Bromine water is a reagent which is used to test for unsaturation in organic compound. It is $2.8~\%$ bromine in water according to Wikipedia. But how is it made?

This link describes the procedure for making bromine water, in which it is stated to

simply decant the bromine vapors in water and it will form a solution which is yellow in colour.

But bromine can also react with water to form a mixture which includes hydrobromic acid ($\ce{HBr}$) and hypobromous acid ($\ce{HBrO}$).

$$\ce{Br2 + H2O -> HBr + HBrO}$$


So, if bromine reacts with water to form a mixture of acids, how can it be called a solution? Components of solution do not react with each other.

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    $\begingroup$ Just ignore the reaction (it is not that fast) and make the solution. $\endgroup$ – Ivan Neretin Dec 24 '15 at 15:20

Much like with chlorine or iodine, the disproportionation of bromine can occur in pure water, but it is not fast. As you can see, the reaction generates acids, so without any base present the equilibrium is pretty close to the bromine side. You do need bases to disproportion bromine in any meaningful amount — much like chlorine which is generally quenched with $\ce{NaOH}$ solution to generate hypochlorite solutions.

So if you just put bromine and water together, they may react very slightly, but not much, and most of the bromine will sink to the bottom (assuming you used a lot) while the lesser part will dissolve in the water first. Since the concentration of $\ce{Br2}$ is much higher (and we’re talking orders of magnitude) that those of $\ce{Br-}$ and $\ce{OBr-}$, it is best to consider this a bromine solution and not a bromide/hypobromite one.


According to the IUPAC Goldbook a solution is:

A liquid or solid phase containing more than one substance, when for convenience one (or more) substance, which is called the solvent, is treated differently from the other substances, which are called solutes. When, as is often but not necessarily the case, the sum of the mole fractions of solutes is small compared with unity, the solution is called a dilute solution. A superscript attached to the ∞ symbol for a property of a solution denotes the property in the limit of infinite dilution.

There is no mentioning of that the solute may not react with the solvent. If you investigate a few common solutions, you will also find that it is indeed quite common, that the solvent reacts with the solution. Take for example the dissolution of hydrochloride in water, which is often expressed with the following equation: $$\ce{HCl (g) + H2O (l) <=> Cl- (aq) + H3O+ (aq)}$$ You can see here, that the proton is transferred onto a water molecule and from one point of view this is a displacement reaction; the chloride is displaced by water.

Another example is the dissolution of acetic acid. It is often described with the following equation: $$\ce{H3C-COOH (s) + H2O (l) <=> H3C-COO- (aq) + H3O+ (aq)}$$ The same happens here. We hide the fact that the dissolution process includes getting the solvent into a highly flexible network of hydrogen bonds in the addendum of $\ce{(aq)}$.

When you look at common salts that get dissolved, it is often that the metal ion reacts as a weak acid, co-coordinating the solvent molecules. In many cases this is so flexible, like for the alkaline metals, that we cannot describe even the first solvation sphere approximately.
For other salts this is quite possible. Take for example the dissolution of copper sulfate. We often write for short $$\ce{CuSO4 (s) <=> Cu^2+ (aq) + SO4^2- (aq)}$$ when we actually mean the formation of a complex $$\ce{CuSO4.5H2O (s) + H2O (l) <=> [Cu(H2O)6]^2+ (aq) + SO4^2- (aq)}.$$ A chemical reaction actually takes place. And since everything is in equilibrium, there are many chemical reactions happening in solutions all the time.

Now the dissolution of bromine in water is no different. In fact the disproportionation $$\ce{Br2 (l) + 2 H2O <=> BrOH (aq) + Br- (aq) + H3O+ (aq)}$$ is quite necessary to dissolve bromine in the first place. While I do not even slightly agree to the causality of having a dipole means being a polar molecule, I can still see, that the bromine molecule is less polar than water. It is however quite well polarizable; it has a quadrupole moment.[1] This is what also leads to the above equation.
The bromine hydrolysis has been investigated a couple of times. $$\begin{align} \ce{Br2 + H2O &<=>[K_1] HBrO + H+ + Br- }& K_1 &= \ce{\frac{[HBrO][H+][Br- ]}{[Br2]}} \end{align}$$ The temperature dependency of $K_1$ has been studied by Liebhafsky as early as 1934.[2] In their studies they also considered the effects of the tribromide equilibrium $$\begin{align} \ce{Br3^- &<=>[K_3] Br- + Br2} K_3 &= \ce{\frac{[Br- ][Br2]}{[Br3^- ]}}. \end{align}$$ And later continue to state that the existence of the pentabromide ion is highly probable. $$\begin{align} \ce{Br5^- &<=>[K_5] Br- + 2 Br2} K_5 &= \ce{\frac{[Br- ][Br2]^2}{[Br5^- ]}} \end{align}$$ From the equilibrium constants you can see, that the main dissolved species is still $\ce{Br2}$.

$$\begin{array}{lrrrrr} T/~\mathrm{^\circ C} & 0 & 10 & 25 & 30 & 35 \\\hline K_1\times10^{-9} & 0.70 & 1.78 & 5.8 & 8.3 & 11.3\\ \end{array}$$

Solutions do not exist without chemical reactions. There are often multiple different species in solution, that have very low concentration, so for various purposes a valid approximation is to neglect those species' concentrations.

  1. Jacek Bieroń, Pekka Pyykkö, Dage Sundholm, Vladimir Kellö, and Andrzej J. Sadlej; Phys. Rev. A 2001, 64, 052507.
  2. Herman A. Liebhafsky; J. Am. Chem. Soc. 1934, 56 (7), 1500–1505.
  • $\begingroup$ Iirc, wasn’t $\ce{CuSO4 . 5 H2O}$ actually $\ce{[Cu(H2O)4(\mu-SO4)2] . H2O}$? If that’s the case, then you should rewrite the corresponding equilibrium where you only discuss the formation of a hexaaquacomplex while you are actually creating a hexaaqua from a diaquabissulfato complex. Or you switch to discussing the dissolution of $\ce{CuSO4}$ (water-free). But anyways, great answer; let me give you $+1$ =3 $\endgroup$ – Jan Dec 25 '15 at 18:12
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    $\begingroup$ @Jan I really had no idea. There you have it again, when you think you pick an easy example, you don't. It always depend how deep you went when you learned it. If I edit it now, I'll mess it up since I am only on mobile and not in the office during the next week, so we'll consider your comment as a fine extension and revisit it next year. Thank you for pointing it out though. $\endgroup$ – Martin - マーチン Dec 25 '15 at 18:28
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    $\begingroup$ " If I edit it now, I'll mess it up since I am only on mobile and not in the office during the next week, so we'll consider your comment as a fine extension and revisit it next year." - It has been almost a year that you posted this answer. You want to revisit the answer for some extension. $\endgroup$ – Nilay Ghosh Nov 22 '16 at 17:17

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