# Question #00257

Mar 8, 2015

The pH of your solution will be $\text{13}$.

A solution's pH is all about the concentration of the hydronium ions, or ${H}_{3}^{+} O$, present. In your case, the concentration of these ions will be directly influenced by the concentration of the hydroxide ions, or $O {H}^{-}$, which are added when solid $N a O H$ is dissolved in aqueous solution.

The first thing to do is figure out how many moles of $N a O H$ you have; this will then lead you to the concentration of the hydroxide ions.

So, if you use sodium hydroxide's molar mass, you can calculate the number of moles by

$\text{2.00 g NaOH" * "1 mole NaOH"/"40.0 g NaOH" = "0.0500 moles NaOH}$

You can assume that the volume of the solution will be the same as the volume of the water, which means the concentration of sodium hydroxide will be

$C = \frac{n}{V} = \text{0.0500 moles"/(500 * 10^(-3)"L") = "0.100 M}$

Since sodium hydroxide is a strong base, it will dissociate completely in aqueous solution to give

$N a O {H}_{\left(a q\right)} r i g h t \le f t h a r p \infty n s N {a}_{\left(a q\right)}^{+} + O {H}_{\left(a q\right)}^{-}$

Since 1 mole of $N a O H$ produces 1 mole of $O {H}^{-}$, the concentration of the hydroxide ions will be equal to that of the sodium hydroxide

${C}_{O {H}^{-}} = {C}_{N a O H} = \text{0.100 M}$

Finally, to determine pH use

$p {H}_{\text{solution}} = 14 - p O H = 14 - \left(- \log \left(\left[O {H}^{-}\right]\right)\right)$

$p {H}_{\text{solution}} = 14 - \left(- \log \left(0.1\right)\right) = 14 - 1 = 13$

Therefore,

$p {H}_{\text{solution}} = 13$