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Thread: Chemistry Help

  1. #1 Chemistry Help 
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    1-The concentration of water,solid substances and precipates should not appear in the equilibrium constant equation Kc , why?I don't understand.Please don't tell me that because their concentration remains constant whatever their quantities. because I have read these words tens of times and I don't really understand them.

    2-according to my text book " if the reactants and the products are in the gasoeus state,the concentration is expressed by their partial pressure" What does this mean?

    3-CH3COOH <----->CH3COO- + H3O+

    let a be the no. of dissociated moles.So if we want to calculate the concentration of CH3COOH why should we say its remaining concentration=C-a,where C is the orginal concentration of CH3COOH before dissociation????.

    4-H2O <---->H+ + OH-
    Kc= [H+] [OH-]/[H2O]=10^-14
    Kw=[H+] [OH-] = 10^-14
    how could the two equations have the same value?why we are totally neglecting water?and what is the importance of calculating Kw?

    I've asked a question here here but it made no sense so please don't repeat the same answer.
    http://answers.yahoo.com/question/in...0130754AASJDiN
    Help me please I'm very confused!
    Thanks in advance.


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  3. #2  
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    1. I really don't know another way to explain this one other than the way it's already been explained. Sorry.

    2. Instead of [Products]/[Reactants], you will use (Partial Pressure of Products)/(Partial Pressure of Reactants) to obtain the constant Kp (similar to Kc, but not always exactly the same).

    3. Because you started with "C" mol of CH3COOH, but "a" mol of it dissociated. That means "a" mol of CH3COOH has become CH3COO-, i.e., IT IS NO LONGER CH3COOH. Therefore, the amount of CH3COOH is the original amount ("C" mol) minus the amount that dissociated ("a" mol). (I believe that you have reduced the volume to 1 L in this particular problem. That's why you can do this.)

    4. See #1 ^^


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  4. #3  
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    1. The concentration of water is so close to 100% that you consider it to be 100%. Hence the concentration of water is 1, and 1 times any number gives you the same number so it can be neglected.
    Solids and precipitates are NO LONGER IN THE SOLUTION. They will not react with the reactants or products in the solution and hence are no longer variables of the equation. Molecules can only react with other molecules that they encounter, and molecules in the solution will only encounter molecules in the solution.

    4.
    Kc= [H+] [OH-]/[H2O]=10^-14
    Kw=[H+] [OH-] = 10^-14

    Well, 10^-14 divided by 1 is still 10^-14. Why do we want to know this number? We want to know how much of the H2O is actually in that form and how much is split up into H+ and OH-.

    Misr's answers for 2 and 3 were good.
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