# What is purpose of sulfuric acid in iodine extraction?

I am trying to extract iodine from kelp following this guide here. It calls for sulfuric acid, but I don't see what this does do help the reaction. I thought the reaction was simply oxidizing iodide to iodine with hydrogen peroxide, so I don't understand the purpose of the acid. I do not have any sulfuric acid, and it is expensive to purchase, I do have a large supply of sodium bisulfate. Could this be used as a substitute to the sulfuric acid if all that is needed is a low pH.

• ‘It is expensive to purchase’ — that is a lie. It is cheaper than (laboratory) sand. – Jan Sep 18 '15 at 8:32
• The acid is not expensive, but the shipping costs are very high for strong acid concentrations – popgalop Sep 19 '15 at 0:08
• Have you seen this experiment in the Journal of Chemical Education? Revisiting History: Encountering Iodine Then and Now. A General Chemistry Laboratory To Observe Iodine from Seaweed. This article has a simpler procedure, if your experiment is for teaching purposes. pubs.acs.org/doi/abs/10.1021/ed086p206 – M. Farooq Jun 22 '19 at 13:40

Reading the guide, you are supposed to use $2\,\mathrm{ml}$ of $1\,\mathrm{M}$ sulphuric acid in $20\,\mathrm{ml}$ of water which will put you in the range of $\mathrm{pH}\, 1$.

It is important due to the equation shown in trb456’s answer that there is acid present, since $\ce{H2O2}$ is usually only an oxidising agent under acidic conditions. You can see why if you deduce the partial equations of both oxidation and reduction to get the complete redox equation.

It doesn’t matter what the counterion to oxonium ($\ce{H3O+}$ is. So any acid that is strong enough will do the trick. Sulphuric acid is one of the cheapest choices, but dilute hydrochloric acid will do the same trick. $\ce{NaHSO4}$ (which I hope is the correct chemical; I never understood the bi- nomenclature and prefer hydrogen sulphate) may work since it is still pretty acidic, but you would need higher concentrations. Acetic acid will probably not work.

Nitric or phosphoric acid should also do the trick, in case you accidentally have them standing around on the shelf. Whatever the acid you use, if you have pH papers, just use one to check the pH. If it is around $1$, you’re good.

Don’t worry about oxidising $\ce{Cl-}$ to $\ce{Cl2}$ with excess $\ce{H2O2}$. For one, the concentration of chlorine will be too low to cause any harm (unless you decide to stick your nose in the beaker and take a deep breath or drink the whole thing). Also, the seaweed itself contains enough chloride ions that this would be a background reaction anyway. Stay in a well-ventilated place (as you always should when experimenting) and you should be fine.

You need a strong acid to provide hydrogen (technically hydronium) ions, which is true for most aqueous oxidation reactions:

$\ce{2I- +H2O2 +2H+ -> I2 +2H2O}$

Unfortunately, bisulfate ion appears to be a weaker acid than water $(K_{\mathrm{a}} = 10^{-2})$, so it may not work unless you use a very concentrated amount to obtain enough hydrogen ions. Try it!

A fairly cheap alternative could be hydrochloric acid, which may be available in hardware stores as muriatic acid, and is stronger than sulfuric acid. But read the MSDS, as hydrochloric acid fumes and is nasty stuff.

Best of luck, it’s a cool experiment!

• I thought that sodium bisulfate was a strong acid as it has a pKa of 1.99. I wanted to avoid using HCl, because I am worried I will also oxidize Cl- to Cl gas with the H2O2. – popgalop Sep 18 '15 at 0:15
• @popgalop: Oxidizing $\ce{Cl-}$ to $\ce{Cl2}$ will not happen because the reaction of $\ce{H2O2}$ with chlorine is much slower than with iodine. As I noted, the Ka of bisulfate is weaker than water. – user467 Sep 18 '15 at 0:43
• But the Ka is still higher than water, because water has a Ka of 10^-14 and bisulfate has a Ka of 10−2 – popgalop Sep 18 '15 at 1:03
• According to wikipedia a solution of NaHSO4 can have a ph less than 1. – popgalop Sep 18 '15 at 1:09
• That's not a big deal anyway. True, you can't push all iodide into the protonated form, but you don't really have to. – Ivan Neretin Sep 18 '15 at 8:51