Weak acids and bases are not completely ionised when dissolved in polar medium like water H rarr H^(+) +A^(-) {:(t_(0),C,O,O),(t_(eq),C-Calpha,Calpha,Calpha):} K_(a)=(C alpha^(2))/(1-alpha)=Calpha^(2),alphasqrt((K_(a))/(C)) :. (alpha_(1))/(alpha_(2))=sqrt((Ka_(1))/(Ka_(2))),(alpha_(1))/(alpha_(2))=sqrt((C_(2))/(C_(1))) ( For two acids at same conc.) (for same acid at diff conc. ) alphaand alpha are in the ratio 1:2 at same conc. Ka_(1), =2xx 10^(-4), what will be Ka_(2),?

Weak acids and bases are not completely ionised when dissolved in pola

1.	Weak acids and bases are not completely ionised when dissolved in polar medium like water H rarr H^(+) +A^(-) {:(t_(0),C,O,O),(t_(eq),C-Calpha,Calpha,Calpha):} K_(a)=(C alpha^(2))/(1-alpha)=Calpha^(2),alphasqrt((K_(a))/(C)) :. (alpha_(1))/(alpha_(2))=sqrt((Ka_(1))/(Ka_(2))),(alpha_(1))/(alpha_(2))=sqrt((C_(2))/(C_(1))) ( For two acids at same conc.) (for same acid at diff conc. ) alphaand alpha are in the ratio 1:2 at same conc. Ka_(1), =2xx 10^(-4), what will be Ka_(2),?
Answer» `8xx10 ^(-4)` ` 2xx10 ^(-4)` ` 4xx10 ^(-4)` ` 1xx10 ^(-4) ` Solution :` (alpha_1)/(alpha_2) = SQRT((Ka_1)/(Ka_2) )RARR (1)/(2)= sqrt((2xx 10 ^(-4) )/(Ka_2)) , Ka_2 = 8xx 10^(-4) `