Calculate the milligrams of Ag2CO3 in 250 mL of a saturated solution of Ag2CO3. The Ksp of Ag2CO3 is 8.1 * 10-12.
The salt dissolves as:
Ag2CO3 <----> 2 Ag+ + CO32-
2s s
Ksp = [Ag+]^2[CO32-]
8.1*10^-12=(2s)^2*(s)
8.1*10^-12= 4(s)^3
s = 1.265*10^-4 M
Molar mass of Ag2CO3 = 2*MM(Ag) + 1*MM(C) + 3*MM(O)
= 2*107.9 + 1*12.01 + 3*16.0
= 275.81 g/mol
Molar mass of Ag2CO3= 275.81 g/mol
s = 1.265*10^-4 mol/L
To covert it to g/L, multiply it by molar mass
s = 1.265*10^-4 mol/L * 275.81 g/mol
s = 3.489*10^-2 g/L
volume = 250 mL = 0.250 L
mass dissolved = s*volume
= 3.489*10^-2 g/L * 0.250 L
= 8.72*10^-3 g
= 8.72 mg
Answer: 8.72 mg
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