soln saturated w/ PbS04 @ 50C (ksp 2.3x10^-8) what mass will precipitate from 815ml when cooled to 25c?

Question

soln saturated w/ PbS04 @ 50C (ksp 2.3x10^-8) what mass will precipitate from 815ml when cooled to 25c? The ksp for 25 deg. C is ( 1.6 x 10^-8) I'm a bit confused as where to start, can anyone give me an idea?

Answer 1

At higher temperature the solubility is higher. You need to find exactly how much it is with the help of Ksp50C=2.3*10^-8.
When tou lower the temperature, solubility also decreases.
Use Ksp25 to find the new value. The difference will show you how much will precipitate per L. From then on you fin out how much for 0.815L and mulitply with the MW (303.26) to convert into mass.
In detail:

If s is the molar solubility, then
.. .. .. .. .. .. PbSO4 <=>Pb+2 +SO4(-2)
Dissolve .. .. .. s
produce .. .. .. .. .. .. .. . .. s .. .. .. . s

Ksp=[Pb+2][SO4(-2)] =s*s =s^2 => s=Squareroot(Ksp)

At 50 C: s50= SQRT(Ksp50)= SQRT(2.38*10^-8) =>
s50= 1.54*10^-4 mole/L

At 25 C : s25= SQRT(Ksp25) =SQRT(1.6*10^-8) =>
s25= 1.26*10^-4 mole/L

Thus you will have precipitating (1.54-1.26)*10^-4 = 2.8*10^-5 mole/L.
You have 0.875 L so you will have precipitating
=0.875*2.8*10^-5 = 2.45*10^-5 mole

mass=mole*MW= (2.45*10^-5)*303.26 = 7.43*10^-3 g