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Calculate the degree of dissociation and concentration of H_(3)O^(+) ions in 0.01 M solution of formic acid. K_(a)=2.1xx10^(-4) at 298 K. |
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Answer» Solution :Formic acid is weakelectrolyte and ionizes inwaterto give `H_(3)O^(+)` ions ACCORDING to the equation : `HCO OH+H_(2)O HARR H_(3)O^(+)+HC O O^(-)` Let `alpha` be the degree of ionization. Then the concentration of the various species present at equilibrium wouldbe as under : `{:(,HCO OH + H_(2)O hArrH_(3)O^(+) +HCO O^(-)),("Initial conc. ",0.01"00"),("Conc. at eqm",0.01(1-alpha)~=0.01""0.01 alpha ""0.01alpha):}` [`:' alpha` is very small and can be neglected in comparison to 1] THUS, `K_(a)=(0.01 alpha xx 0.01 alpha)/(0.01)=0.01 xx alpha^(2) :. 0.01 alpha^(2)=2.1xx10^(-4)[K_(a)=2.1xx10^(-4), "given"]` or `alpha^(2)=(2.1xx10^(-4))/(0.01)=2.1xx10^(-2)` `:.` Degree of ionisation, `alpha = sqrt(2.1xx10^(-2))=0.14` Concentration of `H_(3)O^(+)` ions `=C alpha = 0.14 xx 0.01 = 1.4 xx 10^(-3) "mol" L^(-1)` |
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