help please?

Question

why PbCl2 dissolved when water is added? what was the effect of water on the concentration of Pb2+ and Cl-?

Answer 1

PbCl2 dissolves when water is added because the solvation of the separate ions is a stronger interaction than the attraction of cation and anion.

The concentration of the two species will be dependent on the starting mass of PbCl2 and how much water is added, but the concentration of Cl- will be twice that of Pb2+.

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Answer 2

PbCl2 has low solubility product 1.70×10-5 (ref 1)
Ksp = 1.70x10-5 = [Pb2+]x[Cl-]^2

With solid PbCl2, dissolution due to solvation is small; sparingly soluble. but n(cl-) = 2 x n(Pb2+); [Cl-] = 2 x [Pb2+]

Ksp = [Pb2+] x {2x[Pb2+]^2 = 4[Pb2+]^3 = 1.70x10-5

So [Pb2+](aq) = (cube root) 1.70x10-5 = 0.0257 M
= concentration of PbCl2 solution - quite low.
m(PbCl2) per dm-3 (or litre) of solution is

n=m/M
n = m/278.106
m = 278.106 x 0.0257
7.1473242 g per dm3

So, at 25̊C, 7.14 g of PbCl2 dissolves in a dm3 of water, double the water, another 7.14 g dissolves, so long as there is an excess of PbCl2(s) when equilibrium is reached, where the concentration of Pb2+ and Cl- return to the same values as before the extra water was added.

Now, assuming that you have the equilibrium established, and decant some transparent solution, adding either (or both separately) of a solution of Pb2+ solution or a Cl- solution will result in a cloudy effect which will settle out as PbCl2 solid

Answer 3

I think that this is a question about observing what happens to a saturated solution of the (pretty insoluble) lead(II) chloride when water is added.

Solid PbCl2 is in equilibrium with its aqueous ions:

PbCl2(s) <----> Pb2+(aq) + 2Cl-(aq)

If you add water, the concentration of the aqueous ions on the right hand side of this equation goes down, so, by Le Chatelier's Principle, solid PbCl2 dissolves to raise their concentration again.