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Chemistry Forums for Students => High School Chemistry Forum => Topic started by: JonathanEyoon on February 27, 2008, 08:08:39 PM

Title: Solution Stoichiometry
Post by: JonathanEyoon on February 27, 2008, 08:08:39 PM
I'm having so much trouble with this one. Can someone help me understand what to do and how to go about finding the answer to this problem?


Calculate the concentration (M) of sodium ions in a solution made by diluting 40.0 mL of a 0.474 M solution of sodium sulfide to a total volume of 300 mL.
Title: Re: Solution Stoichiometry
Post by: azmanam on February 27, 2008, 08:14:02 PM
Start by writing the equation for the dissolution of sodium sulfide.  can you balance it?

how many moles of sodium ions are produced for each mole of sodium sulfide dissolved?

how many moles in 40.0 mL @ 0.474 M?

how many moles after you dilute to 300 mL?

how can you use the answer to those questions to find final concentration of sodium ion in solution?
Title: Re: Solution Stoichiometry
Post by: JonathanEyoon on February 27, 2008, 08:23:14 PM
 ??? HUH? How do you write the chemical formula for this?
Title: Re: Solution Stoichiometry
Post by: azmanam on February 27, 2008, 08:27:19 PM
try wikipedia for chemical formula and here for general dissolution reactions

http://dl.clackamas.edu/ch105-03/dissolut.htm
Title: Re: Solution Stoichiometry
Post by: JonathanEyoon on February 27, 2008, 08:28:52 PM
wait do I just use the

m1v1 = m2v2?
Title: Re: Solution Stoichiometry
Post by: Arkcon on February 27, 2008, 08:30:04 PM
Quote from: JonathanEyoon on February 27, 2008, 08:23:14 PM
??? HUH? How do you write the chemical formula for this?

When an ionic solid dissolves, you can visualize it as a decomposition reaction into two charges species -- the cations and the anions.  Try it like that.

Quote
wait do I just use the

m1v1 = m2v2?

Oh yes, you'll need that as well.
Title: Re: Solution Stoichiometry
Post by: azmanam on February 27, 2008, 08:32:26 PM
Quote
wait do I just use the m1v1 = m2v2?

yes, but you first need M of sodium ions

hint M of sodium ions is not the M of sodium sulfide starting material.
Title: Re: Solution Stoichiometry
Post by: JonathanEyoon on February 27, 2008, 08:44:59 PM
So how would I go about to the point where I use the m1v1=m2v2?
Title: Re: Solution Stoichiometry
Post by: azmanam on February 27, 2008, 08:47:35 PM
Start by writing the equation for the dissolution of sodium sulfide.  can you balance it?

how many moles of sodium ions are produced for each mole of sodium sulfide dissolved?

how many moles in 40.0 mL @ 0.474 M?
Title: Re: Solution Stoichiometry
Post by: JonathanEyoon on February 27, 2008, 08:50:47 PM
I honestly do not know HOW to write a "dissolution" formula for Sodium Sulfide. The furthest i'll get is naming Sodium Sulfide

Na2S + ?  =  ? +  ?
Title: Re: Solution Stoichiometry
Post by: azmanam on February 27, 2008, 08:59:58 PM
Take a look at the website I linked to above. Na2S is a strong electrolyte, similar to Na2SO4. 

There's only one compound on the left side of the arrow.

Na2S --> ?  +  ?
Title: Re: Solution Stoichiometry
Post by: JonathanEyoon on February 27, 2008, 09:01:47 PM
right side would be Na + S?
Title: Re: Solution Stoichiometry
Post by: azmanam on February 27, 2008, 09:03:50 PM
Good, now it needs to be balanced.

then, how many moles of sodium ion produced for each mole of sodium sulfide dissolved?
Title: Re: Solution Stoichiometry
Post by: JonathanEyoon on February 27, 2008, 09:07:16 PM
You're speaking of molar ratio right?  So it'd be 2 Na for each Na2S?
Title: Re: Solution Stoichiometry
Post by: azmanam on February 27, 2008, 09:13:11 PM
excellent.

how many moles of sodium ion in 40.0 mL of a 0.474 M solution of sodium sulfide?

how many moles of sodium ion when that solution is diluted to 300 mL?

what is the concentration of sodium ion in that 300 mL solution?
Title: Re: Solution Stoichiometry
Post by: JonathanEyoon on February 27, 2008, 10:05:35 PM
Huh!?  ???


0.0634 M?
Title: Re: Solution Stoichiometry
Post by: AWK on February 28, 2008, 01:07:19 AM
right side would be Na + S?

wrong!

Na2S dissociates forming cations and anions.
Of course, even that problem can be solved correctly in numbers
Title: Re: Solution Stoichiometry
Post by: azmanam on February 28, 2008, 06:09:52 AM
Quote
Na2S dissociates forming cations and anions.

True, good point.  Can't forget about the charges,  my bad.



Quote
0.0634 M?

Not sure what that is supposed to represent.  show us some work?

If it was a guess at the final answer, then you're off by a factor of two.  Remember, moles of sodium ion in 40.0 mL of a 0.474 M solution is for sodium ion, not sodium sulfide.
Title: Re: Solution Stoichiometry
Post by: ARGOS++ on March 05, 2008, 08:52:18 AM

Dear JonathanEyoon, Dear Reader;

(JonathanEyoon: Sorry that you ran already the second time ‘against’ a wall!)

If you feel lost, even with such simple Problems, start with the “Stoichiometry Recipe”:
A.)    Reaction:               Na2S (s)     --->        Na+ (aq)    +        S2- (aq)
B.)    Balancing:           1 Na2S (s)     --->    2  Na+ (aq)    +      1 S2- (aq)
C1.)   MW’s:                   78.1 g/m                   27.0 g/m             32.1 g/m   
For this problem even the line: C1.) is not really required. 

So let’s do the dilution first:
You should remember that 1.0 molar means 1.0 moles per 1.0 Liter Solution, and you have to take only 0.040 Liter, so:    0.474 moles L-1 * 0.040 L = 0.01896 moles Na2S.
But this amount is now put into a total volume of 0.300 L final solution, so that it results in a concentration of:  0.01896 moles / 0.300 L   =   0.0632 molar Na2S.

Finally from line B.) and/or from the Statement in your Reply #13 you know that you get 2  Na+ (aq) for each Na2S, so you should get:  2  *  0.0632 molar  =  0.1264 molar Na+.

For the whole “Stoichiometry Recipe” (incl. Excesses) and future explanations you may visit again:   "Stoichiometry Problem (http://www.chemicalforums.com/index.php?topic=23025.0)”.

I hope it may be of help to you all.

Good Luck!
                    ARGOS++