# 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?

## $\text{a. adding more hydrocyanide;}$ $\text{b. raising the temperature;}$ $\text{c. adding more solvent;}$ $\text{d. adding another acid.}$

Dec 28, 2015

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; $p {K}_{a}$ SHOULD be higher at elevated temperatures, $>$ $298 K$, and indeed it is.

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