Derive K_w=K_axxK_b and K_w=pK_a+pK_b for weak base B and its conjugat

1.	Derive K_w=K_axxK_b and K_w=pK_a+pK_b for weak base B and its conjugate acid BH^+.
Answer» Solution :If weak base B, so equilibrium in its solution is, `B_((aq))+H_2O_((l)) hArr BH_((aq))^(+) + OH_((aq))^(-)` …(i) In above ionic equilibrium of base , the constantis `K_b` and `H_2O_((l))`is taken as constant….. `K_b=([BH^+][OH^-])/"[B]"`...(ii) If this expression is multiplied & DIVIDED by `[H^+]`, `K_b=([BH^+][OH^-][H^+])/([B][H^+])=([OH^-][H^+][BH^+])/([B][H^+])` In it `[OH^-][H^+]=K_w` and `([BH^+])/([B][H^+])=1/K_a` Because , `BH^+` (acid) `hArr B_((aq)) + H_((aq))^(+)` So, `K_b=K_w/K_a` and `K_w=(K_a)(K_b)` According to above, `K_bxx K_a = K_w = 1xx10^(-14)` taking log both the SIDE `THEREFORE (-log K_b)+(-log K_a)=-log K_w= log (1xx10^(-14))` `thereforepK_b+pK_a = pK_w=+14`.....(Eq.-iii)

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