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Answer 1 · 2016-03-05T08:01:23

This is for expression (a).

$K_"h"$ 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 $K_"a"$ ?

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

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

$= K_"h"$