Ka relating to Dissociation of Weak acid and Water
Yes, water autodissociates according to the equation $\ce{2H2O &lt;=&gt; H3O+ + OH-}$. 

But I should point out that the equilibrium constant is written as though the concentration of water itself is a constant, so the reactions becomes: 

$\ce{H2O &lt;=&gt; H+ + OH-}$. 

and the equilibrium is expressed as:

$K_\rm{w} = 1\times10^{-14} = \ce{[H+][OH-]}$

The pH is $\text{-log}\ce{([H+])}$ thus the pH of pure water is 7. 

This also has to do with significant figures. Few pKa's are known to better than two significant figures. So:


For weak acid with pKa &lt; 5 you can ignore water contribution.
For acid with pKa > 9 you can assume all the  $\ce{H+}$ is from water and that all of the acid is essentially present as the HA form. 
For 9 > pKa > 5 you have to consider both ionization of the acid and water.