Question

If 2.00 ml of 0.100 m pb(no_3)_2 are added to 1.00 l of 0.100 m kcl will a precipitate form?

Answer 1

In our reaction 0.1 M PbCl2 is produced. Which is = 27.82 g in 1 L. Thus it is more then the solubility of PbCl2 and hence precipitate will be formed.

Solution:

Pb(NO3)2(aq) + KCl(aq) ⇄ KNO3(aq) + PbCl2(s)

2 ml of 0.1 M Pb(NO3)2 will react with 2 ml of 0.1 M KCl solution,

Remaining of 998ml of 0.1 M KCl will be unreacted.

PbCl2 Ksp= 1.6 x 10-5

PbCl2 has some solubility, hence whatever amount is getting in to solution it will produce some ions of Pb2+ and Cl- .

   PbCl2 ⇄ Pb2+ + 2Cl-

I   1                  0          0

C 1-x                 x             2x

E   1-x               x              2x

Ksp = [Pb2+][Cl-]2

Ksp = 1.6 x 10-5 = [Pb2+][Cl-]2

1.6 x 10-5 = (x)(2x)2 = 4x3

x^3 = 0.4 x 10-5 =

x = 1.6 x 10-2 = [Pb2+]

Thus the solubility of PbCl2 is 0.016 M or we can say that, 0.016 moles/L or (0.016 mol x 278.2 g/mol) = 4.4 g/L of PbCl2

So if we add 4.4 g in 1 L water it will get dissolved completely to make a saturated solution and further addition of PbCl2 will be precipitated.

In our reaction 0.1 M PbCl2 is produced. Which is = 27.82 g in 1 L. Thus it is more then the solubility of PbCl2 and hence precipitate will be formed.

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