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Derive K_a xx K_b=K_w equation. |
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Answer» SOLUTION :`NH_3` (ammonia) is a weak base and conjugate acid of `NH_3` is `NH_4^+` . Following equilibrium is ESTABLISHED in `NH_3`. (i)`NH_(3(AQ)) + H_2O_((l)) hArr NH_(4(aq))^(+) + OH_((aq))^(-)` `K_b=([NH_4^+][OH^-])/([NH_3])` (Suppose `K_b=1.8xx10^(-5)` ) `NH_4^+` is a conjugate acid of `NH_3`.Its equilibrium in aqueous solution act as acid is following (ii)`NH_(4(aq))^(+) + H_2O_((l)) hArr H_3O_((aq))^(+) + NH_(3(aq))` Take ionization constant of weak acid `NH_4^+` is `K_a`, `K_a=([H_3O^+][NH_3])/([NH_4^+])` (Suppose `K_a=5.6xx10^(-10)`) Addition of reaction (i) and (ii) and net reactionis, (i)+(ii) = `2H_2O_((l)) hArr H_3O_((aq))^(+) + OH_((aq))^(-)` This reaction is self equilibrium of water, `K_w=[H_3O^+][OH^-]=1.0xx10^(-14)` `K_b` reaction of (i) x `K_a` reaction of (ii) , `THEREFORE K_axxK_b=([H_3O^+][NH_3])/([NH_4^+])xx([NH_4^+][OH^-])/([NH_3])` `therefore K_a xx K_b= [H_3O^+] [OH^-]= K_w` ....(Eq.-i) e.g., `(5.8xx10^(-10))(1.8xx10^(-5))= 1.0xx10^(-14)` This can be extended to make a generalisation . Thus, one reaction equilibrium constant =`K_1` and SECOND reaction equilibrium constant = `K_2` and reaction (1) + reaction (2) = reaction (3) So, Reaction (iii) `K_3=K_1xxK_2`....(Eq.-ii) |
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