[trio]

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.

[Enthalpy]

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?

[trio]

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.

[Borek]

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 SO₄^2-.

[Enthalpy]

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