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Thread: Very hard chemistry problem. >:(

  1. #1 Very hard chemistry problem. >:( 
    Forum Junior DivideByZero's Avatar
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    Ok. First of all this is a homework problem so please provide your steps on how you solved this.

    Calculate the for the following acids using the given information.

    (a) 0.220 M solution of H3AsO4, pH = 1.50



    so i know the concentration of H+ ions would be 10^-1.5 = 0.0316 M

    I think the dilution of the H3AsO4 is:
    H3AsO4 --> 3H + AsO4

    now what?


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  3. #2  
    Forum Masters Degree organic god's Avatar
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    Well firstly we can say that this acid is a weak acid.

    Therefore we can make the assumption that the dissacotiation of
    the acid is low and therefore
    [H+] = [H2AsO4-]

    Ka = [H+][H2AsO4-]/[H3AsO4]

    therefore using the above assumption

    Ka = [H+]^2 / [H3AsO4]

    you used the correct method to get [H+]

    doing this i get an answer of 4.55 *10^-3 mol dm-3


    everything is mathematical.
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  4. #3  
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    I want to hijack this thread. Sorry.

    How can you calculate the pH of chloroform, for example?
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  5. #4  
    Forum Junior DivideByZero's Avatar
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    Thats exactly what I put for my answer but I keep getting it wrong!

    I'm entering my answer in a website called 'Webassign' where the teacher assigns us homework but it keeps saying my answer is wrong. Its so frustrating.
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  6. #5  
    Forum Junior DivideByZero's Avatar
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    Quote Originally Posted by georginho_juventusygr
    I want to hijack this thread. Sorry.

    How can you calculate the pH of chloroform, for example?
    I think the pH of any substance depends on its concentration.
    For example a very concentrated amount of hydrochloric acid would have a lower pH than a not so concentrated one.
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  7. #6  
    Forum Masters Degree organic god's Avatar
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    Thats exactly what I put for my answer but I keep getting it wrong!

    I'm entering my answer in a website called 'Webassign' where the teacher assigns us homework but it keeps saying my answer is wrong. Its so frustrating.
    try changing the number of significant figures in your answer
    everything is mathematical.
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  8. #7  
    Forum Sophomore oceanwave's Avatar
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    if its a weak acid, i doubt that it would undergo such a great hydrogenation thus i think only 1 ion for the equation would be correct.
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  9. #8  
    Forum Junior DivideByZero's Avatar
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    Quote Originally Posted by oceanwave
    if its a weak acid, i doubt that it would undergo such a great hydrogenation thus i think only 1 ion for the equation would be correct.

    nope still didn't work

    oh well..
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  10. #9  
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    Quote Originally Posted by DivideByZero
    Quote Originally Posted by georginho_juventusygr
    I want to hijack this thread. Sorry.

    How can you calculate the pH of chloroform, for example?
    I think the pH of any substance depends on its concentration.
    For example a very concentrated amount of hydrochloric acid would have a lower pH than a not so concentrated one.
    Aye ... but it seems I can't figure it out. It's chloroform, not a usual acid.
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  11. #10  
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    I get 0.0053

    Ka = [H+] * [H2AsO4-]/[H3AsO4]

    [H+] and [H2AsO4-] both equal 0.0316 M

    [H3AsO4] = .22-0.0316

    so Ka = 0.0316^2/(.22-.0316) = 0.0053

    Edit: I suspect that you are forgetting to subtract the concentration of H+ from the concentration of H3AsO4 when you set up your equilibrium equation.
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  12. #11  
    Forum Junior ArezList's Avatar
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    H3AsO4 --> 3H + AsO4
    so I believe that [H+]=[AsO4,3-] isn't is right

    Yeah, it's a weak acid and there're 3 H in such a acid, then comes 3 stages of ionization, someone just calculate the first two stages which is like:
    Ka' = [H+] * [H2AsO4-]/[H3AsO4]
    Ka'' = [H+][HAsO4,2-]/[H2AsO4-]
    But i think the overall Ka should be

    Ka = [H+][AsO4,3-]/[H3AsO4].
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  13. #12  
    Forum Sophomore oceanwave's Avatar
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    Quote Originally Posted by ArezList
    H3AsO4 --> 3H + AsO4
    so I believe that [H+]=[AsO4,3-] isn't is right

    Yeah, it's a weak acid and there're 3 H in such a acid, then comes 3 stages of ionization, someone just calculate the first two stages which is like:
    Ka' = [H+] * [H2AsO4-]/[H3AsO4]
    Ka'' = [H+][HAsO4,2-]/[H2AsO4-]
    But i think the overall Ka should be

    Ka = [H+][AsO4,3-]/[H3AsO4].
    shldnt a weak acid have a 'double arrow' instead? since Ka means the acidity constant, what we are calculating for would be the constant for a weak acid. if it really did ionise 3 ways which i think is very improbable for weak acids, then the Ka would really be very hard a ascertain.
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  14. #13  
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    Only the first proton comes off.
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