Volume required to dilute solution for a pH change A $\pu{100 mL}$ solution of $\ce{NaOH}$ has a $\mathrm{pH}$ of $13$. What volume of water in $\pu{mL}$ must be added to change the $\mathrm{pH}$ to $11$?


My steps:


Begin by calculating conc. of $[\ce{OH-}]$ for $\mathrm{pH}~13$.  
Find the moles by multiplying concentration and volume ($\pu{0.1 L}$)
Determine new concentration of $[\ce{OH-}]$ for $\mathrm{pH}~11$.
Use $\text{amount}/\text{new concentration}$ to find the total volume.
Use total volume ($\pu{100 mL}$) to find volume needed to be added.


The above doesn't produce the right answer, which is $\pu{9.900 mL}$. Can someone please fix my working out, or provide alternative steps to calculate this?
Since the amount of substance stays equal, you can calculate the volume of your final solution $V_\mathrm{end}$, form your start concentration $c_\mathrm{start}(\ce{OH-})$ and your initial volume $V_\mathrm{start}$ and your target concentration $c_\mathrm{end}(\ce{OH-})$. Your working equation should be:

\begin{align}
c_\mathrm{start}(\ce{OH-})\cdot V_\mathrm{start} 
&amp;= c_\mathrm{end}(\ce{OH-})\cdot V_\mathrm{end} \\
V_\mathrm{end} &amp;= \frac{c_\mathrm{start}(\ce{OH-})}{c_\mathrm{end}(\ce{OH-})}\cdot V_\mathrm{start}
\end{align}

You can work out the start concentration for the initial pH and the target concentration from your final pH. Given (at $T=298.15~\mathrm{K}$, $p=1~\mathrm{atm}$):

\begin{align}
\ce{pH}_\mathrm{start} &amp;= 13 \\
c_\mathrm{start}(\ce{OH-}) &amp;= 10^{-(14-\ce{pH})}~\mathrm{mol/L} = 0.1~\mathrm{mol/L} \\
V_\mathrm{start} &amp;= 0.1~\mathrm{L} \\
\ce{pH}_\mathrm{end} &amp;=11 \\
c_\mathrm{end}(\ce{OH-}) &amp;=10^{-(14-\ce{pH})}~\mathrm{mol/L} = 0.001~\mathrm{mol/L}
\end{align}

Hence

$$
V_\mathrm{end}= \frac{c_\mathrm{start}(\ce{OH-})}{c_\mathrm{end}(\ce{OH-})}\cdot V_\mathrm{start} = 10~\mathrm{L}
$$

and therefore

$$
\Delta V = V_\mathrm{end}-V_\mathrm{start} = 9.9~\mathrm{L}= 9900~\mathrm{mL}.
$$