Getting from pH 2 to pH 6 by mixing with water

In our book we learned that we can get an acid to a higher $\mathrm{pH}$ by mixing it with water in the relation $\frac {1}{10} $.


Now to my problem:
How much water do you need in order to get the $\mathrm{pH}$ value from $2$ to $6$. It is $\pu{250 L}$ of acid.


1. Try: I have $\pu{250 L}$ of acid, I will mix it with water in the relation $\frac {1}{10} $. Then I will have $\pu{2500 L}$ of this mixture, is basically means that I have needed $\pu{2250 L}$ of water to get to $\mathrm{pH}\space 3$. $\mathrm{pH}\space 4$ will have $\pu{25000 L}$ so I needed $\pu{22500 L}$ of water to get to $\mathrm{pH}\space 4$ and so on...
This would mean that I would need $2250+22500+225000+2250000=2499750 \space \mathrm{L} $ to get to $\mathrm{pH}\space 6$.
2. Try
I know that $c (\ce{H3O+})=\mathrm {\frac {mol}{L}}\Rightarrow 10^{-2}=\pu{\frac {mol}{\pu{250 L}}}\Rightarrow m=\pu{2.5mol}$. 
Also $\mathrm{pH} (6)=c (10^{-6})$
$\frac {\pu{2.5mol}}{x\space \mathrm{L}}=10^{-6}\Rightarrow x=2500000\space \mathrm{L}$. This would mean that I need to add $\pu{2500000L}$.


What am I doing wrong, how can a fix my mistake in one of my possible solutions?




In first try, everything is ok. 
In second try, you find the final volume of solution, $\pu{2500000L}$, but you already have $\pu{250L}$ acid solution so in order to reach that, $\mathrm {2500000\space L-250\space L=2499750\space L}$ water is needed.


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