1
$\begingroup$

Why is the reaction between hydroxide and water reversible, when the reaction between oxygen ions and water is not? And is there a reason why the products of reaction 2) are written separately, instead of "2OH"?

1) $\ce{OH- + H2O <=> H2O + OH-}$

2) $\ce{O^{2-} + H2O -> OH- + OH-}$

$\endgroup$

3 Answers 3

2
$\begingroup$

Correct, all reactions are reversible, just to different extents. Think about the combustion reaction in your vehicle. Is that reaction reversible? Sure, in that nothing is 100% irreversible. The reversing of the combustion just might take a while. But it's not reversible to any significant extent in the short and medium-term; i.e. you're not going to be re-capturing your combusted gasoline and be putting that back together as usable fuel for your vehicle.

Also, about the reaction of oxide ion with water - that reaction goes practically 100% because oxide ion is a very strong base. The strongest base supported in high concentration by a medium is the conjugate base of the medium. (Leveling effect). For water, that would be hydroxide ion (remember that conjugate bases lose a proton). Oxide ion is the conjugate base of hydroxide ion. We would expect oxide ion to be an even stronger base than hydroxide ion because of oxide ion's even higher negative charge density - we're talking about a negative two charge here (versus negative one for hydroxide) and we're missing a hydrogen proton (which only further decreases the size of the oxide ion and further increases its negative charge density). If you think about this Coulombically, then of course oxide ion is going to be a stronger base than hydroxide ion - opposite charges attract (recall that water has two partial positive charges/dipoles stabilized by its hydrogen atoms). F=kq1q2/r^2 and one of the q's here is massive.

Also about how the products are written - it doesn't matter how the products are written. Either way you are creating two HO- (hydroxide ions). Write hydroxide with the negative next to the O (so either -OH or HO- are acceptable) because that's what you're doing - heterolytic clevage of the O-H bond in H-O-H, which (in this case) creates a negative formal charge that is mostly stabilized by the oxygen atom (NOT the hydrogen atom). Plus it's the hydroxide ion not the oxyhydride ion (whatever the hell that is).

On second thought, when writing -OH, make sure the "-" is clearly a superscript, lest people think you're talking about a hydroxyl group.

$\endgroup$
1
$\begingroup$

Hydroxide is the strongest base in water (leveling effect). The first posted reaction is symmetric. Given reactant and product sides of the equation, how do you tell them apart?

Oxide is a much stronger base than hydroxide. The second reaction spontaneously proceeds to the right (leveling effect). If you look up the Brønsted basicities of oxide and hydroxide, you can write a simple thermodynamic ratio for their relative abundances at equilibrium.

There is no reason not to write $\ce{2 OH^-}$ or $\ce{2 (OH^{-})}$.

$\endgroup$
0
$\begingroup$

All reactions are reversible but some are more reversible than others. Some reactions are so little reversible that they are practically irreversible, like that reaction of yours between oxide anion and water.

It all comes down to comparing the stabilities of reactants and products. The actual question is, why is oxide anion so unstable compared to hydroxide ions? One approach would be to think how would you make the oxide anion from a water molecule. Clearly, you can remove one hydrogen from a water molecule with a mild base but taking the second one would require a much stronger base. Take a look at the pkb value of hydroxide ion and visualize the equilibrium populations in your head.

Another possible approach would be to compare the electronic structures of the species. Oxide anion has twice the negative load of the hydroxide ion. Hydroxide ion has one hydrogen to relieve the stress.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.