Ksp of PbI2 at 25 degrees celcius and 15 degrees celcius is 7.5×10^-8 and 7.5×10^-9 respectively. (Pb=207 I=127)

Question

1) write an expression for the equilibrium of PbI2

2)write an equation for the Ksp of PbI2

3)calculate the solubility of PbI2 at 25 degrees celcius in gdm^-3

4)what is the mass of PbI2 precipitated when temperature of 1l of PbI2 solution is reduced from 25 degrees celcius to 15 degrees celcius

5)state whether the solubility of PbI2 is endothermic/exothermic

Please tell me the difference between 1st and 2nd questions?

Also I don't need answers for any of these questions...but a little GUIDANCE on each question would be much appreciated :) Thanks.

Answer 1

First things first. Learn how to spell celsius.
1. The equilibrium equation is just the solubility in water.
PbI2(s) ==> Pb^2+(aq) + 2I^-(aq)

2. Surely you know how to write the Ksp.

3. From the equation is (1), solve for the solubility.
.....PbI2 ==> Pb^2+ + 2I^-
I...solid.....0........0
C...solid-x...x........x
E...solid-x...x........x

Ksp = (Pb^2+)(I^-)^2
Substitute the E line into Ksp expression and solve for x = solubility. Be sure and use the correct value of Ksp for the temperature in the problem. The solubility you get is in mol/dm3. Convert to grams/dm3

4.Redo the solubility which you calculated in 3 and calculate solubility for 15 C. Convert solubility to g/dm3. Then subtract the two to find how much will ppt at the lower temperature.

5. First time I've seen a question like this but you just reason it out.
PbI2 --> Pb^2+ + 2I^-
Look at your data for solubility at the two T values. It is MORE solubility at the higher T so your equation could be written as
PbI2 + heat ==> Pb^2+ + 2I^-
Since you know the solubility is increased (adding heat drives the equation equilibrium to the right---from Le Chatelier's Principle) with heat, that makes it endothermic. Right?

Answer 2

.....PbI2 ==> Pb^2+ + 2I^-
I...solid.....0........0
C...solid-x...x........2x
E...solid-x...x........2x

I put 2x considering stoichiometry. Am I wrong here?
Thanks.