Posted by **Sigurd** on Tuesday, April 8, 2008 at 8:21pm.

I really don't understand how to set this up. Any help is really appreciated! I really don't quite understand how to set up

a caluculation to figure out whether a precipitate will form and i'm struggling with the second half too!

**When 25.75 mL of 0.00826 M lead II nitrate is mixed with 75.10 mL of 0.0183 M sodium chloride solution, will a precipitate of PbCl2 form? Show all calculations to prove your answer. The ksp for lead II chloride is 1.7 x 10^-5.**- Ksp/Ka Questions-Chem -
**DrBob222**, Tuesday, April 8, 2008 at 8:47pm
Just remember what Ksp is!!

Ksp, in this case for PbCl2, = 1.7 x 10^-5. What does that mean? It simply means that (Pb^+2)(Cl^-)2 MAY NOT EVER EXCEED 1.7 x 10^-5. So you calculate the product of those concns, (raise each to the power, of course, in the balanced equation) and if that number is higher than Ksp, a ppt will occur. If that number is less than Ksp, no ppt will occur. If that number is the same as Ksp, then the solution is exactly saturated and the next Pb ion and Cl ion added to that solution will ppt (1 molecule of PbCl2).

mols lead (II) nitrate = M x L = ??

mols NaCl = M x L = ??

(Pb^+2) = mols/L (remember to add mL lead nitrate to mL NaCl to find total mL of solution. We assume the volumes are additive.

(Cl^-) = mols/L.

Then (Pb^+2)(Cl^-)^2 = ?? and compare with Ksp.

Let me know if this isn't clear.

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