# 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?

$$\ce{2Na + 2H2O -> H2 + 2NaOH}$$

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?
• Welcome to Chemistry.SE! Take the tour to get familiar with this site. Mathematical expressions and equations can be formatted using $\LaTeX$ syntax. Please note that the proper term for "number of moles" is amount of substance. The former would be the same as referring to the mass as "number of kilograms". Mar 29, 2017 at 12:11

Amount of $\ce{Na} = \frac{0.023}{23} = 0.001$ mole
$[\ce{NaOH}]= \frac{0.001}{0.100} (\mathrm{moleL^-1}) = 0.01$ M
$\downarrow$
$p\ce{OH} = 2 \to p\ce{H} = 14 - 2 = 12$