Question eece7

Mar 5, 2016

This is for expression (a).

Explanation:

${K}_{\text{h}}$ 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

$2 {\text{H"_2"O" rightleftharpoons "H"_3"O"^+ + "OH}}^{-}$

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 $\text{HCN}$

${\text{HCN" + "H"_2"O" rightleftharpoons "H"_3"O"^+ + "CN}}^{-}$

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

So what do you get if you divide ${K}_{\text{w}}$ by ${K}_{\text{a}}$?

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

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

$= {K}_{\text{h}}$