Saved Bookmarks
| 1. |
Explain ionization and ionization constant in di and polyprotic acid . |
|
Answer» SOLUTION :As a example , the ionization of dibasic acid `H_2X` in aqueous solution is represented in TWO step . (i)`H_2X_((aq))+aq hArr H_((aq))^(+) + HX_((aq))^(-)` (ii)`HX_((aq))^(-) + aq hArr H_((aq))^(+) + X_((aq))^(2-)` If EQUILIBRIUM constant of `K_a` (i) and `K_a` (ii) of this both equilibrium (i) and (ii) then, `therefore K_a (i)= ([H^+][HX^-])/([H_2X]) , K_a (ii)= ([H^+][X^(2-)])/([HX^-])` So, `K_a (i) xx K_a (ii) = ([H^+]^2 [X^(2-)])/([H_2X])` but Reaction (i) + Reaction (ii) `H_2X_((aq))+aq hArr 2H_((aq))^(+) + X_((aq))^(2-)` For this , equilibrium constant `K_a` (iii) is, `K_a (iii) = ([H^+]^2[X^(2-)])/([H_2X])` So, For dibasic acid, `K_a (iii) = K_a (i) xx K_a` (ii)..... (Eq. -i) where , `K_a` (i)=FIRST ionization constant , `K_a` (ii) is second ionization constant. For any polybasic and RESPECTIVELY `K_a (i) , K_a (ii)`..... than `K_a = K_a (i) xx K_a (ii) xx` ....(Eq.-ii) Generally `K_a(i) gt K_a(ii) gt K_a (iii)`....as the after formation ion the remove of proton is difficult. |
|