How can I calculate the pH of the solution created by reacting sodium with water? $\pu{23mg}$ of sodium was added to $\pu{100cm3}$ of water. 
  
  
  What is the pH of the solution produced?
  


$$\ce{2Na + 2H2O -&gt; H2 + 2NaOH}$$

So firstly to find the moles, I've divided $\pu{0.023g}$ by $23$ to reach $\pu{0.001 mol}$.

Am I correct that the concentration of $\ce{NaOH}$ would be $\pu{0.01mol\,dm^{-3}}$?
$[\ce{OH-}] = 0.01$ which should be multiplied by $2$ due to the balancing numbers.
So $K_\mathrm{w}$ divided by $0.02$ gives $5\times10^{-3}$
$$
\mathrm{pH} = -\log[\ce{H+}]\\
\mathrm{pH} = 12
$$


Is this correct?
You answer is right, but your reasoning is not.

Amount of $\ce{Na} = \frac{0.023}{23} = 0.001$ mole

$[\ce{NaOH}]= \frac{0.001}{0.100} (\mathrm{moleL^-1}) = 0.01$ M

$\downarrow$

$p\ce{OH} = 2 \to p\ce{H} = 14 - 2 = 12$