Ooo good one. I got that one wrong during my second run through the Destroyer and going to definitely be reviewing it again before my exam because I couldn't remember right away where I went wrong. Firstly, I still solved the problem using ICE tables. Looking at the solution in the back, I remember being confused, too - do I not use ICE??? I'm not sure if Dr. Romano used ICE, but just summarized it into the way he set up the work - or if there is a slightly different (maybe even faster) approach to it? But I didn't want to get out of my comfort zone; I didn't think it was too time consuming to solve it with ICE. Honestly, he pretty much used ICE tables - I imagine setting it up like he did if one is super fast/comfortable with ICE tables...Read my explanation and you'll see what I mean. Look out for when I use "2x" -- that is the same as the [I-] that Dr. Romano uses so when comparing my solution to his, just know that [2x] = [I-].
First, just write the reaction. I always do this because I tend to mess up somehow if I do something in my head. Then, put the ICE table set-up underneath. However, there's a
catch - and this is where I messed up on when I first did the problem myself. Since we are told that the solution of PbI2 "
contains 1x10^-3 M Pb(2+)," this means that "we know we have 10^-3 M of Pb(2+) in the
Equilibrium row of the ICE table.
In terms of how to easier identify such a "tricky" wording and know that this implies E value and when to ignore something - it beats me, unfortunately. I'm sure if I understood it, it would be much easier for me to set up the problem and not be fooled if it came up on the DAT. Hopefully, someone can shed some light on that because that's something I'm weak-ish with, too. Moving on, we use this information in our ICE table...let's recall that the whole point of an ICE table is to focus on the
Equilibrium row - these concentrations can be used to plug into our
Ksp expression. Why? Because K for anything is an
equilibrium constant (Keq, Kp, Ka). Anyways, since we already know our E value of Pb(2+), we can just ignore the I and C rows for Pb(2+).
PbI2 --> Pb(2+) + 2I(-)
I: ---
0
C: ---
+2x
------------------------------------------
E: 10^-3 M +2x
Ksp = [Pb(2+)]*[I-]^2
1.4 x 10^-8 = [10^-3] * [2x]^2
Solve for x...
x = 1.87 x 10^-3 M
But wait... that's not one of the options
That's cause we're not done yet! While yes, we may know what "x" is, let's not forget what the problem is asking for: what is the minimum concentration of I- that must be added to begin precipitate formation? Well, we just need to go back to our
E row in our
ICE table - we already were told how much Pb(2+) we have (aka how much of it forms). How much of the I- do we need to completely react with the 10^-3 M?
2x!
[I-] = 2x = 2 * (1.87x10^-3)
[I-] = 3.7 x 10^-3 M
Note that I may not be
scientifically 100% correct in all my explanation, but I tried the best I could while also just showing you how I went about it. I hope it helped!
