# Calculate pH of 10^-9 M NaOH?

Jan 30, 2016

$p H = 7$

#### Explanation:

Consider the two equilibria happening in solution:

$N a O H r i g h t \le f t h a r p \infty n s N {a}^{+} + O {H}^{-}$

and

${H}_{2} O r i g h t \le f t h a r p \infty n s {H}^{+} + O {H}^{-}$

From $N a O H$: $\left[O {H}^{-}\right] = {10}^{- 9} M$

From ${H}_{2} O$: $\left[O {H}^{-}\right] = {10}^{- 7} M$

Since ${\left[O {H}^{-}\right]}_{{H}_{2} O} \text{ >> } {\left[O {H}^{-}\right]}_{N a O H}$,

The $p O H$ will be determined from the ${\left[O {H}^{-}\right]}_{{H}_{2} O} = {10}^{- 7} M$

$p O H = - \log \left(\left[O {H}^{-}\right]\right) = - \log \left({10}^{- 7}\right) = 7$

Therefore, to calculate the $p H$:

$p H = p {K}_{w} - p O H = 14 - 7 = 7$

Here is a video that explains in details this topic and treats a similar example:

Acids & Bases | pH of a Strong Acid & a Strong Base.