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Thread: Law of ostwald confusion

  1. #1 Law of ostwald confusion 
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    Hello,

    My book says the following about the law of ostwald:

    A + H2O <---> B + H3O+

    "If we increase the concentration of water the equilibrium will shift to the right."
    The concentration of water is supposed to be constant, this is why we cancelled it out! So how are they saying it increases. The concentration of water does not increase it stays constant! It cant increase since the concentration of water is a whole different thing than the concentration for example acid A in water. The concentration of water is calculated by dividing the amount of substance of only H2O by the volume of only the water molecules ( A molecules, B molecules & H3O molecules excluded).


    Thanks in advance!


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  3. #2  
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    Again, if that's a quote from your book I think you need to get a better book. It is explained better here:

    https://ftp.rush.edu/users/molebio/B...0Chpt.%207.pdf

    Startng near the bottom of page 273


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  4. #3  
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    Thank you for the help! I come to the following conclusion from reading this text: Because we cant explain the equilibrium shift to the right with le chateliers principle, we use the law of dilution (ostwald law) to explain the shift of equilibrium to the right.
    In
    If we dilute the solution, c will decrease, but K must always be constant (excluding temperature change of course) so the a (alpha) must increase, meaning that the equilibrium shifts to the right.
    This equilibrium shift could not be explained by le chateliers principle , because dilution does not change the concentration of water (concentration of water has been agreed to be constant) and therefore no equilibrium shift is "produced" by le chateliers law.
    This is why you told me that what my book says "If we increase the concentration of water the equilibrium will shift to the right." is wrong, because le chateliers law does not apply here & does not explain the equilibrium shift to the right. But Ostwald law which only works for weak electrolytes can explain that change. Am I approximately correct?


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  5. #4  
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    Also on a side note regarding le chateliers: the book says that the concentration of water increases therefore the equilibrium shifts to the right. That is not correct according to you. But then we could use volume change to see whether it actually does something according to le chatelier. Because we dilute the solution there is a volume increase. So the equilibrium should shift towards the side with highest sum of coefficients.
    In both sides have the same sum 2 so the equilibrium does not shift according to le chatelier. But the equilibrium does (!) shift when we dilute the solution so we can conclude that le chateliers principle cant explain the shift. This is the reason we use Ostwald's law, because this law explains why there is an equilibrium change!
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  6. #5  
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    I think a lot of the problems you are having is that you are reading too much into Ostwalds Law, Le Chatelier's principle, etc. These are phenomenological "principles" that "sort of" explain what is going on and are taught in lower school so simple predictions can be made without getting too involved in thermodynamics. They are part of what I call the "lies to children until they can handle the maths" school of teaching chemistry. Once you "get" physical chemistry these principles aren't used and it's all about "chemical potential". It's so long since I used these "principles" and not the proper thermodynamics I'm probably not the best person to help you out so I'll bow out and hopefully someone else will give you an answer that makes sense to you.
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  7. #6  
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    I cant agree more. The more I look into thing the more I notice something is not right. When I learn chemistry I think every little block must fall into place like in mathematics. But I think chemistry is not like math since in chemistry 1+1 is not always 2 in my opinion. I will skip this, because I think its not necessary to go deeper actually it could make me more confused, but I thank you for the help, because you have provided me with very useful information.
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  8. #7  
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    Quote Originally Posted by mcfaker123 View Post
    Also on a side note regarding le chateliers: the book says that the concentration of water increases therefore the equilibrium shifts to the right. That is not correct according to you. But then we could use volume change to see whether it actually does something according to le chatelier. Because we dilute the solution there is a volume increase. So the equilibrium should shift towards the side with highest sum of coefficients.
    In both sides have the same sum 2 so the equilibrium does not shift according to le chatelier. But the equilibrium does (!) shift when we dilute the solution so we can conclude that le chateliers principle cant explain the shift. This is the reason we use Ostwald's law, because this law explains why there is an equilibrium change!
    According to my understanding of Le Chatelier's principle, equilibria tend to shift to oppose any change. There are a lot of "principles" like this in science e.g. Lenz's Law in electromagnetism. At school we called it, generically, the Principle of Cussedness.

    It is not necessary to have different numbers of components on either side for Le Chatelier's Principle to operate. I wonder if you may be thinking of gas reactions in which there are different numbers of molecules on either side, leading to pressure being able to shift the equilibrium. In that case you can invoke Le Chatelier's Principle to conclude that increasing pressure favours the side of the equilibrium with fewer molecules. But that's just one application of it that is not relevant here.

    In this case, indeed, adding more water will tend to produce a change that takes up some of the newly introduced water - in this case converting it to hydronium ions. The way I think of it is by statistics: more water means more dilution so the chances of an encounter between a hydronium ion and B to re-form A are lower than they were before. So less likelihood of going from right to left. I'd assume no change to the likelihood of going to the right, as A will be fully surrounded by water already, unless the solution is very concentrated indeed. So the net effect is a shift to the right.
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    Indeed I was talking about the fact that pressure change does shift equilibrium to the side with fewer molecules. I just found out that it actually could be applicable here if we cancelled out the H20:
    volume increase/ concentration decrease/ pressure decrease leads here to a shift to the right because 2>1. But for this to work I think we would have to cancel out the H20. When we did not cancel H20 out it would be 1 vs 1 and that leads us nowhere.


    Regarding ostwalds law in general, the dilution does not affect the Ka value, therefore it is not supposed to affect the acid's strenght. But when we look closer at the definition of a strong acid it means an acid that easily donates a proton.
    Now when we dilute the solution according to the law of dilution the Co will decrease and to maintain a constant K value the a will increase! Now when the a increases it means more acid molecules give away their protons! It gives me the impression that diluting the solution the strenght of the acid is increased, because more acid molecules give away H+ and this contradicts with the constant Ka value! If Ka is constant then the strenght of acid should not change, but it does, or not?
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  10. #9  
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    Quote Originally Posted by mcfaker123 View Post
    Indeed I was talking about the fact that pressure change does shift equilibrium to the side with fewer molecules. I just found out that it actually could be applicable here if we cancelled out the H20:
    volume increase/ concentration decrease/ pressure decrease leads here to a shift to the right because 2>1. But for this to work I think we would have to cancel out the H20. When we did not cancel H20 out it would be 1 vs 1 and that leads us nowhere.


    Regarding ostwalds law in general, the dilution does not affect the Ka value, therefore it is not supposed to affect the acid's strenght. But when we look closer at the definition of a strong acid it means an acid that easily donates a proton.
    Now when we dilute the solution according to the law of dilution the Co will decrease and to maintain a constant K value the a will increase! Now when the a increases it means more acid molecules give away their protons! It gives me the impression that diluting the solution the strenght of the acid is increased, because more acid molecules give away H+ and this contradicts with the constant Ka value! If Ka is constant then the strenght of acid should not change, but it does, or not?
    Be careful. The volume changes involved in reactions that take place in solution are very small indeed and can almost always be neglected. And you can't predict the direction it will go unless you have a lot of knowledge of the space different species take up in solution, which may not be at all obvious due to solvation effects.

    But you are right, in principle, the more dilute a solution of acid is, the more completely it will be dissociated. Obviously that does not mean it has become a "stronger" acid, though, in any meaningful sense. The "strength" of an acid is usually defined by its tendency to dissociate. That's what things like pKa are all about. The degree to which it actually does dissociate, at any given concentration, indeed varies according to the formula for the equilibrium constant that you have given.

    What we habitually call "strong" acids are those that are to all intents and purposed fully dissociated at just about any concentration we are likely to encounter, but in principle they are also subject to these equilibria.
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  11. #10  
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    Perfect explanation. I appreciate your help a lot & thank you for clearing that up!
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