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Derive the equation of ionization constants K_a of weak acids HX. |
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Answer» Solution :A weak acid HX that is partially ionized in the aqueous solution. The equilibrium can be expressed by : `{:("Equilibrium:", HX_((aq)) +H_2O_((l))hArr, H_3O_((aq))^(+)+, X_((aq))^(-)),("Concentration M :", C,0,0),("Change in conc.:", -Calpha,(0+Calpha), (0+Calpha)),("Concentration at Equili. in molarity:",(C-Calpha),Calpha,Calpha),(,=C(1-alpha),=Calpha,=Calpha):}` Where, `alpha`= Extent UPTO which HX is ionized into IONS. SUPPOSE , initial concentration of undissociated acid HX = `Calpha` M `therefore` Ionized HX= `Calpha M` At equilibrium `[HX]=(C-Calpha)=C(1-alpha)` `[H_3O^+]=[X^-] =C alpha M` The equilibrium CONSTANT for the above DISCUSSED acid-dissociation equilibrium: `K=([H_3O^+][X^-])/([HX][H_2O]) therefore K[H_2O]=([H_3O^+][X^-])/([HX])` In `[H_2O]` = constant = K. So, `K[H_2O]`= K x K. new constant `K_a` where, `K_a`=Ionization constant of weak acid. `therefore K_a=([H_3O^+][X^-])/([HX])=([H_3O^+]^2)/([HX])` ....(Eq.-i) `therefore K_a=((Calpha)(Calpha))/(C-(1-alpha))=(Calpha^2)/((1-alpha^2))`....(Eq.-ii) Above equation -(ii) is used for calculation of ionization constant of weak acid HX. |
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