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

Question

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}

Answer 1

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

\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.