I'm confused on when you have to use an ICE table and when you can just set up the equation Ksp=[]*[]
Solubility/Ksp
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Accounting for any initial concentration that may already be in solution, for example.
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Ok, this should help clear this up. I'm studying for the DAT right now also, and I just covered this material in Chad's videos.
AB2 <-> A+ + 2B-
Ionic compound AB dissolves into A+ and B-, in a 1:2 molar ratio respectively. Question asks you to find molar solubility (aka x) or concentration of ions at equilibrium (x or 2x depending on which ion you're looking at).
So, you set up an ICE table as follows:
A+ 2B-
I 0 0
C x +2x
E x 2x
Now you set Ksp equal to x(2x)^2, and solve for X. X is the molar solubility.
Another case when you use ICE table is with the common ion effect....so if you add a solution of .1M AX to the AB2 compound, your ICE table becomes:
A+ 2B-
I .1 0
C x +2x
E .1+x 2x
This simplifies to Ksp = (.1+x)(2x)^2 = (.1)(2x)^2 IF the Ksp is very small...because that means that x is insignificant in comparison to .1.
Now, to answer your question: when do you not need to use the ICE table?
Well,
Let's say the Ksp for compound AB is 1.8E-5, and the concentration of A+ ions in solution is 1.8E-10. What is the maximum concentration of B- ions that can be put into solution without precipitation? This requires no ice table.
Just do: 1.8E-5 = [1.8E-10][B-]^2 and solve for [B-]. Bam. Note: you do not do [2B-]^2 because the 2 inside the concentration bracket refers to the 1:2 A+ to B- ratio that comes from dissociation of the ionic compound. Here, you're adding B- from a different source.
Hope this helps.
AB2 <-> A+ + 2B-
Ionic compound AB dissolves into A+ and B-, in a 1:2 molar ratio respectively. Question asks you to find molar solubility (aka x) or concentration of ions at equilibrium (x or 2x depending on which ion you're looking at).
So, you set up an ICE table as follows:
A+ 2B-
I 0 0
C x +2x
E x 2x
Now you set Ksp equal to x(2x)^2, and solve for X. X is the molar solubility.
Another case when you use ICE table is with the common ion effect....so if you add a solution of .1M AX to the AB2 compound, your ICE table becomes:
A+ 2B-
I .1 0
C x +2x
E .1+x 2x
This simplifies to Ksp = (.1+x)(2x)^2 = (.1)(2x)^2 IF the Ksp is very small...because that means that x is insignificant in comparison to .1.
Now, to answer your question: when do you not need to use the ICE table?
Well,
Let's say the Ksp for compound AB is 1.8E-5, and the concentration of A+ ions in solution is 1.8E-10. What is the maximum concentration of B- ions that can be put into solution without precipitation? This requires no ice table.
Just do: 1.8E-5 = [1.8E-10][B-]^2 and solve for [B-]. Bam. Note: you do not do [2B-]^2 because the 2 inside the concentration bracket refers to the 1:2 A+ to B- ratio that comes from dissociation of the ionic compound. Here, you're adding B- from a different source.
Hope this helps.
