[cliverlong]

Hi,

 I am having difficulty explaining the dissolving of sodium hydroxide in water as an acid-base reaction.

The sodium hydroxide is a strong base.  Therefore when it dissolves in water, the water acts as a weak acid, donating a proton.

H⁺ + OH^- + Na⁺ + OH^- → H₂O + Na⁺ + OH^-  (1)

Now I can't show any change here because the products and reactants are the same, even if I claim a H⁺ has come from water and combined with OH^- from sodium hydroxide, water has reformed.

If I look at HCl (aq) and NaOH(aq) the association of H⁺ + OH^- to form water is apparent.

How do I argue the water has acted as an acid in (1) ?


Thanks

Clive

[nj_bartel]

NaOH + H₂O  Na⁺ + OH^- + H₂O

is how I would write it, as the sodium hydroxide is anhydrous and undissociated until you add it to water.

The reason it's hard to see is because your acid and conjugate acid are the same.

EtOH + NaOEt    Na⁺ + EtO^- + EtOH

If you wanted to show it more explicitly, you could convert the sodium hydroxide hydrogen to deuterium.

[cliverlong]

[nj_bartel]

OD^- + HOH    OH^- + DOH

K_b = [DOH][OH^-] / [OD^-]

This method of determination of K_b couldn't be used in actuality because the DOH would have it's own equilibrium back into H₂O and into D₂O, but I think it might at least be an ok way of visualizing it better?

It is confusing; I hope I'm explaining it well and not confusing the both of us 

[Albert Brady]

Hello ,

  I'd like to give a very simple explanation of this phenomena.  I am also studying this topic, and it really is a good question, right.  Why does Sodium Hydroxide dissolve?  The bond strength is higher than covalent bonds, so how do you know that the hydroxide makes it to solution to become a base, right? 

  Well, the answer is that hydrogen bonding is responsible for its dissolution, when NaOH goes into water, the cage of hydrogen bonds of water is broken, and begins to accomodate Na+ and OH-.  Its clear that Sodium and Hydroxide act as both an electron acceptor and a proton donor, participate in both parts of the hydrogen bonding, so it should have a really extensive solvation shell.  But how do you know if its stronger than water and will dissolve?  It's a good question.

  The Simple answer, that will give you some work to explain why is that you can just look up the HEAT OF SOLVATION of sodium hydroxide. It's in the CRC in the inorganic section.  The heat of dissolution of sodium hydroxide is NEGATIVE.  Therefore, it is a thermodynamically stable reaction for NaOH(s) --> Na+ + OH- (aq).   And, NaOH dissolves.  If you want to explain WHY, take a look at some trends within the hydrogen bonding and see what dissolves and what does not.  This will tell you how to explain the dissociation of salts in water, and then OH-  goes into solution at a really high level, 1M NaOH makes most of a molar of OH- but the equilibrium value is 10^-7, so its a strong base.

  Does that help?

[Albert Brady]

Ooops, actually there is a mistake in this calculation.  You really need to include the Entropy also.
DelG is the real way to determine spontaneity.

Here's some people that have calculated these things.  They say that salts like silver don't
dissolve, but sodium salts do.  Take a look at this website.

http://www.rod.beavon.clara.net/solubility.htm

The summary if you don't find the link is that the Entropy of Solution of Sodium Chloride is about
three times the enthalpy of solution pretty consistently, and positive, so that
DelG of Solution is always negative.

The search term 'why do salts dissolve' and 'Free Energy' gives some hits.