# What is the pH of 1 * 10^-4 M NaOH?

May 28, 2016

$p H = 10$

#### Explanation:

In water the following equilibrium operates:

$2 {H}_{2} O \left(l\right) r i g h t \le f t h a r p \infty n s {H}_{3} {O}^{+} + H {O}^{-}$

As for any equilibrium, we can write the equilibrium expression:

$\frac{\left[{H}_{3} {O}^{+}\right] \left[H {O}^{-}\right]}{\left[{H}_{2} O\right]}$ $=$ $K {'}_{w}$

Since $\left[{H}_{2} O\right]$ is huge, we can simplify this expression:

$\left[{H}_{3} {O}^{+}\right] \left[H {O}^{-}\right]$ $=$ ${K}_{w}$

At $298 K$, $\left[{H}_{3} {O}^{+}\right] \left[H {O}^{-}\right]$ $=$ ${K}_{w} = {10}^{- 14}$

Taking ${\log}_{10}$ of both sides:

${\log}_{10} \left[{H}_{3} {O}^{+}\right] + {\log}_{10} \left[H {O}^{-}\right]$ $=$ ${\log}_{10} \left({10}^{-} 14\right)$ $=$ $- 14$

Now we can define $p H$ $=$ $- {\log}_{10} \left[{H}_{3} {O}^{+}\right]$, and

$p O H$ $=$ $- {\log}_{10} \left[H {O}^{-}\right]$.

And thus $p O H + p H = 14$ $\text{(finally!)}$

Thus $p O H$ of $1 \times {10}^{-} 4 m o l \cdot {L}^{-} 1 \text{ NaOH}$ $=$ $- {\log}_{10} \left(1 \times {10}^{-} 4\right) = 4$

Since $p O H + p H = 14$, $p H = 10$, as required.