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Topic: Precpitation and ionic concentration  (Read 471 times)

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kdbmvp

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Precpitation and ionic concentration
« on: March 01, 2019, 09:45:39 AM »
You have a solution of 0.1M AgNO3 and 0.09M KI, and AgI is precipitated. What is [Ag+] and [I-] in the solution after precipitation?

Given the higher concentration of AgNO3 I understand there must be 0.1 - 0.09 = 0.01 M [Ag+] left in the solution. But what about the [I-]? Shouldn't that be 0, as all the KI was consumed in the precipitation of AgI?

mjc123

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Re: Precpitation and ionic concentration
« Reply #1 on: March 01, 2019, 10:40:51 AM »
Nearly all, but not quite. There must be some I- in solution to be in equilibrium with solid AgI, but the concentration will be very small compared to the original concentration of I-.

Are you familiar with the concept of the solublilty product? The SP of AgI is 8.3 x 10-17 M2. Can you use that to work out the answer?

kdbmvp

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Re: Precpitation and ionic concentration
« Reply #2 on: March 01, 2019, 11:58:19 AM »
Thanks! I am familiar with the solubility product, and I am able to work out an answer. What I don't quite understand is where this "excess" [ı-] comes from?

mjc123

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Re: Precpitation and ionic concentration
« Reply #3 on: March 01, 2019, 07:30:30 PM »
There is no "excess" I-. It was present in the initial solution. It doesn't all precipitate. It very nearly all does, so we say it does to a first approximation, but there must always be some left in solution. Suppose the concentration of I- left is x M. Then what precipitated was 0.09 - x M, and the concentration of Ag+ left is 0.01 + x M. So you have (0.01 + x)*x = Ksp. You can solve the quadratic equation for an exact value of x, but x is so much smaller than 0.01 that you can simplify it by assuming that 0.01 + x ≈ 0.01 without any loss of accuracy. (This is the "small-x approximation", and you should always check that the value you get for x is much smaller than 0.01(or whatever), which it clearly is here. If it isn't, you must solve the quadratic.)