KSP calculations confusing me

[MGA100]

Hey Guys,

I have been going through the destroyer for GCHEM and was just confused as to some calculations for KSP. Ill just provide the examples below as it would be too confusing to try and word what I am struggling with understanding.

In the 2018 Destroyer, Problem 300: A solution of PbI2, ksp = 1 X10^-8 has 1x10^-3 M Pb2+. What is minimum amount of I- needed to cause precipitation to occur?

The solution is as follows: PbI2 --> Pb2+ + 2I- ; then Ksp = [Pb2+] [ I-]^2.....AND SO ON

My question is for the bolded part of the previous line when we are writing the expression for the iodide ions; why are we not writing [ 2I-] instead of [I -]?? I am just confused as to when to include the coefficient of the ions within the ksp equation. Do we only use the coefficient when we are trying to calculate the molar solubility ?

If you could respond at your earliest convenience I would really appreciate it.

--Mike--

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[DAT Destroyer]

Here is a RULE that I give my students......If an ion concentration is given DO NOT USE any x values...simply solve for the given ion. The logic is that the ion amount will far exceed the x values,,,,,,,hope this helps.

Dr. Jim Romano

 

[Elisonen]

Hi Dr. Romano,

So the Pb+2 concentration is [1 x 10^-3]. Shouldn't the [I-] just be double [2 x 10^-3]? Would that be a way of solving this question as well?

Thanks for your help

 

[DAT Destroyer]

Hi Dr. Romano,

So the Pb+2 concentration is [1 x 10^-3]. Shouldn't the [I-] just be double [2 x 10^-3]? Would that be a way of solving this question as well?

Thanks for your help

If you are given an ion concentration, you CANNOT use x and 2x. Since extra ion is added, we must use that concentration only. Ksp is very slightly soluble salts......in this case Pb is x and iodide would be 2x,,,,,,BUT.....if lead is added,,,that will far far exceed the value of the x term. The bottom line is this.....if you are given a concentration, simply solve for the needed ion without using any x values.

Hope this helps.

Dr. Jim Romano