# balanced equation for the reactions of..?

aminoethanoic acid with dilute sulphuric acid

and

aminoethanoic acid with nitrous acid

Relevance

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(+)