calculate the ksp of Ag2SO4 with a solubility of 1.20x10-6 g/L.

Question

so my prof told me that ksp is in mol/l what i did is convert it to mol/l by dividing 1.20x10-6 to mw of ag2s04 which is 312. (=3.85x10-9)
ag2s04= 2ag+so4+2 (i balanced it)
Ag2SO4= [2(3.85x10-9] 2(squared) [3.85x10-9]
is my work correct?
please help me

Answer 1

and please help me with this.. the ksp of Cr(OH03) at room temperature is 1.00x10-30, what is its molar solubility? please help me with this. i don't know where to start because we havent tackle this thing yet.. i don't know what to do with its temperature and molar solubility.. help

Answer 2

........Ag2SO4 ==> 2Ag^2+ + SO4^2-
I.......solid.......0..........0
C.......solid.....2*3.85E-9..3.85E-9
E.......solid.....7.7E-9.....3.85E-9

Ksp = (Ag^+)^2(SO4^2-)
Ksp = (7.7E-9)^2(3.85E-9)
Ksp = ?
You were right to calculate solubility in mols/L as you did and you have the right answer. Also, you multiplied 2*M to gind (Ag^+) which is correct also. Actually the equation you have of [2(3.96E-9]^2 is almost right. That's the (Ag^+)^2 but you must multiply by (SO4^2- also.

Answer 3

You would do well to post just one problem to the post. This one is done just the reverse of the Ag2SO4.

.........Cr(OH)3 ==> Cr^3+ + 3OH^-
Initial..solid........0.......0
change...solid........x.......3x
equilibrium solid.....x........3x

Ksp = (Cr^3+))(OH^-)^3
Substitute Ksp and the E line into the Ksp expression and solve for x = solubility Cr(OH)3 in mols/L.