In which aqueous system is PbI2 least soluble and why?
[A] 0.5 M HI [B] 0.2 M HI [C] 0.8 M KI
I understand that I- is the common ion. I don't understand why [C] 0.8 M KI is the correct answer.
Thanks for telling me what the problem is. That helps a lot.
This is the common ion effect.
HI is a strong acid. On ionization it give a (I^-) = 0.5 M for A and 0.2 M for B and 0.8 M for C. So here's the way it works, and I'll use the 0.8 M KI for the reasoning.
.......................PbI2(s) ==> Pb^2+ + 2I^-
I........................solid.............0...........0
C.......................solid.............x...........2x
E........................solid.............x...........2x
So, if I want to know the solubility of PbI2 in a solution, you see it is
Ksp = (Pb^2+)(I^-)^2 or (x)(2x) = 4x^3 = Ksp.
But, when you add a common ion, in this case I'm using KI, the KI (its true the other HI too) ionizes completely and it dissolves completely to give
..................KI(s) ==> K^+ + I^-
I..................0.8 M.......0........0
C.................-0.8 M.....0.8M.....0.8 M
E....................0.........0.8..........0.8 M
So now that Ksp expression looks this way:
Ksp = (Pb^2+)(I^-)^2
For Pb you substitute x.
For I^- you substitute x from PbI2 and 0.8 from the KI so the total is x+0.8
Ksp = (x)(x+0.8) so now (neglecting the quadratic because x is small) you have x = solubility = Ksp/0.8 = much larger number than above. The bottom line is this.
.......................PbI2(s) ==> Pb^2+ + 2I^-
I........................solid.............0...........0
C.......................solid.............x...........2x
E........................solid.............x...........2x
Adding the I^- as a common ion, you INCREASE the 2I^- on the right which forces the equilibrium to the left to reduce the I^- anyway it can and that makes the PbI2 solid much less soluble. C is the answer because the I^- in KI is the largest of the three.
Explanation is long because of the typing but it's all there. Follow up if necessary. I'm here for another few hours.
oops.
I didn't proof when I should have proofed and this
"For Pb you substitute x.
For I^- you substitute x from PbI2 and 0.8 from the KI so the total is x+0.8
Ksp = (x)(x+0.8) so now (neglecting the quadratic because x is small) you have x = solubility = Ksp/0.8 = much larger number than above." should have been this.
"For Pb you substitute x.
For I^- you substitute x from PbI2 and 0.8 from the KI so the total is x+0.8
Ksp = (x)(x+0.8)^2 so now (neglecting the quadratic because x is small) you have x = solubility = Ksp/(0.8)^2 = much larger number than above when solving for x = solubility PbI2." Sorry about that.
To determine which aqueous system will result in the least soluble PbI2, we need to consider the common ion effect. PbI2 is less soluble in the presence of a common ion, which in this case is the iodide ion (I-).
In options [A] and [B], we have varying concentrations of HI, which provides the common ion (I-) necessary for the common ion effect to occur. Meanwhile, option [C] contains KI, where KI provides the potassium ion (K+) but not the common ion (I-).
The common ion effect suppresses the solubility of a salt by introducing additional amounts of the common ion into the solution. As a result, the equilibrium position of the dissolution reaction is shifted towards the solid salt, reducing its solubility.
Since KI in option [C] does not provide the common ion (I-) required for the common ion effect to occur, it will not have a significant impact on the solubility of PbI2. Thus, option [C] 0.8 M KI is the correct answer as it will result in the least soluble PbI2.
To determine in which aqueous system lead iodide (PbI2) would be least soluble, we need to consider the common ion effect. In this case, the common ion is iodide (I-). The solubility of a compound decreases when it shares a common ion with another compound in the solution. Therefore, the aqueous system with the highest concentration of the common ion will result in the least soluble lead iodide.
Let's compare the three options provided:
[A] 0.5 M HI: Here, we have hydroiodic acid (HI), which completely dissociates into H+ and I- ions in water. The concentration of I- ions in this solution is 0.5 M.
[B] 0.2 M HI: Similar to option [A], the concentration of I- ions in this solution is 0.2 M.
[C] 0.8 M KI: In this case, we have potassium iodide (KI), which also dissociates in water, releasing K+ and I- ions. The concentration of I- ions in this solution is 0.8 M, which is higher than the I- concentrations in options [A] and [B].
Since option [C] has the highest concentration of the common ion (I-), it will result in the least soluble lead iodide (PbI2). The increased concentration of I- ions in the solution reduces the solubility of lead iodide, leading to its precipitation.
Therefore, option [C] 0.8 M KI is the correct answer because it provides the least soluble environment for PbI2 due to the higher concentration of the common ion, I-.