balanced equation for the reactions of..?
aminoethanoic acid with dilute sulphuric acid
aminoethanoic acid with nitrous acid
- 1 decade agoFavorite Answer
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