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in which of these solutions would silver carbonate have the lowest molar solubility? for silver carbonate ksp =8.5 x10^-12?
A) 0.03 M H2SO4
B)0.1 M AgNO3
C)0.01 M AgNO3
D) 0.01 M Na2CO3
e) pure water
3 Answers
- pisgahchemistLv 74 years agoFavorite Answer
Le Chatelier's principle.....
Ag2CO3(s) <==> 2Ag+ + CO3^2- ................ Ksp = 8.46x10^-12
............................. 2x ........ x
Ksp = [Ag+]²[CO3^2-]
8.46x10^-12 = 4x³
x = 1.28x10^-4
[Ag+] = 2.57x10^-4M
[CO3^2-] = 1.28x10^-4M
Adding either silver ion (from 0.1M AgNO3) or carbonate ion (from 0.01M Na2CO3) will cause a shift to the left according to Le Chatelier's principle.
Ag2CO3(s) <==> 2Ag+ + CO3^2- ................ Ksp = 8.46x10^-12
.................... ........0.1 ......1.28x10^-4 .......... initial
................... ........-2x ....... -x ........................ change
................... .......0.1-2x ... 1.28x10^-4 - x .... equilibrium
8.46x10^-12 = (0.1-2x)²(1.28x10^-4 - x)
x = 1.28x10^-4
Essentially all of the carbonate ion is removed leaving only an excess of silver ion. That means that silver carbonate is insoluble in 0.1M Ag+.
Ag2CO3(s) <==> 2Ag+ + CO3^2- ................ Ksp = 8.46x10^-12
.................. .... 2.57x10^-4.. 0.01 .................. initial
.................. .... -2x .............. -x ...................... change
.................. 2.57x10^-4 - 2x ..0.01-x ............. equilibrium
8.46x10^-12 = (2.57x10^-4 - 2x)²(0.01-x)
x = 1.14x10^-4
[Ag+] = 2.93x10^-5 M
[CO3^2-] = 0.01 M
There is still silver carbonate present in the solution (granted, not much) with the addition of 0.01M carbonate ion, but the concentration is greater than when silver ion is added.
Therefore, Ag2CO3 is least soluble when 0.1M silver ion is added.