Question

Calculate the milligrams of Ag2CO3 in 250 mL of a saturated solution of Ag2CO3. The Ksp of Ag2CO3 is 8.1 * 10-12.

Answer 1

Ag2CO3 <--> 2Ag+ + CO3-2
2s s

Ksp = [Ag+]^2*[CO3-2]
8.1*10^-12 = (2s)^2*s
8.1*10^-12 = 4*s^3
s^3 = 2.02*10^-12
s = 1.26*10^-4

[Ag2CO3] = [CO3-2]
= s
= (1.26*10^-4) M

number of mole of Ag2CO3 = (molarity of Ag2CO3)*(volume of Ag2CO3 in L)
= 1.26*10^-4*0.250
= (0.314*10^-4) mole

mass of Ag2CO3 = (number of mole of Ag2CO3)*(molar mass of Ag2CO3)
molar mass of Ag2CO3 = 275.7g/mol
so,
mass of Ag2CO3 = 0.314*10^-4*275.7
= (8.66*10^-3) g
= 8.66 mg

Answer : 8.66 mg