Gem Chem Solubility

[liveoak]

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Kind of a general question...bear with me and help me learn

Abbreviated passage:

The ksp for lead (II) iodine is 1.3 X 10 ^-3

How many grams of lead (II) iodine exist in .1M of 100 ml of the saturated solution?

The answer simply takes the ksp, finds the # of moles, and converts to grams.

So, my question is what does the 1.3 X 10 ^ -3 represent?

The equation is pBI2 - > PH + 2I

correct? Thus, in a saturated solution, how much of each should exist?

thanks.

 

[Preciouzgeek]

I'm not the best at explaining things, but here it goes lol 😛...

Ksp is synonymous to Keq, it's used to find solubility equilibrium (when rate of dissociation = precipitation, thus a saturated solution).

Dissociation reaction: (including ions)
PbI2 --> Pb^(2+) + 2I^(-1)

Dissociation reaction: (w/o ions)
PbI2 --> Pb + 2I

As a result:
PbI2 --> Pb + 2I
Ksp = [Pb]
Ksp = [x][2x]^2

If the question were asking for the concentration of Pb or I, it would be x or 2x, respectively.

But, the question is asking for the concentration of PbI2.
Since Ksp = saturation (dissociation = precipitation), PbI2 = Ksp.

That's why in the answer they just converted the Ksp, which is the concentration of PbI2 in M, to grams.

Hope that helped 🙂

 

[liveoak]

is the PbI2 in the solution a precipitate or not?

i guess that's what is tripping me up.

 

[liveoak]

i am a loser. they gave me the solubility, not the ksp. i read it wrong.