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?

  • $\begingroup$ Is this really just a simplifying approximation. Using the same logic, if I diluted this to a total volume of 1000 liters, I would have a pH of 9, and diluting it to 100,000 liters would yield a pH of 7. If I diluted it to a total volume of 1,000,000 liters then the pH would be 6. I'm sure that no matter how much you dilute it, it would always be basic, or at least never acidic. So is the formula that you showed just an approximation that works for smaller changes in pH, but doesn't extrapolate for larger changes? $\endgroup$ – Itsme2003 Oct 15 '17 at 5:27

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}


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

and therefore

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

  • $\begingroup$ instead of 10 L i kept writing 10 mL at the end.. -.- $\endgroup$ – confused Mar 26 '14 at 7:24
  • 3
    $\begingroup$ Ah! You're confused. $\endgroup$ – MaxW Mar 21 '16 at 17:22

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.


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