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 


$$\ce{HA <=> H+ + A-}$$


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 -


$$\ce{H\_2O(l) + H\_2O(l) <=> H3O+ + OH-}$$


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](https://en.wikipedia.org/wiki/Self-ionization_of_water) according to the equation $\ce{2H2O <=> H3O+ + OH-}$. 


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: 


$\ce{H2O <=> H+ + OH-}$. 


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 < 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.


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I think MaxW's answer covers the gist of the question from the calculation side, but its also worth noting something else about your question. $\ce{H^+}$ and $\ce{H3O^+}$ are often used interchangeably for the acidic proton, but $\ce{H3O^+}$ is more accurate; free protons are unlikely to exist in solution. $\ce{H3O^+}$ is one example of how the proton could be passed into the solution and so is used to represent the general case of the proton being solvated by a water cluster. For more information on these water clusters, go the solvation section of <https://en.wikipedia.org/wiki/Hydronium>. 


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