[QuickerQuestion]

Ok I can't figure this out for the life of me so hopefully someone can help me out here.


Determine the molar concentration of each ion in solution obtained by reacting 400 ml of 2 M HCL with 150 ml of 4 M NaOH.


The answer is 1.45 M Cl. 1.09 M Na, .364 M H

I know that NaOH is the limiting reagent, more specificaly OH. But I don't know how to do the rest.



Any help is greatly appreciated.
I feel that I can solve the problem but I'm missing some key element (no pun intended). I've tried working backwards but to no avail.

[Eric]

can you show me your process?

[QuickerQuestion]

Unfortunately I don't have a process yet.

I've got it down to HCl + NaOH > NaCl + H2O

.8 moles HCl, .6 moles NaOH which goes to 29.2 g HCl and 24g NaOH.

That's as far as I've gotten. I think the molarity of NaCl is 1.1. I'm basing this off of NaOH being the limiting reagent thus only .6 moles could be used thus you'd have .6 moles of NaCl.

I'm not sure if that's true or not (I'm almost certain it is but it's 1:30 AM sooo...)

I just don't know how to find the molarity of each individual ion. The groups is not a problem.

[AWK]

Write down a balanced reaction
calculate moles of excessing HCl
calculate concenration of NaCl and HCl in a new volume - this will give you concentration of Na+ and H+ (why?)
Concentration of Cl- is a sum of both concentrations (why?)

[Eric]

Yes,  .6 moles OH-, Na+, .8 moles H+, Cl-.  now i see that H+ is going to be in excess, and sodium and chloride are spectator ions.  they remain .6 and .8.  you will be left with .2 moles H+ and no OH- .  now we divide moles by Liters and get Molarity.  but what are the liters? remember you added .15 L to .4 Liters.  you need to use the total volume to find the final concentration.  once again, you are left with .2 moles H+, .6 moles Na+ (spectator ion), .8 moles Cl- (spectator ion).

all reactants present will 100% dissociate, so in the solution HCl, NaOH , NaCl exist as H+ and Cl-, Na+ and OH-, Na+ and Cl-.  only the H+ and the OH- react to form water.  

Does that explain it well enough?

[QuickerQuestion]

Sort of. I tried .2 / .55 for Hydrogen on a whim and got that part right. I'll work on finding Na and Cl off the information ya'll gave now.

[QuickerQuestion]

Ok I did .2 / .55 for H and .6 / .55 for Na and .8 / .55 for Cl but I'm still a bit confused on how you know that you have .6 of Na and .8 of Cl. I realize the moles of HCl was .8 but I didn't know it stayed as .8 on the other side of the equation. Unless Cl is given off in excess too?

And if so, how do you know that OH is the limiting reagent? I know how to tell that NaOH is the limiting reagent but not too sure about OH specifically. Best I can guess is because Na is present in the completed reaction and OH isn't (well it is in water but it's not OH any more).

Edit: By the way do ya'll know where to find a good solubility table online? I typed it into google but didn't get any good ones. They were either incomplete or hard to read.

[Borek]

but I'm still a bit confused on how you know that you have .6 of Na and .8 of Cl

These are spectator ions - their amount can not change, that's mass balance.

And if so, how do you know that OH is the limiting reagent? I know how to tell that NaOH is the limiting reagent but not too sure about OH specifically.

Write net ionic reaction, then it should be obvious.

[QuickerQuestion]

Ok I moved on to some other problems and am getting them right now so I guess I understand it (or at least well enough for the test).

Thanks.