# Basic behaviour of bromide ion with sulfuric and phosphonic acid

I  recently came across these two reactions:

\begin{align}\ce{2KBr + 3H2SO4 &-> 2KHSO4 + Br2 + SO2 + 2H2O}\tag{1}\\[0.6em] \ce{KBr + H3PO4 &-> KH2PO4 + HBr}\tag{2}\end{align}

I understand that the bromide ion being the conjugate base of a strong acid $\ce{HBr}$ is a very weak base and therefore it cannot take $\ce{H+}$ from a very strong acid $\ce{H2SO4}$ in the first reaction. But the second reaction is an acid-base reaction.

I checked the $\mathrm{p}K_\mathrm{a}$ values of $\ce{H2SO4}$ and $\ce{H3PO4}$ to find that $\ce{H2SO4}$ is a way better acid than $\ce{H3PO4}$, then why is that bromide ion could take $\ce{H+}$ from $\ce{H3PO4}$ but not from $\ce{H2SO4}$?

The first reaction is a redox reaction and I understand that redox reactions take place when acid-base reactions are not possible because acid-base reactions are quick. Then why is the first reaction not an acid base reaction?

• en.wikipedia.org/wiki/Hydrobromic_acid In first reaction you'd need sufficiently concentrated acid and in second too, I guess, as in very diluted solutions nothing would separate. – Mithoron Dec 6 '16 at 22:57

A common misconception is that the more acidic component will always protonate the less acidic componend and displace it from its salts. That misconception would be exemplified by equation $(1)$

$$\ce{HBr + KH2PO4 -> KBr + H3PO4}\tag{1}$$

This is incorrect. There is a constant acid-base equilibrium and even bromide ions will be protonated to a non-neglegible extent. The principal difference between $\ce{HBr}$ and $\ce{H3PO4}$ is that the former is a gas at room temperature and standard pressure. Thus, in an open vessel, any $\ce{HBr}$ in solution is again automatically in equilibrium with $\ce{HBr}$ in the gas phase. Thus, the equilibrium should better be written in form $(2)$:

$$\begin{multline}\ce{KBr (aq) + H3PO4 (aq) <<=>\\ <<=> HBr (aq) + KH2PO4 (aq) <=> \\ <=> HBr (g) ^ + KH2PO4}\tag{2}\end{multline}$$

As you can see, the gaseous $\ce{HBr}$ can diffuse away. This reduces the partial pressure of $\ce{HBr}$ above the solution, which affects the right-hand side equilibrium by Le Chatelier’s principle, reducing the concentration of $\ce{HBr}$ in solution, which in turn affects the left-hand side equilibrium, reducing the concentration of $\ce{KBr}$. Thus, if left to equilibrate in a well-vented area, large portions of $\ce{HBr}$ will diffuse away leaving a solution of potassium phosphate or potassium hydrogen phosphate.

The principle alluded to in the first sentence is thus correct, if rather than using acid strength it is quoted as:

The less volatile acid will displace the more volatile acid from its salts.

It is worth noting that an acid-base equilibrium $(3)$ also is observeable in your first reaction. However, the redox reaction is faster than the removal of hydrogen bromide. It is also irreversible, hence those are the final products obtained.

$$\ce{KBr + H2SO4 <=> HBr + KHSO4}\tag{3}$$

• You said the redox reaction in the case of H2SO4 occurred faster than acid base reaction. But why is that, sulphur is in its most stable +6 oxidation state in H2SO4 and it has to go to +4 oxidation state in the redox reaction ? There should not be a great driving force for this redox reaction. – Arishta Dec 7 '16 at 3:19
• @Blue I did not say that the redox reaction be faster than the proton transfer. I said that the redox reaction is faster than diffusion of HBr out of the solution. Big difference. Also, do not fool yourself into thinking a ‘most stable oxidation state’ exists; these are only the most stable within given circumstances. What counts is the overall $\Delta G$ term, which, for redox reactions, is negative if and only if $\Delta E$, the potential difference, is favourable. – Jan Dec 7 '16 at 13:29