Question #eece7
1 Answer
This is for expression (a).
Explanation:
At equilibrium, we know that the expression
#frac{ ["HCN"] ["OH"^{-}] }{ ["CN"^{-}] } = K_"h"#
is a constant. But what is it equal to?
Firstly, you must know that
#2 "H"_2"O" rightleftharpoons "H"_3"O"^+ + "OH"^-#
has equilibrium constant of
Also you must know the dissociation of
#"HCN" + "H"_2"O" rightleftharpoons "H"_3"O"^+ + "CN"^{-}#
has equilibrium constant of
So what do you get if you divide
#frac{K_"w"}{K_"a"} = frac{["H"_3"O"^+]["OH"^-]}{(frac{["H"_3"O"^+][ "CN"^{-}]}{["HCN"]})}#
#= frac{ ["HCN"] ["OH"^{-}] }{ ["CN"^{-}] }#
#= K_"h"#