If the solubility of Ag2CO3 is 1.3x10^-4 mol/L, what is its Ksp?
(the answer is said to be 8.8x10^-12)
Ag2CO3(s) -----> 2 Ag+ (ag) + CO3 2-
Ksp= [Ag+]^2 [CO3^2-]
= [1.3x10^-4]^2[1.3x10^-4]
=2.197x10^-12
thats what I did, but I found that if you first multiplying the 1.3x10^-4 by 2 which gives you 2.6x10^-4 and then you insert it into the expression and multiply it by 1.3x10^-4 you get the answer. But it doesn't make sense to x2 it and then square it. Does anyone know why or if their answer is wrong and I'm right?
1 answer
A typical student error. Yes, the answer is correct. For every 1 molecule of Ag2CO3 that dissolves you get TWICE the Ag^+ and ONE TIMES for the carbonate. So if Ag2CO3 is 1.3E-4 then Ag^+ is twice that and CO3^2- is 1.3E-4.