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
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.
50mL of a 0.002M Pb(NO3)2 solution are mixed with 50mL of a 0.004M KCl solution. Will...
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