How is NaOH considered a strong base does it even accept protons?

[mcfaker]

Hello,

Ive been seeing it everywhere" NAOH is a strong base".
But in order for it to be strong, it has to meet the requirements which are:

-The Kb of the reaction 6qvtA.pngmust be high. This means that base strenght is not determined by the dissociation % of a substance.
Now I ve seen a lot of books say " NaOH is a strong base because it dissociates completely in water" But the dissociation of NaOH in water is not the same reaction as mentioned above right?So this is not a requirement for it to be a strong base!

Can anyone help me out please because I am stuck.

Thanks in advance!

 

[SteamKing]

Your post is garbled. What does '6qvtA.png' mean?

In any event, yes, NaOH is considered to be a strong base. There are several types of base compounds in chemistry. NaOH is an Arrhenius base.

http://en.wikipedia.org/wiki/Base_(chemistry)

 

[adjacent]

What do you mena by NAOH accepting PROTONS?

 

[Dadface]

What do you mena by NAOH accepting PROTONS?

Acids can be described as proton donors because the can form H+ ions and alkalis proton acceptors because they can form OH- ions. Together they can neutralise(crudely speaking alkalis accept the H+ ions which are protons) to form water(and a salt)

 

[mcfaker]

sorry for the image: It should be

 

[mcfaker]

The Kb for NaOH must be very high right? I found this online:

The concentration of BOH what do they mean with that? The concentration of dissolved BOH or the concentration of solid BOH? Which phase does BOH represent?

 

[SteamKing]

The Kb for NaOH must be very high right? I found this online:

The concentration of BOH what do they mean with that? The concentration of dissolved BOH or the concentration of solid BOH? Which phase does BOH represent?

You seem to be missing a fundamental concept here. The NaOH dissociates because it is in aqueous solution. Solid NaOH does not normally dissociate.

Read the link in Post #2.

 

[mcfaker]

But the problem is that NaOH(aq) does not even exist! Thats not even a molecule, it doesn't represent anything, it does not represent NaOH in aqueous solution since it dissolves in ions there! The value of NaOH(aq) will always be zero since it never existed & will never exist. So the only equilibrium you could think of is the solubility equilibrium here meaning there's a solid (s) involved! And I know this is not supposed to represent the solubility equilibrium. Also all hydroxides are strong electrolytes meaning there is no equilibrium so why is there an equilibrium arrow then?

 

[SteamKing]

The only thing I know fer sure is if you stick your hand in a concentrated NaOH solution, you'll get burned just as surely as if you stuck it in concentrated HCl solution.

You could make a similar argument about HCl in solution as NaOH. Your way of looking at things is IMO 180 degrees opposite of how every other chemist looks at acids and bases.

 

[hjelmgart]

I am not sure, I understand the problem?

As mentioned earlier, the notition follows the Arrhenius definition
http://en.wikipedia.org/wiki/Acid–base_reaction#Arrhenius_definition

The molecule has to split up in ions for it to even to be considered as an acid or a base. Ionic bonds may define molecules just like covalents bonds may do.

It also makes perfect sense, if you look into the functionality of an etchant for instance.

 

[Borek]

But the problem is that NaOH(aq) does not even exist! Thats not even a molecule, it doesn't represent anything, it does not represent NaOH in aqueous solution since it dissolves in ions there! The value of NaOH(aq) will always be zero since it never existed & will never exist.

Don't be so sure. pKb for NaOH is 0.2. Compare http://www.chembuddy.com/?left=FAQ

NaOH is a source of OH^- and it is OH^- that is the base present in the solution and ready to accept protons. Actually it is the strongest base that can exist in water (every stronger one will react with water, stripping it of protons and leaving just OH^-). However, you can't introduce just OH^- into the solution, you need a counterion.