[Fizzley]

Hi. I'm performing an experiment wherein I do a redox titration of copper (I) chloride (CuCl) with nitric acid (HNO3).

I'm placing 20mL of 1M copper (I) chloride in a beaker below the burette, and titrating it with 1M nitric acid. My question is, in order to completely oxidize the 20mL of 1M copper +1 to copper +2, how much nitric acid would you expect to need to add?

The two half-cells and complete reaction are:

3(Cu⁺ + Cl^- --> Cu²⁺ + e^- + Cl^-)
NO₃^- + 4H⁺ + 3e^- --> NO(g) + 2H₂O

3Cu⁺ + 3Cl^- + NO₃^- + 4H⁺ --> 3Cu²⁺ + 3Cl^- + NO(g) + 2H₂O

So, there are, as I see it, two possible ratios. The first, if NO₃ is considered the limiting reagent, would be 3:1 mL of copper to mL of nitric acid. But on the other hand, it could be H⁺ which limits it, in which case the ratio would be 3:4.

However, I have heard that the 4H⁺ in the redox half-cell only means the presence of acid, and it is therefore not a limiting reagent.

Any thoughts/answers would be greatly appreciated. :-)

[Borek]

Strange idea, IIRC nitric acid doesn't react stoichiometrically as oxidizing agent, depending on concentration and pH it gives a mixture of oxides as products, at least during metallic copper dissolution.

If there is an excess of other acid H⁺ shouldn't matter. If the titrated solution is neutral, H⁺ will be limiting reagent.