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Topic: CuSO4 solution  (Read 4557 times)

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Offline Rutherford

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CuSO4 solution
« on: May 16, 2012, 04:33:17 AM »
Through 50cm3 of CuSO4 electricity is passed until the blue colour dissapears. To neautralize the solution 30cm3 of NaOH (c=0.105mol/dm3) is spent. Calculate the c of the CuSO4 solution.

What happens when the electricity is passed?
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Offline Hunter2

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Re: CuSO4 solution
« Reply #1 on: May 16, 2012, 04:39:04 AM »
Keywords electrolysis, electro plating, Faraday laws
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Offline Rutherford

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Re: CuSO4 solution
« Reply #2 on: May 16, 2012, 04:44:30 AM »
I guess:
2H+ :rarrow: H2
What is neutralised with NaOH?
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Offline Borek

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Re: CuSO4 solution
« Reply #3 on: May 16, 2012, 04:45:57 AM »
You guess wrong.
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Offline Rutherford

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Re: CuSO4 solution
« Reply #4 on: May 16, 2012, 07:01:40 AM »
Then it is water. I got the result 0.0315mol/dm3, is it correct?
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Offline Hunter2

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Re: CuSO4 solution
« Reply #5 on: May 16, 2012, 07:12:11 AM »
No its not water, but the concentration is correct.

« Last Edit: May 16, 2012, 07:31:50 AM by Hunter2 »
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Offline Rutherford

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Re: CuSO4 solution
« Reply #6 on: May 16, 2012, 07:27:41 AM »
K-:Cu2++2e- :rarrow: Cu
A+:2H2O :rarrow: 4H++O2+4e-?
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Offline Hunter2

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Re: CuSO4 solution
« Reply #7 on: May 16, 2012, 07:30:25 AM »
Yes so you get at the end sulfuric acid, because sulfate SO42- is still present. This acid gets neutralized by NaOH.
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Offline Rutherford

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Re: CuSO4 solution
« Reply #8 on: May 16, 2012, 07:39:35 AM »
Ok, thanks.
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Offline Hunter2

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Re: CuSO4 solution
« Reply #9 on: May 16, 2012, 07:45:15 AM »
But how did you calculate it? If you thought its water?
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Offline Rutherford

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Re: CuSO4 solution
« Reply #10 on: May 16, 2012, 08:00:48 AM »
I've meant that water takes part in the electrolisys.
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