[kdbmvp]

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]

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]

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]

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 = K_sp. 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.)