How to calculate the volume of a sodium hydroxide solution necessary to add to water, to raise the pH to a certain value? 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 $12.42$?

This is my attempt at a solution:
$$\ce{NaOH + H2O-&gt; Na+ + OH- + H2O}$$
\begin{align}
\mathrm{pH}(\text{initial}) &amp;= 7.00 \\
\mathrm{pH}(\text{final}) &amp;= 12.42 \\
\mathrm{pH} &amp;= -\log{\ce{[H+]}} \\
\ce{[H+]}_{\text{intial}} &amp;= 10^{-7} \\
&amp;= \pu{10^{-7} ions} \\ \hline
V_1&amp;= 0.016L \\
C_1 &amp;= (0.300)(10^{-7})= 3\cdot10^{-8} N\\
\mathrm{pOH} &amp;= 14-12.42  \\
[\ce{H+}]&amp;= 10^{-1.58} \\  
&amp;=\pu{ 0.02630 ions} \\
C_2 &amp;= (0.300)(0.02630) \\
&amp;=7.89\cdot10^{-3} N \\
C_1V_1&amp;=C_2V_2 \\
(3\cdot10^{-8})(0.016)&amp;= (7.89\cdot 10^{-3})(V_2) \\
V_2&amp;= 6.08\cdot10^{-8} L \\
\end{align}
$$\ce{NaOH + H2O-&gt; Na+  +OH- + H2O}$$
\begin{align}
\mathrm{pH}_1 = 7.00 &amp;\implies\mathrm{pOH}_1 = 7 \\
\mathrm{pH}_2 = 12.42 &amp;\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.