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

  1. #1 How is NaOH considered a strong base, does it even accept protons? 
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    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 Attachment 263503 must 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 im really stuck here.


    Thanks in advance!






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  3. #2  
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    NaOH is an alkali (i.e. a soluble substance that is basic in solution) it dissociates completely in water to give Na+ and OH-, it is the OH- that "accepts protons" and makes it basic via the reaction:

    H+ + OH- ----> H2O

    It is considered a strong base because all of the available "[OH]" in the NaOH is present in solution as [OH]- and available to accept protons.

    The idea of Kb is only really useful for compounds that do not completely dissociate (it is an equilibrium constant that is based on the concentrations of base and OH- at equilibrium. If the base is completely dissociated it is not defined (equivelant to dividing by zero). For example for NaOH, Kb would be equal to



    As the dissociation is complete the equlibrium concentration of [NaOH] is zero and so Kb is undefined.


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    The attachment should be

    But why are you using Kb for the reaction NaOH----> Na+ + OH- . Kb is for a substance it water that takes up a proton from H2O as in the above image^^.
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  5. #4  
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    Not necessarily, you are only considering Bronsted-Lowry bases, it can also be used for Arrhenius bases (i.e. the dissociation of BOH).

    You may find the link below useful:

    Chemistry Tutorial : Base Dissociation Constants (Kb)
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    But then it would not be called Kb. Kb is the base constant as defined in my book. You need a reaction where a substance (bronsted lowry base,not arrheneus) takes up protons from water. Then you will be able to calculate the base constant. The is not a base constant value because NaOH is not a bronsted Lowry base! it is a K value however, but to calculate base strenght we need to see the K of the reaction that takes up protons. NaOH does not take up protons. If we calculate the K (which is not Kb) of NaOH--> Na+ + OH- we get an impression of how many molecules have dissolved.
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    Then I think you need to get a better book. Did you read the link I provided? Kb is not restricted to only Bronsted bases.
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    Thank you so much you provided me with very useful information. I cant believe it: Ive been trying to figure this out on other forums, irc, ... for almost 2 weeks & I didnt find anything untill now (I got pretty desperate). Sorry I forgot to click on the link when I read everything. Damn what an awful book I have: It didnt even mention that! This got me thinking that NaOH, KOH, Ca(OH)2, Sr(OH)2, Ba(OH)2 were all of same base strenght , because I thought since none of these accept protons we have to take the OH- as a proton acceptor from water in reaction to calculate Kb which would be the same for NaOH, KOH, Ca(OH)2, Sr(OH)2, Ba(OH)2 since we calculated the Kb for OH-.

    Does this mean that in:

    BOH is in phase (aq) ?? Because if it was in solid phase (s) , the concentration would not be zero as you say.
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  9. #8  
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    No problem, glad to have been able to help and yes, it is the concentrations in solution that you should use.
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    couldn't have explained it any better than phdemon tbh

    op: if you have any more questions feel free to inbox me
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  11. #10  
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    @bio your damn right.
    Can I ask 2 more things please?

    If the solubility of an ionic compound is low, would that ionic compound still be considered a strong base?I know it would be considered a strong electrolyte for sure! Although there would be very little OH- in solution (because of the low solubility) , the % of dissociation of the dissolved "molecules" would be very high, meaning its a strong electrolyte (even though the little amount of ions we would get, again because of the low S). And strong electrolyte means strong base!

    Also I have found a table ( of acid / base conjugates) where I see NaOH in a very weird form. It seems hydrated to me:

    I could believe that the Na+ ion is hydrated, but I dont know about NaOH. Do they mean that NaOH is always hydrated in solid form? Because according to what I know molecules of NaOH dont exist, so there is no possibility of a "molecule" of NaOH hydrated by water.
    I dont know if this is supposed to represent the normal NaOH (they had to add water to it to confuse us).

    I analyzed the data from the image above & made up the following reaction, I dont know whether its right or wrong though:

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    I get the impression that this Na(OH)(H2O)(n-1) is not the same as NaOH, because as you said when 100% dissociates the base concentration becomes zero so the value of Kb is extremely high (10^9 range), but in the above image we can see that Na(OH)(H2O)(n-1) has a Kb value of only 1 which is not even slightly high. But I am not sure if my thoughts are correct.
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  13. #12  
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    If the solubility is low this is an indication that it does not readily dissociate into ions (if it readily dissociated it would be thermodynamically favourable for it to dissolve due to to the enthalpy of solvation of the ions) so insoluble compounds are not strong bases.

    The images you have shown are for the acid/base behaviour (or hydrolysis) of the solvated sodium ion (not NaOH), each ion will have a cluster of water molecules hydrogen bonded around it, the images you have shown are for the loss and gain of a proton from this solvation shell. This would occur in even non-basic sodium salts like NaCl. Also remember even in neutral solutions there are always H+ and OH- ions present due to the autodissociation of water.
    Last edited by PhDemon; January 20th, 2014 at 10:14 AM. Reason: typo
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    Quote Originally Posted by mcfaker123 View Post
    I get the impression that this Na(OH)(H2O)(n-1) is not the same as NaOH, because as you said when 100% dissociates the base concentration becomes zero so the value of Kb is extremely high (10^9 range), but in the above image we can see that Na(OH)(H2O)(n-1) has a Kb value of only 1 which is not even slightly high. But I am not sure if my thoughts are correct.
    You're right it isn't the same, it is a solvated Na+ ion that has lost a proton from one of the water molecules in it's solvation shell.
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    If the solubility is low this is an indication that it does not readily dissociate into ions (if it readily dissociated it would be thermodynamically favourable for it do dissolve due to to the enthalpy of solvation of the ions) so insoluble compounds are not strong bases.
    So an ionic compound with low solubility would not be considered a strong base, but it would definately be a strong electrolyte according to the present definition right?

    The definition of strong electrolyte has changed over time. In the past people used to define a strong electrolyte a substance that dissolves in water into ions & conducts the electric current well (meaning it could make a lamp burn, if the substance would be dissolved into solution). That meant a high concentration of OH- in case it would be a base.
    Now the definition has changed into: A strong electrolyte is a substance that when dissolved in water completely/almost completely ionizes into ions. So this means that even if you have low solubility as long as all that dissolves ionizes into ions, we still would have a strong electrolyte even though it doesnt conduct the current well.

    I have some proof that I found on the internet to back this definition change up:

    To take an extreme case: if a solid were only soluble at 1 gram per liter but all of it in solution was dissociated, it would be a strong electrolyte. While if a solid were soluble at 400 g/L but only 20% ionized, it would be a weak (or moderate) electrolyte - even though a saturated solution of it would be more conductive than of the saturated "strong electrolyte" solution. Concentration is not strength.
    Ionization is not solubility.
    The strong electrolyte definition chage is also mentioned on wikipedia!:


    I have found other sources indicating this definition change. It always tells that nowadays a strong electrolyte doesnt have to conduct the current well, as long as the "dissolved part" ionizes 100% / completely. You have said that its not a strong base, but the Kb would be high. And according to the link you provided a big Kb means a strong electrolyte!

    In case of the low solubility ionic compound: The Kb will be large. Of the dissolved part 100% would dissociate as mentioned in ^a. . This would also indicate to me that it would be a strong base. What is your opinion on this?
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  17. #16  
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    According to the definitions you have given a sparingly soluble compound could be considered a strong electrolyte if the small amount that does dissolve is fully dissociated (where did the unatrributed quote come from just out of interest), but this is not what I remember from my school days (which granted were a long time ago).

    As for your question about Kb, if the compound dissociates 100% Kb is not large, it is undefined (you can't divide by zero). Because of this as I said earlier the concept of Kb is only useful for compounds that don't dissociate fully (and is also only really useful for compounds that are reasonably soluble), as you are finding if you try and extend it to strong electrolytes (or not very soluble compounds) it doesn't really work!

    In reality acid/base strength is described using a combination of Ka/Kb, pH/p(OH), solubility and extent of dissociation etc. Each of these has their uses (but also limitations as you are discovering!)
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  18. #17  
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    Thank you for explaining. Regarding the source of the quote, I found it on Are all strong acids and bases strong electrolytes? - Yahoo Answers
    2 people are confirming it there. Wikipedia is telling the same thing as it mentions that the old definition where they looked at the conductivity of the solution does not apply anymore. It has been replaced by this new definition. Here is another very clear indication of the change of definition from a book:


    This leads me to believe that if the definitions have changed & a strong electrolyte doesnt have to conduct the current well ( or let a lamp burn) that the definition of strong base has also changed because everywhere is written "strong bases are strong electrolytes" meaning that a base (ionic compound) with low solubility will also be a strong base.
    Do you think that because of this change of definition a low soluble ionic compound base is now a strong base?

    The link you gave me is also referring this (although it doesnt say anything on solubility):


    You have mentioned that we use a combination of factors, but solubility is excluded as the definitions have changed. But I dont know about pH values how they can affect strenght of base. They didnt mention anything about that in the book. pH to my knowledge tells you whether the solution is basic or acidic, if its basic we have more OH- than H+ ions. If its acidic we have more H+ than OH- ions. That has nothing to do with the Kb or Ka which determines & gives meaning to acid/base strenght. So ph tells me if the solution is acidic or basic. On the contrary I think the opposite might be true (?): If we have a weak / strong base, we know OH- is getting released into the solution, and this release determines the pH so we dont need the pH to determine whether its a strong base or not, because we have the Kb which tells us the information (strong or weak base).
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  19. #18  
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    The idea of a strong base is one that can very easily accept protons to neutralise an acid, if it is sparingly soluble there will not be a large amount of OH- in solution to neutralise H+ even if it dissociates fully and so even if a sparingly soluble salt is a strong electrolyte I would not consider it a strong base.

    For aqueous solutions, pH, Ka and Kb are all just ways of allowing the calculation of [H+] or [OH-] in a solution of known concentration. Which parameter is most useful depends on how strong the acid/base is as I've been saying. For strong acids and bases [H+] or [OH-] equals the concentration of acid or base in solution (with appropriate stoichiometric multipliers), Ka and Kb are irrelevant as dissociation is complete and you know how "strong" the acid or base is purely from how concentrated it is and pH tells you what you need to know.

    pH = -log[H+] = 14 + log[OH-]


    For weak acids/bases [H+]/[OH-] is not equal to the concentration of the acid or base (with appropriate stoichiometric multipliers) and so the concentration of the solution does not tell you how strong the acid is. In this case Ka and Kb must be used to calculate [H+] and [OH-].
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  20. #19  
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    Im sorry, but I think its wrong, because the strenght of for example an acid is determined by "whether the acid gives away a proton easiliy or not". When the Ka is very high that indicates that the dissolved molecules always donate protons. This results in complete dissociation of the dissolved acid molecules meaning its a strong acid (!) even though the solubility is low. The low solublity may cause the solution to have not too many H+ ions & therefore the pH value will be slightly under 7 & that tells us that its a slightly acidic solution, but the acid is still a strong acid (because of the high Ka) ! This tells us that we dont look at the pH to determine acid/base strenght. It only tells us whether the solution will be basic or acidic. Of course it can sometimes predict the strenght of an acid/base, but not always as it doesnt work with low soluble high Ka acids for example. In this case the pH would be slightly under 7, maybe 6.8 which does not mean weak acid! The Ka is very high so its a strong acid. What is your opinion on this? I have found other sources saying the same.
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  21. #20  
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    I'd just be going around in circles and repeating what I've already posted. I'll leave it for someone else to have a go at answering you. Hopefully you will get an answer that makes sense to you.
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    Hehe, I have proof to back up my claims though: Mg(OH)2 is slightly soluble, but its still considered a strong base:

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    Well, that is yahoo answers, you might or you might not take what they say there as granted.

    Now, let me put it in an other way. I do not know your level of understanding in Le Chatelier's principle, but here is my argument:

    A precipitate always keeps equilibrium with a saturated solution, eg.:

    Mg(OH)2(solid) <---> Mg(OH)2(dissolved)

    Then, the dissolved species keeps equilibrium with its' hydrated ions:

    Mg(OH)2(dissolved) <---> Mg(2+) + 2 OH(-)

    These are our 2 equilibria. Now, let's suppose that Mg(OH)2 is a strong base. This implies, by definition, that the second equlibrium does not exist, it is shifted all the way to the right, no backreaction. This means that all the Mg(OH)2 precipitate that is dissolved, readily dissociates. Le Chatelier says then, the first equilibrium must be shifted to the right, continuosly replacing all the dissolved Mg(OH)2, eventually dissolving all the precipitate.

    And as I have already mentioned, precipitate only exists under a concentrated solution. If Mg(OH)2 does not exist in the solution (only in ionic form), then the precipitate can not exist.

    Therefore, a precipitate can only be a weak base.
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  24. #23  
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    Does this mean that pH indeed always correctly determines the strenght of a base & the strenght of a base is also dependant on solubility?
    While the strenght of an electrolyte (officially) is not dependant on solubility?
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  25. #24  
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    Indeed, if you have an 1 M solution of a strong base (which has one equivalent OH-, like NaOH), it will be pH 14. If your base is not strong (like NH3), your 1 M solution MUST have less than 14 pH. If you look at it this way, pH determines the strength.

    ps.: If you knew the concentration of your NH3 solution, and the pH of the solution, you could calculate Kb.
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  26. #25  
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    Ok thank you all for the clarification! I didnt know it was that complicated, but I got it now.
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