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

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

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

........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.

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.

To calculate the Ksp (solubility product constant) of Ag2SO4, you have correctly converted the given solubility from grams per liter to moles per liter. However, the balanced equation you provided is incorrect.

First, let's clarify the balanced equation for the dissolution of Ag2SO4:
Ag2SO4 (s) ↔ 2Ag+ (aq) + SO4^2- (aq)

Now, let's proceed with the correct calculation:

1. Convert the given solubility to moles per liter:
Given solubility = 1.20x10^-6 g/L
Molar mass of Ag2SO4 = 2(Ag) + (S) + 4(O) = 2(107.87 g/mol) + 32.07 g/mol + 4(16.00 g/mol) = 311.80 g/mol

Solubility in moles per liter = (1.20x10^-6 g/L) / (311.80 g/mol)
= 3.85x10^-9 mol/L

2. Write the Ksp expression using the balanced equation:
Ksp = [Ag+]^2 * [SO4^2-]

3. Plug in the values into the Ksp expression:
Ksp = (2 * 3.85x10^-9 mol/L)^2 * (3.85x10^-9 mol/L)
= 2.97x10^-51 mol^3/L^3

Therefore, the correct Ksp value for Ag2SO4, calculated from the given solubility, is 2.97x10^-51 mol^3/L^3.