Question #eece7

1 Answer
Mar 5, 2016

Answer:

This is for expression (a).

Explanation:

#K_"h"# is not defined but I assume it to be the equilibrium constant of the hydrolysis reaction.

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 #K_"w" = ["H"_3"O"^+]["OH"^-]#, which is around #10^{-14}# at room temperature.

Also you must know the dissociation of #"HCN"#

#"HCN" + "H"_2"O" rightleftharpoons "H"_3"O"^+ + "CN"^{-}#

has equilibrium constant of #K_"a" = frac{["H"_3"O"^+][ "CN"^{-}]}{["HCN"]}#.

So what do you get if you divide #K_"w"# by #K_"a"#?

#frac{K_"w"}{K_"a"} = frac{["H"_3"O"^+]["OH"^-]}{(frac{["H"_3"O"^+][ "CN"^{-}]}{["HCN"]})}#

#= frac{ ["HCN"] ["OH"^{-}] }{ ["CN"^{-}] }#

#= K_"h"#