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    H2CO3(aq) + H2O <--> HCO3-(aq) + H3O+(aq)

    Given that Ka = 2.9*10^-4 moldm^-3, calculate the pH of 0.010moldm^-3 carbonic acid

    Ka =  \frac{[HCO3^-][H3O^+]}{[H2CO3]} Am I correct in saying H2O(l) doesnt effect Ka ? And that [H3O+] = [H+] ? (why is this please?)

     Ka = 2.0 \times 10^{-4}moldm^{-3} = \frac{[HCO3^-][H^+]}{0.010}

    Im not sure how to work out [HCO3-]...theres no volume given so I cant use c=m/v...or can you assuem the reaction uses exactly 1mol of H2CO3(aq) and therefore use v = m/c..or would that only give the volume of H2CO3- ?

    Please help!

    Thank you

    P.S. sorry im not sure how to do subscripts on here
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    For a weak acid such as this: HA ---> H+ + A-

    and so [A-] = [H+] so Ka = [H+]2/[HA]

    how to do subscripts and superscripts: http://www.thestudentroom.co.uk/anno...hp?f=130&a=581
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    yeah beacuse H+ and A- have the same concentrations, then you can treat them as the same value. So you just square the concentration of H+ (well thats the generally accepted method of doing it but it still works mathematically if you do it the other way, obviously).
 
 
 
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