Calculate the solubility in grams per liter (note that Ksp is in terms of moles and liters) of Ag2CO3?

Question

Ksp = 8.10 x 10−12 (Neglect the basic properties of CO3−2).

Answer 1

The equilibrium is
Ag2CO3 <> 2 Ag+ + CO32-

let x = moles/l that dissolve in water : we would get x moles/l CO32- and 2x moles/l Ag+

Sustitute these in Ksp expression

8.10 10^-12 = (2x)^2 (x) = 4 x^3

x= 0.000126 M

total Ag2CO3 dissolved =

= 0.000126 mol/L ( 1L)(275 74 g/mol)= 0.0347 g

Answer 2

Ag2CO3 --> 2Ag+ + CO3 2-

Ksp= 8.10e-12= [Ag+]^2[CO3 2-]
since there are 2 Ag+ ions for every CO3 2- ion, the concentrations in the Ksp equation are:
[Ag+]= [2x] which then gets raised to the second power
[CO3 2-]= [x]

therefore:

8.10e-12= [2x]^2[x]
8.10e-12= 4x^3
x^3= 2.025e -12
x=1.265e-4

1.265e-4 mol/L x 275.75 g Ag2CO3/mol = .034885 g/L

Answer 3

Ag2CO3---> 2Ag+ +CO3-

[Ag]^2[CO3]= 8.1 10^-12

lets x be the concentration of Ag2CO3
the concentration of Ag+ = 2x , that of CO3 =x

so (2x)^2 *x =8.1 10^-12 or 4x^2 *2x =8.1 10^-12

8x^3 = 8.1 10^12

x = 1.004 10^-4M

molecular weight of Ag2CO3 = 216+12+48=276

you find 0.0277g/l

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Answer 5

Ag2CO3=2Ag+ + CO3-2
Ksp=8.10 x10^-12
Ksp=[Ag+]2 [CO3-2]
lets S be the concentration of Ag2CO3
the concentration of Ag+ = 2S , that of CO3 =xS
=[2S]2 x[S]
=[4S]^3
8.10 x10^-12 =[4S]^3
S=(8.10 x 10^-12/4)1/3
S=2.009 x 10^-4mol/liter
S=2.009 x10^-4 grams/a.weightxliter
S=7.2789x10^-7 grams/liter