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Offline JonathanEyoon

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Solution Stoichiometry
« 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.

Offline azmanam

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Re: Solution Stoichiometry
« Reply #1 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?
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Offline JonathanEyoon

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Re: Solution Stoichiometry
« Reply #2 on: February 27, 2008, 08:23:14 PM »
 ??? HUH? How do you write the chemical formula for this?

Offline azmanam

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Re: Solution Stoichiometry
« Reply #3 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
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Offline JonathanEyoon

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Re: Solution Stoichiometry
« Reply #4 on: February 27, 2008, 08:28:52 PM »
wait do I just use the

m1v1 = m2v2?

Offline Arkcon

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Re: Solution Stoichiometry
« Reply #5 on: February 27, 2008, 08:30:04 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.
Hey, I'm not judging.  I just like to shoot straight.  I'm a man of science.

Offline azmanam

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Re: Solution Stoichiometry
« Reply #6 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.
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Offline JonathanEyoon

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Re: Solution Stoichiometry
« Reply #7 on: February 27, 2008, 08:44:59 PM »
So how would I go about to the point where I use the m1v1=m2v2?

Offline azmanam

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Re: Solution Stoichiometry
« Reply #8 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?
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Offline JonathanEyoon

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Re: Solution Stoichiometry
« Reply #9 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 + ?  =  ? +  ?

Offline azmanam

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Re: Solution Stoichiometry
« Reply #10 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 --> ?  +  ?
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Offline JonathanEyoon

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Re: Solution Stoichiometry
« Reply #11 on: February 27, 2008, 09:01:47 PM »
right side would be Na + S?

Offline azmanam

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Re: Solution Stoichiometry
« Reply #12 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?
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Offline JonathanEyoon

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Re: Solution Stoichiometry
« Reply #13 on: February 27, 2008, 09:07:16 PM »
You're speaking of molar ratio right?  So it'd be 2 Na for each Na2S?

Offline azmanam

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Re: Solution Stoichiometry
« Reply #14 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?
Knowing why you got a question wrong is better than knowing that you got a question right.

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