How to calculate the volume of a sodium hydroxide solution necessary to add to water, to raise the pH to a certain value?


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> What volume of $\pu{0.300N}\, \ce{NaOH}$ would be needed to bring $\pu{16.0 mL}$ of distilled water that has a $\mathrm{pH}$ of $7.00$ to a new $\mathrm{pH}$ of $12.42$?
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This is my attempt at a solution:


$$\ce{NaOH + H2O-> Na+ + OH- + H2O}$$


\begin{align}
\mathrm{pH}(\text{initial}) &= 7.00 \\
\mathrm{pH}(\text{final}) &= 12.42 \\
\mathrm{pH} &= -\log{\ce{[H+]}} \\
\ce{[H+]}\_{\text{intial}} &= 10^{-7} \\
&= \pu{10^{-7} ions} \\ \hline
V\_1&= 0.016L \\
C\_1 &= (0.300)(10^{-7})= 3\cdot10^{-8} N\\
\mathrm{pOH} &= 14-12.42 \\
[\ce{H+}]&= 10^{-1.58} \\ 
&=\pu{ 0.02630 ions} \\
C\_2 &= (0.300)(0.02630) \\
&=7.89\cdot10^{-3} N \\
C\_1V\_1&=C\_2V\_2 \\
(3\cdot10^{-8})(0.016)&= (7.89\cdot 10^{-3})(V\_2) \\
V\_2&= 6.08\cdot10^{-8} L \\
\end{align}




$$\ce{NaOH + H2O-> Na+ +OH- + H2O}$$


\begin{align}
\mathrm{pH}\_1 = 7.00 &\implies\mathrm{pOH}\_1 = 7 \\
\mathrm{pH}\_2 = 12.42 &\implies\mathrm{pOH}\_2 = 1.58
\end{align}


so $[\ce{OH-}]\_1 = 10^{-7}$ and $[\ce{OH-}]\_2 = 10^{-1.58}$


So, initially, you have $16^{-3}\times 10^{-7}$ mol of $\ce{OH-}$ in your water. Your $\ce{NaOH}$ solution is $\pu{0.3 M}$ so you'll bring $\pu{0.3\times 10^{-3} mol}$ each $\mathrm{mL}$.


Then, the resulting concentration will be $$C = \frac{n\_\text{initial} + (0.3\times 10^{-3})\times x}{V\_\text{initial} + x\times \pu{1mL}}$$


You know $V\_\text{initial}$ to be $\pu{16 mL}$  

You know $n\_\text{initial}$ to be $\pu{1.6\times 10^{-9} mol}$
And you know the target concentration to be $10^{-1.58}$


I let you do the solving for $x$ where $x$ is the volume of $\ce{NaOH}$ solution to be added ;) be careful with the units.


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