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What volume of 0.300N $\ce{NaOH}$ would be needed to bring $16.0~\mathrm{mL}$ of distilled water that has a pH of 7.00 to a new pH of 12.42?

This is my attempt at a solution:

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

$pH = 7.00$
$pH= 12.42$

$pH= -\log{\ce{[H+]}}$
$\ce{[H+]}= 10^{-7}$
$= 10^{-7}$ ions

$0.300N$ (ions/L)
$V_1= 0.016L$

$C_1= (0.300)(10^{-7})= 3.10^{-8} N$

$pOH= 14-12.42$
[$\ce{H+}$]= $10^{-1.58}$
$= 0.02630$ ions

$C_2= (0.300)(0.02630)$ $=7.89.10^{-3} N$

$C_1V_1=C_2V_2$
$(3.10^{-8})(0.016)= (7.89. 10^{-3})(V_2)$

$V_2= 6.08.10^{-8} L$

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

$\mathrm{pH}_1 = 7.00$ then $\mathrm{pOH}_1 = 7$
$\mathrm{pH}_2 = 12.42$ then $\mathrm{pOH}_2 = 1.58$

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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