Gen chem question

[txlonghorn]

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This is from Chad quiz #4, problem 2

It states, how many grams of NaOH would be required to make a 10L of a 2M solution?

Ok, so why would it not be 400g? The answer is 800g.

My reasoning is that 1 moles of NaOH would ionize into 2 moles - 1 mole Na and another mole OH. So 1 mole of NaOH = 2M solution. So you only need half of the moles that are in the solution. When you get 20 moles in the solution (2M * 10L = 20 moles), you divide that by 2 to get 10 moles. Then multiply 10 moles by 40 (molar mass of NaOH), and you got 400g.


Tell me where I went wrong!

 

[RUpremed]

ok this is how i approached the problem...

you know that molarity=moles of solute/Liters of solution so...

2 M= x/ 10L (x being the unknown amount of moles)....

so when we multiply we get.. x=20 moles

the mass of NaOH is about 40 g/mol.

so when we convert from moles to grams we get.. 20 moles* 40g/mol (the moles cancel out) and we are left with 800 grams.

the NaOH is the solute. the solution is assumed to be something else. it seems that you're assuming that the NaOH is the solution?

hope this helps.

 

[txlonghorn]

So just as you assume that the solution is not NaOH, could you not assume that the solution is NaOH?

 

[cm8a12321]

This is how you should approach this question:

Molarity (M)= mol/L

when 2.0M soln and 10.0L

All you have to do is apply it to the equation:

2.0=mol/10.0; mole=20. mol NaOH

To find grams of NaOH, you know that FW of NaOH is 40. g/mol, so you just do just dimensional analysis and you would get this:

20.mol(40. g/ (1 mol))= 8.0 *10^2 gram (or simply 800 gram if you ignore significant digits)

----

What you have done: "1 moles of NaOH would ionize into 2 moles - 1 mole Na^+ and another mole OH^-. So 1 mole of NaOH = 2M solution. So you only need half of the moles that are in the solution. When you get 20 moles in the solution (2M * 10L = 20 moles), you divide that by 2 to get 10 moles <Na^+>. Then multiply 10 moles by 40 (molar mass of NaOH), and you got 400g."

is actually your way to find one mol of Na^+, but not find moles of NaOH.

 

[TexasDDS]

From my understanding your getting NaOH to dissociate into 2 ions..as it is a Ionic bond..

so:

NaOH ---> Na+ + OH-

well your right you have 1 solution becoming 2 moles.. but if you think about it if you had 2 moles with 10L.. the Molarity would be .2M not 2M for NaOH

but if it was 2moles in 1L solution then it would have been 2M for NaOH

but the only thing you need in this problem just like RuPremed explained

that 2M = x moles/10L.... which is (2M) * (10L) = x moles.. x moles = 20 moles

then 20moles * 40g/mol = 800grams for NaOH

now the M would be different had they asked for M for Na or for OH...

i"ll confirm this on chad's page see if has anything to say personally.

 

[txlonghorn]

Thanks bud! I think you are missing my point. Even though 1 mole of NaOH would be .2M in 10L, they are telling you that you have a 2M solution already in 10L. So if you work in reverse, you would still have to divide the 20moles by 2, so that you get 10 moles of NaOH.

 

[cm8a12321]

Thanks bud! I think you are missing my point. Even though 1 mole of NaOH would be .2M in 10L, they are telling you that you have a 2M solution already in 10L. So if you work in reverse, you would still have to divide the 20moles by 2, so that you get 10 moles of NaOH.

Like I mentioned above, you are actually finding the Na^+ and not NaOH. Do not overthink it.


You reasoning is right about the ions, but when you start combining Na^+ and OH^-, you will get the correct answer.

Anyway, .2M in 10L WOULD NOT give you 1 mol, but instead of 2 mol of NaOH (1 mol of Na^+ and 1 mol of OH^-)

 

[txlonghorn]

I still don't understand how I am finding the Na ion? Ok so you agree that if I add 10 moles of NaOH into 10L of solution, I would get a 2M solution that is made of 10 moles Na+ and 10 moles OH-. If I now just get the moles from the 20M solution, I get 20 moles, which is 10 moles Na and 10 moles OH. However, if you look at the rxn, NaOH ----> Na + OH. So 1 mole Na = 1 mole NaOH, using stiochiometry. So 10 moles Na = 10 moles OH, so it should be 400g.

 

[RUpremed]

an easy way to think of it is..

you're looking at Na+ in a given solution of 10 L.

 

[cm8a12321]

I still don't understand how I am finding the Na ion? Ok so you agree that if I add 10 moles of NaOH into 10L of solution, I would get a 2M solution that is made of 10 moles Na+ and 10 moles OH-. If I now just get the moles from the 20M solution, I get 20 moles, which is 10 moles Na and 10 moles OH. However, if you look at the rxn, NaOH ----> Na + OH. So 1 mole Na = 1 mole NaOH, using stiochiometry. So 10 moles Na = 10 moles OH, so it should be 400g.

Awg! Let me simplify a little: 1 M of NaOH is equal to 1 mole of Na^+ and 1 mole of OH^- (These two do not add up to 2 because it is mol of NaOH you are counting, not the number of ions; you might be confusing yourself with the Van Hoff factor; if you are, please ignore for these type of example) in 1 L of solvent. 2 M of NaOH= 2 mol of Na^+ and 2 mole of OH^-, etc.

When you want to find the moles of a 2 M NaOH solution of a 10 L, it is not finding the mole of Na^+ (which you are doing), but we are finding the moles of Na^+ combined with the moles of OH^- .

Let's do the way you do it: 2M NaOH= 2 mol Na^+ and 2 mol OH^- in 1 L

So if it is 10 L, we should get 20 mol Na^+ and 2 mol OH^-. Thus, we get 20 mol of NaOH, not 40 mol of NaOH (40 mol of NaOH is clearly saying that you have 40 mol Na^+ and 40 mol OH^-).

Hope this helps!

 

[txlonghorn]

Thanks! Ok, so i was approaching it as a van hoft factor problem. Lesson learned: apply van hoft only to colligative property problems, normality problems can also be multiplied, but nothing else ionized for the DAT. Lol

 

[UCB05]

If you want 10L of a 2M solution, that would require 20mols of solute. I think we can all agree on that. The confusion seems to be about misuse of the van hoft factor. When you make a x Molar solution, you add x mols of solute per liter. The fact that the solute in question ionizes is irrelevant. Chad asked for 10L of 2Molar NaOH solution, not 10L of 2Normal solution. Yes, when you make the 10L of 2M NaOH, it will have an ionic concentration of 4N, but it's still only 2M of NaOH