Question

50mL of a 0.002M Pb(NO3)2 solution are mixed with 50mL of a 0.004M KCl solution. Will a precipitate form? Why or why not? PLEASE SHOW WORK

Answer 1

Solution:

Lead nitrate reacts with potassium chloride to form lead chloride as,

Pb (NO3)2 + 2KCl = PbCl2 + 2KNO3

For precipitation, the ionic product (Qsp) must be greater than solubility product (Ksp).

Number of moles of KCl = molarity x volume in L

= 0.004 M x 0.050 L = 0.0002 mol

Number of moles of Pb(NO3)2 = molarity x volume in L

= 0.002 M x 0.050 L = 0.0001 mol

Since, 1 mol of Pb(NO3)2 is completely reacted with 2 mol of KCl to form 1 mol of PbCl2.

Hence, 0.0001 mol of Pb(NO3)2 is completely reacted with 0.0002 mol of KCl to form 0.0001 mol of PbCl2.

Thus, molarity of PbCl2 = number of moles / Total volume in L

= 0.0001 mol / 0.050 L + 0.050 L = 0.0001 mol / 0.100 L

= 0.001 M

PbCl2 is dissociated as,

PbCl2    = Pb2+ + 2Cl-

0.001 M ---0.001 M -----2 x 0.001 M

Qsp =( Pb2+) x (2 Cl-)^2

= (0.001) ( 0.002)^2 = 4 .0 x 10^-9

Ksp for PbCl2 = 1.6 x 10^-5

Since, Ksp > Qsp, hence precipitate will not formed.