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rushwr · Answer

hold on i'm finding my log book

rushwr · Answer

I think I got it wrong But what's the answer given ? I got it as the 1st one !

australopithecus · Answer

Where,

aA ⇌ cC + dD

You get:
$$K_{sp} = [C]^c[D]^d$$

Where Ksp is just the solubility constant

You know the solubility constant is: 7.1 x 10^-9
You know that PbI2(s) ⇌ 2I-(aq)  + Pb+2(aq)


This is kind of an ICE table question.

The solubility of PbI2 depends on the following relation:

[PbI2] = [Pb2+] = [I-]
  x             x          2x

Where x is the solubility of PbI2

So you use the equation I provided you with earlier using x

7.1 x 10^-9 = [x][2x]^2

BUT WAIT!

Notice that there is already 0.4 M KI in solution so,

KI ⇌ K+ + I-

Therefore you have 0.4 M I- in solution already before you dissolve the PbI2 this is important you need to include this in your problem it is going to have an effect on PbI2 solubility. All you need to do is add!

We know that there is 0.4M of I- in solution already from the dissociation reaction

so,

7.1 x 10^-9 = [x][2x]^2

becomes:

7.1 x 10^-9 = [x][2x+0.4]^2

Solve for x and you have your answer

australopithecus · Answer

Did you go through how I got the formula for the solution?

australopithecus · Answer

http://www.wolframalpha.com/input/?i=7.1+x+10^-9+%3D+ [x][2x%2B0.4]^2

australopithecus · Answer

cut and paste it to your address bar