How do you write an expression for the acid ionisation constant \[({K_a})\] for $HF$?

Hint: As we know that there are weak acids and strong acids on the basis of their ionisation in water and strong acids generally completely ionises in water whereas weak acids are partially soluble in water.

Complete step-by-step answer:
As we are well aware with the ionisation constant of acids and bases are basically the equilibrium constant for ionisation of an acid in water which are usually determined by the ratio of the concentrations of products of a compound to the reactant. So, for a given acid such as $HA$, we know that the ionisation will be given as:
$\Rightarrow HA(aq) \to {H^ + }(aq) + {A^ - }(aq)$
Hence, the expression for the ionisation constant of this acid will be as given below:
$\Rightarrow {K_a} = \dfrac{{[{H^ + }][{A^ - }]}}{{[HA]}}$
Similarly, the ionisation constant for hydrogen fluoride or hydrofluoric acid can be given where we know that it will ionise in water and therefore, the ionisation will be given as:
$\Rightarrow HF \to {H^ + } + {F^ - }$
The ionisation constant expression will be given as:
$\Rightarrow {K_a} = \dfrac{{[{H^ + }][{F^ - }]}}{{[HF]}}$
The value of ionisation constant for hydrofluoric acid is found to be $7.1 \times {10^{ - 4}}$.

Additional information: An ionisation constant informs us about the strength of an acid whether the acid is weak or strong. Also, generally the ionisation constant values are given for weak acids only as they are partially soluble in water and the concentration of reactants is not zero which is not the case for a strong acid.

Note: Always remember that the weak acids possess a higher value of ionisation constant are considered stronger than that of the acids with lower values of ionisation constant whereas the strong acids easily and completely ionises in water and therefore the concentration of reactant becomes zero thereby making the ionisation constant reach infinity.