Question

If 2.00 mL of 0.100 M Pb(NO3)2 are added to 1.00 L of 0.100 M KCl will a precipitate form? What will be the precipitate?

Answer 1

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, rest of 998ml of 0.1 M KCl will be unreacted.

PbCl2 Ksp= 1.6 x 10^-5

PbCl2 has some solubility, hence whaever amount is getting in to silution 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

x³ = 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 desolved comletly to make a saturated solution and further addition of PbCl2 will be precipiteted.

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.