Common Ion effect

[jmart]

Ok, so after studying tons of chemistry my brain is rather overloaded and becoming confused.

I remembered that when in gen chem there was a circumstance were you would have the concentration being (x-.1) and you could just drop the x.


Here is an example problem I found online.


Example #1: AgCl will be dissolved into a solution with is ALREADY 0.0100 M in chloride ion. What is the solubility of AgCl?

By the way, the source of the chloride is unimportant (at this level). Let us assume the chloride came from some dissolved sodium chloride, sufficient to make the solution 0.0100 M. So, on to the solution . . .

The dissociation equation for AgCl is:

AgCl (s) <===> Ag+ (aq) + Cl¯ (aq)
The Ksp expression is:

Ksp = [Ag+] [Cl¯]
This is the equation we must solve. First we put in the Ksp value:

1.77 x 10¯10 = [Ag+] [Cl¯]
Now, we have to reason out the values of the two guys on the right. The problem specifies that [Cl¯] is already 0.0100. I get another 'x' amount from the dissolving AgCl. Of course, [Ag+] is 'x.'

Substituting, we get:

1.77 x 10¯10 = (x) (0.0100 + x)
This will wind up to be a quadratic equation which is solvable via the quadratic formula. However, there is a chemical way to solve this problem. We reason that 'x' is a small number, such that '0.0100 + x' is almost exactly equal to 0.0100. If we were to use 0.0100 rather than '0.0100 + x,' we would get essentially the same answer and do so much faster. So the problem becomes:

1.77 x 10¯10 = (x) (0.0100)




Ok so what is confusing me here is why we are able to just drop out the x in part of the equation. I don't understand why x is so small that it is not significant to the second part??

Is this always the case for common ion effect that you just drop out x and use the common ion?

Thanks

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[Sunny29918]

Technically the X from Cl- still exists. It is dropped out to estimate an answer that would be close enough in most cases of chemistry. Think of it this way, you know AgCl is already pretty insoluble, the large amount of Cl- in the solution before hand is so much bigger than the tiny bit from the AgCl that it is like adding 0.0100 + 0.00001 = 0.01001 That value is so close to the original that it is essentially negligible in comparison. The main amount of 0.01 estimate is going to get you close enough to your answer to make you not need to worry about the X amount of Cl-.