# Question #091ee

May 27, 2015

The pH of your solution will be equal to 13.0.

Sodium hydroxide is a strong base, which means that it dissociates completely in aqueous solution to produce sodium cations, $N {a}^{+}$. and hydroxide anions, $O {H}^{-}$, according to the following chemical equation

$N a O {H}_{\left(a q\right)} \to N {a}_{\left(a q\right)}^{+} + O {H}_{\left(a q\right)}^{-}$

Notice that 1 mole of sodium hydroxide produces 1 mole of $N {a}^{+}$ and 1 mole of $O {H}^{-}$ ions. This means that the concentration of the hydroxide ions will be equal to

$\left[O {H}^{-}\right] = \left[N a O H\right] = \text{0.1 M}$

You can use this concentration to determine the pOH of the solution

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

$p O H = - \log \left(0.1\right) = 1.0$

Therefore, the pH of the solution will be equal to

$p {H}_{\text{sol}} = 14 - p O H = 14 - 1.0 = \textcolor{g r e e n}{13.0}$