If anyone could offer tutoring for atomic weights, isotopes, and the like, I'll love you foreverrrr.

THANKS!!!  I'll message you now.

Thanks, Yasha. That does help a bit.  It'll get easier when I keep drilling it into my head too.

For the reactants and type of reaction listed, write out the correctly balanced chemical equation. Potassium iodide plus lead nitrate react in a double replacement reaction.

HALP

Thank you so much gpink!

Also, sorry for never emailing you like I said I would; I've always been pressed for time in this class ><

gpink wrote:

This is a fairly common basic chem reaction demonstration. Search Youtube.
Chemical involved are KI and Pb(NO_3)_2 Reaction is double displacement both compounds must swap an element or ion. Ions normally are considered a monolithic block.

Reaction looks like this as a chemical equation.
KI --> Pb(No_3)_2

This is not quite right.  Sorry -- I'm not a chem major, just a high school tutor, but here's how things should look.

In order for KI to combine with Pb(NO3)2, you have to add them; they are both reactants.  So the setup looks like this:

KI + Pb(NO3)2 --> ???

Now, this is a double displacement reaction.  Notice that each molecule has two kinds of ions: potassium iodide has K and I, while lead nitrate has Pb and NO3.  In a double displacement reaction, the ions switch partners.  So our products should be potassium nitrate and lead iodide, pairing K with NO3, and Pb with I.

KI + Pb(NO3)2 --> KNO3 + PbI2

We still have to balance this: we have twice as much nitrate on the left as on the right, and vice versa for iodide, so let's put a 2 in front of the molecules we don't have enough of and see where we are.

2KI + Pb(NO3)2 --> 2KNO3 + PbI2

And now the equation balances.  This is the correct, final answer.

I'm skipping over some details, because in my imagination this is due tomorrow and you don't have time to worry about it.  I'll write out something a little more complete in a minute.

I wasn't in time to help   But you can always fix it later, I guess!  I don't know if the details matter or if you're all done now, but I already wrote out this long post and I'm damn well not deleting it now!

Potassium iodide plus lead nitrate react in a double replacement reaction.

Chem, like math, is a foreign language, and it's best to translate it one word at a time.  Or one phrase, anyway.

Potassium iodide . . .

Let's start with iodide.  That word comes from "iodine," which is written I, but the "ide" on the end is Chem-language for "this is an ion."  If you know your common ions, you'll know that when iodine is an ion, it takes a -1 charge, so iodide is I(-1).  The -1 would be written in "superscript," above and to the right of the I.

Now, potassium iodide must be potassium (which is K) bonded to the iodide.  But since the I has a negative charge, the K must have a positive charge to balance it out.  If you look at your common ions again, you'll see that when potassium ionizes, you get K(+1).  +1 and -1 balance each other out perfectly, so potassium iodide is just KI.

. . . plus . . .

Just what it looks like!  You're adding these together, so "plus" will become + in the equation.

. . . lead nitrate . . .

The "ate" in "nitrate" is another ion marker.  Look up nitrate (or memorize it the way your chem teacher probably told you to ) and you'll see that this one is a little trickier than iodide: nitrate is NO3(-1).  Even though nitrate is named after nitrogen, you'll see that it has some oxygen attached to it.  "Ate" ions are all like that: they all have bonus oxygen on top of whatever element they're named after.

Now lead.  Since it's binding to the negatively-charged nitrate, it must have a positive charge.  But what positive charge?  Check your ion table and you'll see that lead comes in +2 and +4 varieties, but +2 is more common; we'll assume (correctly) that it's +2.  So the lead is Pb(+2).  But wait -- +2 seems too high!  It more than cancels out the -1 we got from the nitrate.  You can't have Pb(+2) bind with NO3(-1); you'd still have a positive charge at the end.

To solve this problem, we need more than one nitrate ion.  If we have two NO3(-1), then the nitrates will give us a total of -2 charge.  That is perfect to cancel out the lead's +2 charge!  So lead nitrate is Pb(NO3)2.  This way of writing things shows how the molecule has only one lead ion, but two nitrate ions.

Let's review our translation so far.  We have:

KI + Pb(NO3)2 . . .

Okay!  Even though it feels like we haven't moved much from where we started, we're actually about half done.  (And the steps we went through above get a lot quicker, like you don't even have to think about it, the more you practice chem; do your homework and I promise combining ions won't always be this painful.)

Now, the problem says:

. . . react . . .

"React" (or "combine") is Chemspeak for -->, the arrow that separates reactants from products in a chemical equation.  Everything to the left of the arrow is a "reactant;" it's what you start with, just as you start with potassium iodide and lead nitrate in this problem.  Metaphorically, it's two pieces of bread and a slice of ham.  Everything to the right of the arrow is a "product;" it's what you end up with (a ham sandwich).  Our job now is to figure out what we should end up with in this reaction.  How do we do that?  By decoding the rest of the problem:

. . . in a double replacement reaction.

This kind of reaction, which is also called a double displacement reaction, works like a slow night on the dance floor: after the dance, you trade partners with the only other couple dancing.  Right now we have K(+1) paired off with I(-1), and Pb(+2) paired off with a couple of NO3(-1)s.  We need to switch these couples' partners!  But in the oppressive, heteronormative world of chemistry, pluses can only dance with minuses, and vice versa.  So the only legal way for these couples to switch partners is to pair K(+1) with NO3(-1), and to pair Pb(+2) with I(-1).  The resulting pairs will be called potassium nitrate and lead iodide.  (See how the words just switched places from the original problem?)

Now, K(+1) is a match made in heaven for NO3(-1): you have a +1 and a -1, and they cancel out perfectly, so KNO3 will be one of our products.  Pb(+2) and I(-1) is a little bit messier.  Lead is too much of a man* to date just one iodide; +2 more than cancels out -1.  Remember how we solved this problem before?  Yeah, it's going to take two I(-1) to match Pb(+2)'s charge.  It's a menage-a-trois for the lucky lead ion.  Lead iodide needs to be PbI2: one lead, two iodide.

* Or woman.  This metaphor is getting a little strained.

Well hey, look at that: we've figured out how to write both our products!  Now we have an almost finished chemical equation:

KI + Pb(NO3)2 --> KNO3 + PbI2

As you can see, the reactants we started with are on the left, and the products we should end up with are on the right.

At this point it's tempting to say "okay, done!"  But you should get in the habit, before you say that about a chem equation, of saying "wait, is it balanced?"  And to our dismay, we find that this one is not balanced!  The left side has only one I versus the two on the right side; the right side has only one NO3 compared to the two on the left.

Sometimes balancing is tricky, but if there's an obvious thing to do, try that first.  In this case the obvious thing (obvious with practice!) is to double how much I you have on the left by writing 2KI instead of just KI, and double how much NO3 you have on the right by writing 2KNO3 instead of just KNO3.  This gives:

2KI + Pb(NO3)2 --> 2KNO3 + PbI2

Let's count.  I see two K on the left and 2 K on the right.  Two I on the left (from the two KI molecules), and two on the right (in the single PbI2 molecule).  One Pb on the left, one Pb on the right.  Two NO3 on the left, two NO3 on the right.  Hey, it balances!  We're done!  Someone call for the motherfucking champagne!



---

Before I add one more note, let me say that based on the students I've tutored, this stuff pretty much terrifies everyone the first few times they see it.  Totally normal to be a little wtf at this point.  But in my experience, this gets easier so fast you don't even know.  You'll be solving these in a minute flat in a couple months, if you keep at it.  (And you'll need to, since these reactions are often building blocks of larger, harder problems.)  The important things to practice are A) turning the name of a compound like lead iodide into a chemical formula like PbI2, and B) using the types of reactions, like double displacement, to predict the products of a reaction.

Last thing, and then I promise I'll leave you alone   A lot of students get confused about the difference between putting a 2 in front of a molecule (like 2KI, in the final reaction above) and using 2 as a subscript of an atom (as in PbI2).  The difference is that the 2 in PbI2 indicates that there are two iodides, but still only one lead; by contrast, the 2 in 2KI indicates that you have two entire KI molecules, so there are two potassiums and also two iodides.  Let me try that again.



You use the subscript (like PbI2) when you're building a molecule out of ions; this is the phase where we figured out that it takes two iodides, each with a -1 charge, to match the +2 lead ion.  But you use the coefficient (like 2KI) when you're balancing an equation.  By the time you're balancing the equation, it's too late to change the subscripts; you have the molecules you have, and all you can change is how many of them you use.

If you (or any other frustrated chem students!) have any questions or anything, ask away, or shoot me a PM!

winksniper wrote:

THANK YOU LOADS, SAT!

My pleasure.  Always happy to help -- anything other than work on my term papers.  (I think that from the standpoint of a complete novice, understanding the ideas in the chem problem we just did is probably harder than understanding the ideas in my grad-school term papers, but the term papers take longer to write.)

@winksniper
Sorry about that. I wasn't thinking too clearly. I saw the problem and jumped on it. Hope I didn't hurt your grade too much.

*Sorry to Double Post*

Alright, friends, my last few assignments in Chemistry have been graded and I received a 46% (x__x) on the Classifying Chemical Reactions assignment with the potassium iodide.  I take all the credit for the bad grade, however, because 2KI+Pb(NO3)2-->PbI2 + 2KNO3 was counted correct! 

I am now going back and making the effort to raise my final grade (62% ;___; ) by re-doing a lot of my assignments.

Now the assignment I got the absolute lowest grade EVER on (an ASTONISHING 7%; I am in complete shock that I got such a low grade when I usually get As and Bs - this just goes to show how much I don't belong in an online college prep chemistry course.  ><) is one about Ionic and Net Ionic equations.

They gave two websites to learn from:
www.chem.vt.edu/RVGS/ACT/notes/Notes_on … _rxns.html
&
web.fccj.edu/~ksanchez/1032/wksheet/IonicEq.html

From just these two sites, I clearly couldn't get enough of a grasp on the subject.

Can someone please help clear this up for me?

I can eve give an example problem, if that would help.

Silver is added to aqueous copper(II) sulfate. For this reaction, write the molecular equation, the full ionic equation, and the net ionic equation. [Hint: Be sure to watch the physical state of each element or ion!]

satyreyes wrote:

Wow, you're learning how to do this stuff from vaguely worded random websites your teacher googled.  Man, no wonder this is hard.  Okay!  I can't get to a place where I can write this up in a helpful way until this evening, but in the meantime, could you answer three questions for me, so I can understand how much detail you need?

1) Do you understand the concept of an ion?  For example, do you know the difference between Na and Na(+1)?

2) Do you have a table of common ions?  If I say "phosphate," do you have somewhere you can go to find out what phosphate is?  (Even Wikipedia is better than nothing, but it tends to go into much too much detail; I really mean something more like this.)

3) And specifically, for the example you gave, do you know (or can you guess) how to write "silver is added to aqueous copper(II) sulfate" using chemical symbols?

Thanks!  I'll explain what's up with problems like this one later.

Yeah, it suuuucks.  DX

1) Yes.  Na is just sodium whereas Na(+1) is sodium with a positive charge.

2) Yes.

3) Uh, I can guess.
Ag + Cu2S(aq)?

Hiraku is spot-on with the abridged version   I'll go over it in a little more depth, from scratch.

Wink, you're right that silver is just Ag.  There's no indication in the problem that it's an ion, so it doesn't need any + or -.  Copper sulfate, though...

Copper (II) sulfate

Copper (II) means the copper ion with a +2 charge, or Cu(+2).  (The number in Roman numerals gives you the positive charge of certain metals.)  Sulfate, as your common-ion table will tell you, is SO4(-2).  (Note that it's not just S; a sulfur ion by itself is called sulfide, not sulfate.)  So we have Cu(+2) and SO4(-2).  +2 and -2 cancel, so CuSO4 is the complete formula for copper (II) sulfate.

Now we have the equation:

Ag + CuSO4 --> ???

First let's finish the molecular equation -- that is, figure out what molecules will be our products.  Spoiler warning: we can only end with the atoms we started with!  The ion SO4(2-) needs to stay attached to something, and if it's not copper, it must be the only other element in the equation: silver.  So silver sulfate must be one of our products, and copper must be the other one, since it's all that's left.

Ag + CuSO4 --> copper + silver sulfate

Now, how do we write these products?  Copper is easy: it's just Cu.  But wait a minute: wasn't it an ion, Cu(+2), a second ago?  Yes, it was!  During this reaction, copper has lost its +2 charge and now has zero charge.  It's just copper metal.

What about silver sulfate?  Well, silver had zero charge when it was a reactant, but it must have become an ion in order to bond with sulfate.  Silver ions, as your table will tell you, have a +1 charge: Ag(+1).  We need to bond that with sulfate, SO4(-2).

Quiz: based on those ions, would you expect silver sulfate to be AgSO4, Ag2SO4, or Ag(SO4)2?  I encourage you to work it out before reading on!

Okay, done?  We need the total positive charge of the silver to match the total negative charge of the sulfate, right?  The sulfate's charge is -2, and silver is only +1 apiece.  We need two silvers to get a total of +2, which will match the sulfate's -2.  So the right answer is Ag2SO4.

Ag + CuSO4 --> Cu + Ag2SO4

Balance:

2Ag + CuSO4 --> Cu + Ag2SO4

And that's the molecular equation.  So far, everything we've done is pretty similar to that double displacement problem we did last night.

Now, a full ionic equation is the same thing, only you split out all of your ionic compounds into the ions they're made of.  For instance, instead of writing copper (II) sulfate as CuSO4, you write Cu(+2) + SO4(-2).  Unlike molecular equations, ionic equations tend to have pluses and minuses in them.  In this case, split up the compounds copper (II) sulfate and silver sulfate, and you get:

2Ag + Cu(+2) + SO4(-2) --> Cu + 2Ag(+1) + SO4(-2)

Notice that the silver ion in the products got a 2 in front of it, because we needed two silvers to balance the sulfate earlier.  This is your full ionic equation.

Finally, you need a net ionic equation.  That's just the full ionic equation, but with anything that didn't change between the reactants and products step crossed out.  Take a look at the full ionic equation: can you spot the ion that didn't change?

SO4(-2), the sulfate ion, is the same on the left as it is on the right.  The net ionic equation should leave it out.  So the net ionic equation is:

2Ag + Cu(+2) --> Cu + 2Ag(+1)

This equation illustrates how silver went from 0 charge to +1 charge during the reaction, while copper went from +2 charge to 0 charge.

I think that should answer your question.  Let me know if you've got any more