How to change pH of CuSO4? I'm currently very desperate to know this. I'm doing an experiment about adsorption/chelation of Cu ions in CuSO4 using banana peels using the UV-Visible spectroscopy. How do I change the pH of 0.1M or 0.05M of CuSO4? My main problem is with making the CuSO4 more alkaline/basic or whatever u call it. My base pH of the CuSO4 is about 4.0 or 4.1. I've tried making a pH 5 and pH 6 buffer solution then adding it to the CuSO4 solution, but so far it's only made it more acidic.
These are the calculations:
pH 5
Weigh 5.10575g of potassium hydrogen phthalate(C8H5KO4)
Dissolve it in a 100cm3 beaker
Transfer the solution into a 250cm3 volumetric flask and fill it with distilled water
Measure 5cm3 of 2M NaOH and add it to a 100cm3 volumetric flask and fill it with distilled water
Add 100cm3 of 0.1M potassium hydrogen phthalate and 45.2cm3 of 0.1M NaOH into a 250cm3 beaker
pH 6
Weigh 3.40225g of KH2PO4
Dissolve it in a 100cm3 beaker
Transfer the solution into a 250cm3 volumetric flask and fill it with distilled water
Measure 5cm3 of 2M NaOH and add it to a 100cm3 volumetric flask and fill it with distilled water
Add 100cm3 of 0.1M KH2PO4 and 11.2cm3 of 0.1M NaOH
But I realised the phosphate buffers will cause a precipitate. So I tried using a citric acid + sodium citrate buffer. I think the calculation was something like 88.5cm^3 of citric acid + 11.5 cm^3 of sodium citrate (or it could be the other way around). But I don't think that was a buffer solution, cus when I added the slightest amount of deionized water, the pH increased. I got it from https://www.sigmaaldrich.com/life-science/core-bioreagents/biological-buffers/learning-center/buffer-reference-center.html#citric2.
Someone please help this is experiment is basically graded, so I'm really desperate.
edit: I made the copper sulfate using CuSO4.5H2O if that helps
Alas, your problem has no solution, because of the risk of producing $\ce{Cu(OH)2}$ at pH $5$ and $6$. The solubility product of $\ce{Cu(OH)2}$ is $2.2·10^{-20}$. At pH $5$ and $6$, the concentrations of the ion $\ce{OH-}$ are respectively $10^{-9}$ and $10^{-8}$ M. This means that, at pH $5$ and $6$, the maximum molar concentrations of the ion $\ce{Cu^{2+}}$ are $2.2·10^{-2}$ M and $2.2·10^{-4}$ M respectively. So you will never be able to obtain a $0.1$ M solution of $\ce{CuSO4}$ at pH $5$ or $6$.