The solubility of lead(II) chloride is 0.45 g/100 mL of solution. What is the Ksp of PbCl2?a. 8.5 ? 10^-6b. 4.2 ? 10^-6c. 1.7 ?
1 answer:
The value of of
is 1.7×
.
So, option C is correct one.
Calculation,
Number of moles of = mass/molar mass = 0.45/278.1 = 1.618×
moles
Concentration of and
ion
Molarity of = Number of moles of solute /volume in lit
Molarity of = 1.618×
moles /0.1 lit = 0.0162 M
→
+ 2
For certain amount of , twice this amount of
ion formed.
The concentration of ion formed = 2× 0.0162 M = 0.0324 M
Now,
of
= [
]
= 0.0162 ×
= 1.7×
.
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