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Author Topic: aqueous solutions  (Read 175 times)
Willow18
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« on: February 07, 2010, 07:48:00 PM »

If 25.0 mL of an aqueous solution of H2SO4 requires 19.7 mL of 0.72 M NaOH to reach the end point, what is the molarity of the H2SO4 solution?  I have no idea how to do this.


Given the following values of (OH) calculate the correspondig value of (H2O+) for each solution?  a.  10-10 M
b. 10-2 M    c. 10-7 M   d. 10M

No idea how to do this.   
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kingchemist
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« Reply #1 on: February 08, 2010, 02:48:57 AM »

Moles NaOH = volume in L x molarity

2NaOH   + H2SO4 -----> Na2SO4  + 2H2O
2 mol        1mol

So moles H2SO4 needed is half of the number of moles of NaOH present.

Then you know moles H2SO4 and its volume.
Molarity = moles/volume in L
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Hertzinger
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« Reply #2 on: February 12, 2010, 09:44:08 AM »

This is how I did it tell me if I am wrong.

-log[x/44.7] + -log[.72/44.7]= mols= 3.44328255
M= mols/liter= 3.44328255/44.7 then do ur sig fig rules.
I might be wrong I'm trying a new theory I made
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