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Topic: Molarity problem  (Read 6311 times)

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nemzy

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Molarity problem
« on: April 22, 2004, 01:21:21 AM »
If you put 25 mL of 1.6 M acetic acid, then, what is the molarity of acetic acid?

Is it just 1.6 M x .025 L, or is it .160 M / .025 L?

And..if you have 10 mL of 0.160 M sodium acetate , what is the moles of acetate in the solution?  

Is it .160 M X .01 L,

im getting confused with the concepts involved in here

Offline gregpawin

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Re:Molarity problem
« Reply #1 on: April 22, 2004, 01:40:06 AM »
You've just gotta keep track of those units and it'll all become obvious to you:

First molarity is moles over L, so just replace all M's with mol/L, and change mL into L.  Whenever you see the word "of" in any kind of problem involving math, it means multiply.  So if you're using 25mL of 1.6M of acetic acid, you're going to multiply them together.  0.025L X 1.6mol/L... notice how the units cancel; when dealing with these equations you're never going to get a mol2, L2, or anything similar unless its length, and even then you're probably doing something wrong if you have to do that... so cancel cancel cancel units!

Notice how you're not going to get molarity, but moles and there's no way around it with the information you gave.

Here's just a few reminders about conversions in chemistry:

Concentration X Volume = Mass(or Moles depending on concentration convention)
Moles X Molar Mass = Mass
Mass/Molar Mass = Moles

All of the understanding lies within keeping track of the units and canceling them out; if its not canceling, just multiply by its inverse.
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Offline Donaldson Tan

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Re:Molarity problem
« Reply #2 on: April 27, 2004, 08:13:01 AM »
Acetic acid dissociate partially. You'll need to look up the acid dissociation constant of acetic acid to work out the amount of acetate present and the amount of acetic acid at equilibrium.
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Offline AWK

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Re:Molarity problem
« Reply #3 on: April 30, 2004, 01:37:07 AM »
If you put 25 mL of 1.6 M acetic acid, then, what is the molarity of acetic acid?

Is it just 1.6 M x .025 L, or is it .160 M / .025 L?


Molarity is 1.6M.

That's another horse of color if you want to calculate  a concentration of species in this solution (H+, CH3CCO- and undissociated CH3COOH) as Geodome suggests.

And..if you have 10 mL of 0.160 M sodium acetate , what is the moles of acetate in the solution?  

Is it .160 M X .01 L,


OK

« Last Edit: April 30, 2004, 01:38:25 AM by AWK »
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Offline jdurg

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Re:Molarity problem
« Reply #4 on: April 30, 2004, 09:03:16 AM »
If you put 25 mL of 1.6 M acetic acid, then, what is the molarity of acetic acid?

Is it just 1.6 M x .025 L, or is it .160 M / .025 L?


Molarity is 1.6M.

That's another horse of color if you want to calculate  a concentration of species in this solution (H+, CH3CCO- and undissociated CH3COOH) as Geodome suggests.

And..if you have 10 mL of 0.160 M sodium acetate , what is the moles of acetate in the solution?  

Is it .160 M X .01 L,


OK



Heh.  That second question can be a real doozy if you start taking into account the ability of the acetate ion to go and pull a hydrogen atom off of a water molecule and form acetic acid in solution.  ;)
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