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:

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

\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.