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:


1. Begin by calculating conc. of $[\ce{OH-}]$ for $\mathrm{pH}~13$.
2. Find the moles by multiplying concentration and volume ($\pu{0.1 L}$)
3. Determine new concentration of $[\ce{OH-}]$ for $\mathrm{pH}~11$.
4. Use $\text{amount}/\text{new concentration}$ to find the total volume.
5. 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} 
&= c\_\mathrm{end}(\ce{OH-})\cdot V\_\mathrm{end} \\
V\_\mathrm{end} &= \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} &= 13 \\
c\_\mathrm{start}(\ce{OH-}) &= 10^{-(14-\ce{pH})}~\mathrm{mol/L} = 0.1~\mathrm{mol/L} \\
V\_\mathrm{start} &= 0.1~\mathrm{L} \\
\ce{pH}\_\mathrm{end} &=11 \\
c\_\mathrm{end}(\ce{OH-}) &=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


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


 4

Here is a different approach:


$\ce{NaOH}$ is a strong base and will be completely dissociated in water. A solution of $\ce{NaOH}$ with $\mathrm{pH} = 13$ will contain $\pu{0.10 M}$ $\ce{NaOH}$, since


$$[\ce{OH-}] = 0.10 \implies \mathrm{pOH} = 1.0 \implies \mathrm{pH} = 14.0 - 1.0 = 13.0.$$


What will the $\mathrm{pH}$ be of a $\pu{0.0010 M}$ solution of $\ce{NaOH}$?


We estimate the $\mathrm{pH}$ by assuming the water autoprotolysis will not significantly affect the $\mathrm{pH}$.


$$[\ce{OH-}] = \pu{0.0010 M} \implies \mathrm{pOH} = 3.0 \implies \mathrm{pH} = 14.0 - 3.0 = 11.0.$$


So, I suggest you dilute the original solution of $\pu{100 mL}$ $\ce{NaOH}$ $100$ times, i.e. you should add $\pu{9900 mL}$ of water.


 1

Using the definition of $\mathrm{pH}$ (which is a log value) i.e. the difference of $\mathrm{pH}~1$ value is 10 folds difference in real terms.


You have $\pu{100 ml}$ of $\ce{NaOH}$ @ $\mathrm{pH}~13.$ If you dilute by $10$ times, then the new $\mathrm{pH}$ will be $12,$ so you'll need to dilute it another $10$ times to reach $\mathrm{pH}~11.$ If you start with $\pu{100 ml},$ you'll need to end with $10$ litres in total ($100$ folds). You already have your $\pu{100 ml},$ so you'll need to add $\pu{10000 ml} - \pu{100 ml} = \pu{9900 ml}.$


Obviously, if the required final $\mathrm{pH}$ had fractions this simple mental arithmetic will not work. 


 1