Question

Question #eece7

Answer 1

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"`