Using results of problems #7 and #8, compare the solubility of PbI2 in .10 M NaI
ID: 1012064 • Letter: U
Question
Using results of problems #7 and #8, compare the solubility of PbI2 in .10 M NaI with solubility of PbI2 in pure water, and rationalize why there is such a difference.
(#7 Lead (II) Iodide, PbI2, has Ksp=7.9*10^-9. Using this information,
a)calculate the solubility of PbI2 in water (mol/L)
KEY: 0.0013 M
b)Convert the solubility of PbI2 to g/L
KEY: .60 g/L
#8 Lead (II) iodide, PbI2, is now being dissolved in .10 M solution of NaI. The Ksp of lead(II) iodide is still = 7.9*10^-9.
a)set up the ICE table that corresponds to equilibrium in solution
—confused how to do this one.
b)determine the molar solubility of PbI2 in .010 M NaI.
KEY: x=7.9*10^-7 )
Thank you
Explanation / Answer
7a) The Ksp for PbI2 is 7.9*10-9. PbI2 undergoes dissociation in water as per the equation
PbI2 (s) <=====> Pb2+ (aq) + 2 I- (aq)
-- x 2x
Since PbI2 is sparingly soluble in water, we can write down the equilibrium constant as
Ksp = (x).(2x)2 [the concentration of a solid species is taken as unity]
===> 7.9*10-9 = x.4x2
===> 7.9*10-9 = 4x3
===> x3 = 1.975*10-9
===> x = 1.254*10-3 1.3*10-3 = 0.0013
Now, the molar ratio of PbI2:Pb2+:I- is 1:1:2; therefore, the molar solubility of PbI2 in water is the same as the molar solubility of Pb2+ in water. Therefore, the molar solubility of PbI2 in water is 0.0013 M (ans).
b) The molar mass of PbI2 is 461.01 gm/mol. Now 0.0013 M PbI2 solution = 0.0013 mole PbI2 in 1 L water = (0.0013 mol)*(461.01 gm/1 mol) = 0.599 gm 0.6 gm PbI2 in 1 L water.
Thus, the solubility of PbI2 in water is 0.6 gm/L (ans).
8a) The Ksp is the same in this case also, but now there is a common ion, I-. The concentration of NaI is 0.1 M. This concentration of NaI is much higher than the concentration of I- furnished by the dissociation of PbI2; hence the concentration of I- is 0.1 M. We now set up the ICE chart (this is a mathematical relation between the initial and the equilibrium concentrations of PbI2).
PbI2 (s) <======> Pb2+ (aq) + 2 I- (aq)
initial x 0 0.1
change -x + x + 2x
equilibrium 0 x (0.1 + 2x)
This is the ICE chart.
b) Now, we have, Ksp = (x).(0.1 + 2x)2
We need to simplify the expression by noting that x<<0.1 M; thus (0.1 + 2x) 0.1.
Therefore,
Ksp = x.(0.1)2
===> 7.9*10-9 = x.(0.01)
===> x = (7.9*10-9)/0.01 = 7.9*10-7
As above, the molar ratio of PbI2:Pb2+ is 1:1 and thus, the molar solubility of PbI2 in 0.1 M NaI is 7.9*10-9 M (ans).
The ratio of the solubility of PbI2 in water and 0.1 M NaI is
[PbI2]W/[PbI2]NaI = (0.0013)/(7.9*10-7) = 1645.569 1645.6
The ratio of the solubility is high and its easily seen that PbI2 is much more soluble in water than in 0.1 M NaI. The reason for this can be explained in terms of Le Chaitilier principle. As per Le Chatilier, if the concentration of a substance at equilibrium is altered, the equilibrium modifies or alters itself in such a way that the effect of the change is nullified. We must realize that solubility product, Ksp is an equilibrium constant and hence, it is only temperature dependent. Since the temperature of the experiment is not changed in problems 7 and 8, the value of Ksp remains the same. Now, in 0.1 M NaI, we are adding a huge excess of I-. Thus, the second term in the Ksp expression above increases. To keep Ksp constant, the first term of the expression must decrease. This leads to a decrease in the molar solubility of PbI2 (since PbI2:Pb2+ = 1:1).
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