Question

what volume of water is needed to completely dissolve 4.0 g of Ag2CrO4 (Ksp = 8.0 x 10^-12, MM = 331.8 g/mol)

Answer 1

At equilibrium:

Ag2CrO4 <----> 2 Ag+ + CrO42-

   2s s

Ksp = [Ag+]^2[CrO42-]

8*10^-12=(2s)^2*(s)

8*10^-12= 4(s)^3

s = 1.26*10^-4 M

Molar mass of Ag2CrO4 = 331.8 g/mol

Molar mass of Ag2CrO4= 331.8 g/mol

s = 1.26*10^-4 mol/L

To covert it to g/L, multiply it by molar mass

s = 1.26*10^-4 mol/L * 331.8 g/mol

s = 4.18*10^-2 g/L

Now use:

Volume = mass / s

= (4.0 g) / (4.18*10^-2 g/L)

= 95.7 L

Answer: 96 L