Solution.
For weak acid:
Ka=[H+][A−][HA]Ka = \frac{[H^+][A^-]}{[HA]}Ka=[HA][H+][A−]
For weak base:
Kb=[B+][OH−][BOH]Kb = \frac{[B^+][OH^-]}{[BOH]}Kb=[BOH][B+][OH−]
Final reaction:
A−+B++H2O=HA+BOHA^- + B^+ + H2O = HA + BOHA−+B++H2O=HA+BOH
Kh=[HA][BOH][A−][B+]Kh = \frac{[HA][BOH]}{[A^-][B^+]}Kh=[A−][B+][HA][BOH]
So, we get:
Kh=KwKa×KbKh = \frac{Kw}{Ka \times Kb}Kh=Ka×KbKw
Because:
Kw=[H+][OH−]Kw = [H^+][OH^-]Kw=[H+][OH−]
Answer:
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