Let's begin by finding the solubility of lead (II) chloride
PbCl2 in water from its solubility product constant
(KSP).
The dissolution of lead (II) chloride can be represented by the following equation:
PbCl2(s)⟶Pb2+(aq)+2Cl−(aq)For the equilibrium:
KSP=[Pb2+][Cl−]2Let
s be the solubility of
PbCl2 in
mol∕L. From the dissociation equation, it can be seen that the concentration of
Pb2+ ions in solution is
s and the concentration of
Cl−ions is
2s. Insert these into the solubility product expression:
KSP=(s)(2s)2=4s3Given that
K sp=3.2×10−8, we can solve for
s :
4s3=3.2×10−8s3=s3=8×10−9 s=3√8×10−9=2×10−3mol∕LThis is the molar solubility of
PbCl2. Next, calculate the amount of water required to dissolve
0.2g of
PbCl2.
The molar mass of
PbCl2 is given as
278g∕mol. To find the number of moles of
0.2g of
PbCl2 :
n==≈7.194×10−4molSince the solubility of
PbCl2 is
2×10−3mol∕L (from the previous calculation), the volume of water
V needed to dissolve these moles can be calculated by:
V==| 7.194×10−4mol |
| 2×10−3mol∕L |
=0.3597LConvert this to milliliters:
0.3597L=359.7mLThus, the volume of water needed to dissolve
0.2g of
PbCl2 is approximately
359.7mLSo, the correct answer is Option B -
359.7mL.