# How does the Hammett acidity function work and how to calculate it for [H2SO4] = 1,830? Any references?

I'm modelling the concentrations in the Belousov Zhabotinsky reaction. For this model, I need to convert [BrO3-] to [HBrO3]. I discovered in several papers that the following correlation applies:

$$[\ce{HBrO3}] = \frac{h0}{h_0+0.2} * [\ce{BrO3^-}]_0$$

See for example this dissertation on page 28 or this paper on page 2.

I have a solution which has $$[\ce{H2SO4}]$$ = 0.915M (I assume $$[\ce{H+}]$$ = 1,830 M).

My $$[\ce{BrO3^-}]$$ was 0,2411 M.

I researched the Hammett Acidity function, but I still don't completely understand the concept. What I do know is that $$h_0$$ substitutes for $$[H+]$$ and that $$h_0$$ has the unit of M.

Can anyone explain the concept of this function in a bit more detail and explain in what way it exactly relates to pH (and therefore to H+)?

On the internet I found this paper, which says that $$h_0$$ = 3 when $$[\ce{H2SO4}]$$ = 1.5M. Are there any other tables or references with $$h_0$$ related to the concentration of $$\ce{H+}$$ or $$\ce{H2SO4}$$? There is one more question on SE about Hammett acidity. This question links to tables with $$h_0$$ for certain % of H2SO4. Can I perhaps use that? How does $$[\ce{H2SO4}]$$ relate to a % of H2SO4?

Is there a way to calculate the $$h_0$$ when $$[\ce{H2SO4}]=1.830 M$$?

• Just plug pH-meter in there and used the values you get - H0 is approx. = pH for diluted sol. Nov 7, 2020 at 20:15
• Oke, thanks @Mithoron! So in this case, I assume H3O+ = 1.830 M, my pH = -log(1.830) = -0.262 and I can assume h0 = -0.262? Because when h0 = -.262, the [HBrO3] would be 1,01 and that's about four times the amount of BrO3- I added... But well, you're perhaps the wrong person to ask this, whilst you didn't devise the formula😉 Thanks anyway! Nov 9, 2020 at 11:48
• As Karsten mentioned, you assume wrong, more so H2SO4 is not the only thing affecting pH in such mixture. Nov 9, 2020 at 15:34
• Hi Mithoron, thanks for pointing that out again. The formule here is about the initial concentrations, so you kind of cancel out the other things affecting pH in such a mixture. I'm just secondary school and there we assume that H2SO4 completely becomes H+. As this thing isn't at a very advanced level, I'll for now just assume H0=[H+]. Nov 9, 2020 at 15:40

I have a solution which has $$[\ce{H2SO4}]$$ = 0.915M (I assume $$[\ce{H+}]$$ = 1,830 M).
While $$\ce{H2SO4}$$ is a strong acid, $$\ce{HSO4-}$$ is not ($$\mathrm{p}K_\mathrm{a}$$ around 2). So your pH should be about zero, and you might not need to use the Hammett acidity at all.