Asked by Michelle
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?
(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?
Answers
Answered by
DrBob222
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.
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