mol/0.0.5 L = 3 M & [Cl(-)] = 0.030 mol/0.0.5 L = 6 M. as a result, Ksp = 3 M x (6 M)^2 = 108 M^3.