# Question #4af57

Jan 12, 2015

Sodium hydroxide is a strong base, which means it dissociates completely in aqueous solution into ${\text{Na}}^{+}$ and ${\text{OH}}^{-}$ ions.

This tells you that the concentrations of the ${\text{Na}}^{+}$ cations and of the ${\text{OH}}^{-}$ anions will be equal to each other, and to the initial concentration of $\text{NaOH}$.

Since you start with ${10}^{- 5}$ $\text{M}$ $\text{NaOH}$, in aqueous solution you'll end up with

$\left[{\text{Na}}^{+}\right] = {10}^{- 5}$ $\text{M}$ and $\left[{\text{OH}}^{-}\right] = {10}^{- 5}$ $\text{M}$

The equation to use in this case is

$p O H = - \log \left(\left[O {H}^{-}\right]\right) = - \log \left({10}^{- 5}\right) = 5$

This implies a pH of

$p H = 14 - p O H = 14 - 5 = 9$