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Topic: Ions in Aqueous Solutions and Colligative Properties  (Read 8752 times)

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Offline Kimyko

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Ions in Aqueous Solutions and Colligative Properties
« on: April 16, 2007, 01:57:01 AM »
Well, I think that's what it is. At least, that's what it says in the heading of my homework. I think it's also a mix of several other things we learned in the past...'Cause while, the title says 'Ions in Aqueous Soluations and Colligative Properties,' I see a lot of other things, but I don't think I could have fit it all into the title. Uhm, yeah, anyway, I simply have a couple of problems that I don't really understand; there are a lot of them, but obviously, you don't have to answer them all or anything. I just thought I'd put it up there, and maybe, in the process of writing out my process, I can figure it out on my own...

Anyway, here goes nothing, I guess...

1. 1.0 mol of magnesium acetate is dissolved in water.

a) Write the formula for magnesium acetate.


My answer: Mg(CH3COO)2

Process: I don't really know, that's just what I got. Writing out these things have never really been my strong point, along with everything else in regards to Chemistry, so, to be honest, I just guessed.

b) How many moles of ions are released into solution?

My answer: 3

Process: I wrote out what I think was the equation for it (though, obviously, it's probably wrong), which turned out to be.

Mg(CH3COO)2 ----> Mg2+ + 2(CH3COO)-

So, add them all up, there would be three moles, right?

c) How many moles of ions are released into a solution made from 0.20 mol magnesium acetate dissolved in water?

My answer: ?

Process: ...I really don't know. I don't know what's it's asking, or how to get there. I've just been staring at it, and feeling incredibly stupid because I'm sure there's a really easy answer that I've been overlooking because I'm thinking too hard or something.

2. In the following two precipitation reactions, write the formula for the precipitate formed:

a) Combining solutions of magnesium chloride and potassium phosphate.


My answer: Mg3(PO4)2

Process: So, like everything else, I'm not really quite sure, so I stumbled through this rather awkwardly. First, I wrote out the equation because I felt like I had to and got this:

3MgCl2 + 2K3PO4 ----> Mg3(PO4)2 + KCl

(I assumed it was a double replacement reaction, so...)

And then I looked at the chart we were supposed to memorize and found that phosphates were insoluable and chlorides were soluble, and then I looked in my book at an example problem, and since the insoluble one formed the precipitate, I guessed that it was the magnesium phosphate was the answer.

b) Combining solutions of sodium sulfide and silver nitrate.

My answer: No solution.

Process: Same as above. I just took the equation I got, which was:

Na2S + AgNO3 ----> Ag2S + NaNO3

But when I looked at the chart, it said that both of were insoluable, so then, I got stuck and decided it was a no solution...

Though, I know for sure that that is wrong.

5. a) Write the net ionic equation for the reaction that occurs when solutions of lead(II) nitrate and ammonium sulfate are combined.

My answer: ?

Process: Pb(NO3)2 + (NH4)2NO3 ---->

...And then I froze. I know now you're supposed to do something where you write out each of the ions, but...I'm not really sure how you do it, and everything that I did just seemed really strange.

b) What are the spectator ions in this system?

My answer: ?

Process: Well, seeing as how I never got an answer to the previous one, I don't really know the answer to this. I know what spectator ions are...I just can't seem to find it for this, particular equation.

6. The following solutions are combined in a beaker: NaCl, NabPO4, and Ba(NO3)2.

b) Will a percipitate form when the above solutions are combined? If so, write the name and formula of the precipitate.


My answer: Uh...No, because all of the solutions are...insoluble?

Process: Well, as it seems, I am completely useless when it comes to all things chemistry. I thought that in order to get the answer, you would write some kind of equation, but there are three of them, and it completely threw me off because I really don't know how to do ones like that. So I guessed. That was pretty much my process.



Okay, that's pretty much all I have out so far. There are so many other questions I need to ask, but, I haven't actually put in enough effort in figuring them out, so I feel bad about posting them right now, but I might later (on this thread, of course.) I know I put up a lot of questions, and I don't expect them all to get answered, but I would really appreciate all the help I can get.

Maybe, if someone could please explain to me the ionic equation things as well, I would really appreciate it. I mean, I understand what they are, what spectator ions are, I just can't seem to apply it to problems...I guess. Anyway, thanks again for all your help...

Offline AWK

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Re: Ions in Aqueous Solutions and Colligative Properties
« Reply #1 on: April 16, 2007, 02:58:59 AM »
1, a b - OK
1c see 1b and find scale factor
2a 6KCl on the right side
2b unbalanced
5a uncorrect ammonium sulfate formula, solid PbSO4 ia formed
6 Solid Ba3(PO4)2 can be formed
AWK

Offline Kimyko

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Re: Ions in Aqueous Solutions and Colligative Properties
« Reply #2 on: April 16, 2007, 06:55:52 AM »
Scale factor? I don't even vaugely recall learning that...I'm sorry, do you mind explaining what it is? Maybe I learned it under a different name, or I forgot of something.

Uhm, about the unbalanced thing, I was wondering if that mattered when it comes to determining the precipitate. Obviously, it matters because the formula is incorrect if you don't balance it, but if you're just trying to figure out if it's a precipitate or not, is it as important?

Anyway, thank you. ^^;

Offline Dan

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Re: Ions in Aqueous Solutions and Colligative Properties
« Reply #3 on: April 16, 2007, 07:27:42 AM »
Scale factor? I don't even vaugely recall learning that...I'm sorry, do you mind explaining what it is? Maybe I learned it under a different name, or I forgot of something.

A scale factor is very logical and not mysterious.

Ok.
If you have 5 apples, and you chop them all into 3rds, you have 15 3rds.
So, how many 3rds do you get from one apple? The answer is obvious, but consider that if you start with 1/5 of the amount of apples in the first example, you get 1/5 the number of 3rds when you chop it up.
so If you have 1 apple and chop it into 3rds, you have 15/5 3rds = 3 3rds.

Ok. so now we have 1 mol of magnesium acetate, we dissolve it and get 3 mol of ions.
So what do we get if we dissolve 0.2 mol? Well, 0.2 is 1/5 of 1, so use your apple logic to solve this...

Quote
Uhm, about the unbalanced thing, I was wondering if that mattered when it comes to determining the precipitate. Obviously, it matters because the formula is incorrect if you don't balance it, but if you're just trying to figure out if it's a precipitate or not, is it as important?

You're right, it's not necessary to answer the question, but it's good practice. Whenever you write an equation, balance it - you will get better at balancing equations.

To add to what AWK said,

2b) Sodium Nitrate is soluble.
5a) Take it one step at a time. Write the full formula for what is happening.
Then identify the solvated ions. Remove the ions that do not do anything, ie. the ones that do not form the precipitate (these are the spectator ions).

Heres an example

NaCl(aq) + AgNO3(aq) --------> NaNO3(aq) + AgCl(s)

Ok, so all the aqueous compounds are solvated ions:

Na+(aq) + Cl-(aq) + Ag+(aq) + NO3-(aq) ----> Na+(aq) + AgCl(s) + NO3-(aq)

Ok. The sodium and nitrate ions are the same on both sides, so they are the spectator ions. We remove them from the above equation, which gives us the net ionic equation:

Cl-(aq) + Ag+(aq) ------> AgCl(s)
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Offline Kimyko

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Re: Ions in Aqueous Solutions and Colligative Properties
« Reply #4 on: April 17, 2007, 04:37:40 AM »
Oh, God, I hate apple logic...If I get it wrong, I have no excuse.

Uhm, let's see. 0.6? Maybe?

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