Question

For the dissociation of hydrogen cyanide in aqueous solution, i.e. `HC-=N(aq) + H_2O(l) rightleftharpoons H_3O^+ + ""^(-):C-=N`... which scenario will change the equilibrium constant?

`"a. adding more hydrocyanide;"`
`"b. raising the temperature;"`
`"c. adding more solvent;"`
`"d. adding another acid."`

Answer 1

Only (b) will (possibly) change the equilibrium constant.

Explanation:

Hydrocyanic (prussic) acid undergoes the acid-base reaction as follows:

`HC-=N + H_2O rightleftharpoons ""^(-)C-=N + H_3O^+`

As with any equilibrium reaction, we can write the acid-base dissociation expression, `K_a` `=` `([""^(-)C-=N][H_3O^+])/[H_2O]`, which simplifies to `K_a` `=` `[""^(-)C-=N][H_3O^+]`.

Now I don't know the value of `K_a`; I am too lazy to look it up (you shouldn't be!); `K_a` will be reasonably small. `K_a` is quoted for conditions of standard temperature and pressure. I suspect that raising the temperature will increase `K_a` somewhat. Why? Because the above is a bond breaking reaction, and raising the temperature should allow greater bond cleavage. This question is also typically asked of the autoprotolysis of water; `pK_a` SHOULD be higher at elevated temperatures, `>` `298K`, and indeed it is.

I take it that this is a 1st year problem?