Ka relating to Dissociation of Weak acid and Water I am studying $K_a$ currently and there is a passage in the book that doesn't appear to make sense. 

It states that from the equation 

$$\ce{HA &lt;=&gt; H+ + A-}$$

that $$K_a = \frac{[H^+][A^-]}{[HA]}$$

The following passage says When dealing with weak acids, you can assume that dissociation of the acid is much greater than the dissociation of water.

It continues to say that you can assume all the H+ ions come from the acid.

hence $$K_a = \frac{[H^+]^2}{[HA]}$$

What I don't understand is the dissociation of water. 

The book gives the equation of dissociation of water as the following -

$$\ce{H_2O(l) + H_2O(l) &lt;=&gt; H3O+ + OH-}$$

but where does the H+ come from from this dissociation - it's made a Hydroxonium ion?

Is it because the $H_3O^+$ can act as the conjugate acid and release the proton?

Please help!
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.