How to calculate the volume of a sodium hydroxide solution necessary to add to water, to raise the pH to a certain value?
$$\ce{NaOH + H2O-&gt; Na+  +OH- + H2O}$$
\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.