Why is water ignored in the ionic equilibrium of HF? While trying to understand the solution of a problem given in my textbook, I realized I'm having some difficulty with the solution. The problem is as follows:


  The ionization constant of $\ce{HF}$ is $3.2 \times 10^{-4}$. Calculate the degree of dissociation of $\ce{HF}$ in its $\pu{0.02 M}$ solution. Calculate the concentration of all species present $\ce{H3O+}$, $\ce{F-}$ and $\ce{HF}$ in the solution and its PF.


In the solution of this problem, the equation is given as 
$$\ce{HF + H2O &lt;=&gt; H3O+ + F-}$$

The concentration at the time of equilibrium are given as:

\begin{align}
\ce{[HF]}   &amp;= 0.02 - 0.02x, &amp;
\ce{[H3O+]} &amp;= 0.02x, &amp;
\ce{[F- ]}  &amp;= 0.02x
\end{align}

I have the following questions:


Why are we not adding the contribution of water to the $\ce{H3O+}$ ions?
Why are we assuming that the value $0.02x$ is dissociated from $\ce{HF}$ and not just $x$?
Why are we not adding the contribution of water to the $\ce{H3O+}$ ions?
  



Write down the dissociation constant equation with and without $\ce{H2O}$.
What changes? 
Can you find a formulation of the dissociation constant where it seems natural not to include $\ce{H2O}$? 
Read up on "activity".



  
  Why are we assuming that the value $0.02x$ is dissociated from $\ce{HF}$ and not just $x$?
  


Write down and solve the equation with $0.02 x$ and with $x$ only. What is the difference?