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aqueous solutions
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Topic: aqueous solutions (Read 175 times)
Willow18
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Posts: 9
aqueous solutions
«
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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Posts: 2585
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Re: aqueous solutions
«
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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'Chemistry is not just the study of matter; Chemistry is the study which matters!' - Kingchemist
Hertzinger
Labrat
Posts: 13
Re: aqueous solutions
«
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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