$\pu{23mg}$ 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?
  • $\begingroup$ 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". $\endgroup$ – Martin - マーチン Mar 29 '17 at 12:11

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


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

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