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  1. #1
    Senior Member euclid2's Avatar
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    chemistry

    Calculate the hydroxide ion concentration in a 0.25 mol/L HBr(aq) solution.
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  2. #2
    Member TheMasterMind's Avatar
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    Quote Originally Posted by euclid2 View Post
    Calculate the hydroxide ion concentration in a 0.25 mol/L HBr(aq) solution.
    HBr(aq) ---> H+(aq)+Br-(aq)

    entities are H+(aq),Br-(aq),H2O(l)

    Kw=[H+(aq)][OH-(aq)]

    [OH-(aq)]=Kw/[H+(aq)]

    now simply make the necessary substitutions
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  3. #3
    Senior Member euclid2's Avatar
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    Quote Originally Posted by TheMasterMind View Post
    HBr(aq) ---> H+(aq)+Br-(aq)

    entities are H+(aq),Br-(aq),H2O(l)

    Kw=[H+(aq)][OH-(aq)]

    [OH-(aq)]=Kw/[H+(aq)]

    now simply make the necessary substitutions
    I'm not sure where you're getting this from can you expand please?
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  4. #4
    Member TheMasterMind's Avatar
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    Quote Originally Posted by euclid2 View Post
    I'm not sure where you're getting this from can you expand please?
    [OH-(aq)]=Kw/[H+(aq)]

    =1.00*10^-14/0.25

    =4.0*10^-14 mol/L

    [OH-(aq)]=4.0*10^-14mol/L

    Does this make sense?
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  5. #5
    Senior Member euclid2's Avatar
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    Quote Originally Posted by TheMasterMind View Post
    [OH-(aq)]=Kw/[H+(aq)]

    =1.00*10^-14/0.25

    =4.0*10^-14 mol/L

    [OH-(aq)]=4.0*10^-14mol/L

    Does this make sense?
    It seems you are ignoring the amount of [OH-(aq)]
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  6. #6
    Member TheMasterMind's Avatar
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    Quote Originally Posted by euclid2 View Post
    It seems you are ignoring the amount of [OH-(aq)]
    The amount of OH is insignificant produced by the ionization of water in presence of Br
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  7. #7
    Member TheMasterMind's Avatar
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    Quote Originally Posted by euclid2 View Post
    I'm not sure where you're getting this from can you expand please?
    Note: since Kw=[H+(aq)][OH-(aq]

    then [H+(aq)]=Kw/[OH-(aq)]

    and [OH-(aq)]=Kw/[H+(aq)]
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