In: Chemistry
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?
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.