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Derive the equation of ionization constant (K_b) of weak base. |
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Answer» Solution :The general EQUATION of weak base is MOH. The ionization of base MOH can be REPRESENTED by equation. `{:("Equilibrium:",MOH_((AQ))+(aq)HARR, M_(aq)^(+)+,OH_((aq))^(-)),("CONCENTRATION M:",C M, 0,0),("Change in conc.:", -C alpha, +C alpha, +C alpha),("Concentration at equili. in molarity:",C-Calpha,(0+Calpha),(0+Calpha)),(,=C(1-alpha),=Calpha,=Calpha):}` where , C=Initial concentration of base in molarity . `alpha`= The ionization degree of base `therefore` Amount of ionization MOH = `Calpha` `therefore` The concentration decrease in base when equilibrium is reached =`Calpha M` `therefore` At equilibrium concentration of MOH= `(C-Calpha)=C(1-alpha)` Equilibrium `[M^+]=[OH^-]=Calpha M` On equilibrium reaction in solution of MOH base. `K=([M^+][OH^-])/([MOH])`=[Ionization constant `K_b` of weak base ] `therefore K_b=([M^+][OH^-])/"[MOH]"=([OH^-]^2)/([MOH])` ....(Eq.-i) `[OH^-]=sqrt(K_b. [MOH])` ...(Eq. ii) So, `K_b=((Calpha))/(C(1-alpha))=(Calpha^2)/(1-alpha)` ...(Eq.-iii) This equation -(i,ii) and (iii) are ionization constant equations of ionic equilibrium of weak base. |
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