balanced equation for the reactions of..?

aminoethanoic acid with dilute sulphuric acid

and

aminoethanoic acid with nitrous acid

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  • 1 decade ago
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    you can tell which of the compounds is going to have to keep the proton and which one is going to have to take in a proton by observing their Ka values ([H+][A-]/[HA]). Those with larger Ka values want to give away their protons ( and stay as A- )while those compounds with smaller Ka values will want to keep their protons (and stay as HA).

    aminoethanoic acid has a Ka of 10^(-2.35)

    sulfuric acid has a Ka of 10^10

    nitrous acid has a Ka of 10^(-3.3)

    as you can see, sulfuric acid really doesnt want those protons while nitrous acid really wants them. aminoethanoic acid is somewhere in between. Now for the equations.

    since sulfuric acid has a larger Ka than aminoethanoic acid, it looks like aminoethanoic acid will have to take that proton

    2C2H5NO2 + H2SO4 -> 2C2H6NO2(+) + SO4(2-)

    since aminoethanoic acid has a larger ka than nitrous acid, it looks like nitrous acid will have to take the proton

    C2H5NO2 + HNO2 -> C2H4NO2(-) + H2NO2(+)

  • 1 decade ago

    NH2CH3COOH + H2SO4 -------->NH2SO4 + CH3CH(OH)2

    dont know the 2nd one

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