# DAT Destroyer GC #285

#### jeromedr

Hi all, could someone explain in question 285 of GC (2016 version) where the equation 2x=1 came from in the ice table, i am getting lost at this step...

another quick question does anyone have an easy way of determining when an ICE table should be used and plugged into a KSP equation and when not to use one, it seems like half the problems i do i should have plugged in a (2x) into the Ksp equation instead of just x or vice versa.

any help is appreciated

Last edited:

#### moose786

2+ Year Member
2x=1 comes from the "e" portion of the ICE table - specifically referring to the product that formed (2NH3).
Original Equation is N2 + 3H2 <--> 2NH3
Once you set up your ICE table, you should end up with:
"4-x" under "N2"
"4-3x" under "3H2"
"+2x" under "2NH3"

When we say 2x = 1, we are referring to the last bit of information that was given in the question: "2 moles of NH3 are formed". Since we are in a 2 liter flask, this means that we can simply divide our 2 moles of NH3 by the volume: 2 mol NH3 / 2 L = 1M NH3

Remember that we are almost always talking about molar concentrations in our ICE tables. Now, just put our 2 bits of information together to help us figure out what the "x" value is in our ICE table. Since our ICE table tells us that we will end up with "+2x" molar of NH3, and we used our given to figure out that there is 1M NH3 formed in the end, we just set 2X = 1M (since they are the same thing).

x = 0.5

Now, use this value to plug in for the other expressions in your ICE table. All simply plug/chug algebra from there. Let me know if that was clear.

---------------

As for the Ksp question you had, I think you might have a bit of a confusion between Keq expressions vs. Ksp expressions (something I'm not super awesome at either, but I can usually get by using some little tips). Whenever I see Ksp in the problem/question, I immediately think ICE tables. I think mostly what I've come across for Ksp has involved ICE tables. Also, Ksp just involves the products (hence, solubility product constant) - and the reason for this is important to understand because you won't ever ask yourself again if you should "include reactants on the bottom, too" in your Ksp expression. That is something that I used to have trouble on. Anyways, Ksp for our purposes is used for compounds that are insoluble. Hence, when we write the Ksp expression, we should keep in mind that the compound we are writing the reaction for is insoluble. However, everything is a little soluble. This is why we set up the equation in the first place to show that a little bit of the "insoluble" compound can actually dissociate. In addition to this, I also keep in mind an "exception" for Ksp in terms of setting up the ICE table - it is to pretty much always set up the ice table so that I end up "doubling twice," for lack of better wording. I'm talking about the whole Ksp = [x]^a*[y]^z. And how we usually have the problem set up in such a way that you have to do Ksp = (2x)^2*(x) = 4x^3. To be honest, I always forget the legitimate explanation behind why we have to pretty much do (2x)^2 for a lot of the Ksp problems. Whenever I come across the explanation, it makes sense and all but for some reason, I always forget it lol. Hopefully, you get what I'm referring to, though. And hopefully, I helped answer a bit of your question? My thoughts on Ksp are pretty cluttered and it's not a strong suit of mine so I can't really help explain it in other terms. Someone else, please explain with better wording, if possible haha. I may have confused you more from this, so definitely ignore it if it didn't help. No hard feelings from me lol

OP
J

#### jeromedr

thank you! that does help , i have a specific example that could possibly help clarify,

#264 GC

Ksp of PbI2 = 1.4E-8 and concentration Pb2+ = 1E-3

Now next i would do:
PbI2----> Pb2+ + 2I-

straight into the Ksp expression

1.4 x 10^-8 = (1 x 10^-3) [i-]^2

my question is why is no ice table used, i would have thought that the value used for the concentration of I- would have been "2x^2" coming from an ice table..?

Last edited:

#### moose786

2+ Year Member
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!

#### moose786

2+ Year Member
Ahhhh I put lots of spaces and made a neat little table in my original post, but I guess SDN automatically gets rid of all those extra spaces. Hopefully you can still make sense of the ICE table and under which products the variables I put go

OP
J

#### jeromedr

ahhh thank you that totally makes sense, so you pretty much are using 2x in the ksp equation because you have to multiply by 2 at the end to account for the 2I- , i understand now , the only thing that could still throw me off is when you would plug that "2x" into the ksp equation and square it, in some problems explained in destroyer it becomes 4x^2 ... i hope this is not repetitive question
maybe I'm making a math mistake??? but when you would solve for x in the Ksp it would become 4x^2 correct???

(1.4 x 10^-8)/(1 x 10^-3) = 4x^2

i may be making a stupid mistake but for some reason this is how i would do it if i were going about it and this doesn't seem to be correct

all of the help is greatly appreciated!

DAT is in 2 weeks from today i have done boot camp and destroyer so much that it is getting to the point where i have 90 percent of the destroyer book pretty much memorized lol ... any other suggestions on preparing in these last two weeks i am just reviewing biology chapters and continuing practice problems and full length practice tests.

would love to hear other ideas for last 2 weeks

Last edited:
OP
J

#### jeromedr

ahh nvm im dumb I'm making a math mistake idk why i got so tripped up thank you for the help it totally makes sense now

#### moose786

2+ Year Member
Bro, mine is in a week and you're in a much better boat than me LOL. I'm so scared for the Bio...so so scared. I know luck is definitely a factor many times in Bio - I just hope I get really lucky. Hopefully, like in Slumdog Millionaire where I specifically recall learning the fact for each Bio question during my studies.

orgoman22