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Author Topic: help with solutions in lab for a personal project.  (Read 816 times)

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trio

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help with solutions in lab for a personal project.
« on: December 07, 2018, 11:09:45 AM »

I need to make a solution containing 45 g/L of Cu and 2 g/L of H2SO4. All of this in 1L of volume. I´ve calculated the amount of CuSO4*5H2O to add and the amount of acid at 98% purity. The problem comes when the samples are sent to chemical analisis the amount of H+ is waaaay higher than expected. Any feedback on why this is happening and how to fix it?? maybe the disolution of the solid is generating some acidity idk.


pd:this is a personal project on my hs and the analysis is donde by an external entity so idk what test the do to determine the concentrations of H+ and Cu.
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Enthalpy

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Re: help with solutions in lab for a personal project.
« Reply #1 on: December 07, 2018, 11:29:42 AM »

Welcome, trio!

I don't understand how you could dissolve Cu (in water supposedly). CuSO4 is soluble but here you pretend to dissolve more of metallic Cu (or Cu2+ ions?) and less of the acid. The you also tell CuSO4 and H2SO4 in water, which would mean more acid moles than metal moles.

Could you double-check the figures, tell more clearly the masses or moles, and whether you mean Cu ions?
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trio

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Re: help with solutions in lab for a personal project.
« Reply #2 on: December 07, 2018, 12:00:50 PM »

indeed every time I talked about Cu I meant Cu2+.

The task is to simulate a solution from a industrial process (which i dont want to go into much detail). This solution has the characteristic concentration of 45g/L Cu2+ and 2g/L H2SO4 (data that was researched).

 So in order to do this I added 176grams of CuSO4*5H2O and roughly 2grams of H2SO4 on to a liter of solution. But then the analysis came with waay more acid concentration than expected.
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Borek

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Re: help with solutions in lab for a personal project.
« Reply #3 on: December 07, 2018, 01:01:06 PM »

How was the amount of acid measured?

Solution that is just 45 g/L in Cu (be it Cu+ or Cu2+, doesn't matter) and 2 g/L in H2SO4 is impossible. You need an additional counterion to keep the solution electrically neutral. When you use CuSO4 this counterion is SO42-.
« Last Edit: December 07, 2018, 10:02:16 PM by Borek »
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Enthalpy

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Re: help with solutions in lab for a personal project.
« Reply #4 on: December 10, 2018, 10:43:01 AM »

Could it be that 2g/L H2SO4 count only the fraction that does not come from CuSO4?
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