# What is the pH of a 1.0 * 10^-3 M sodium hydroxide solution?

Dec 28, 2016

$p H = 11$

#### Explanation:

We examine the equilibrium:

2H_2O(l) rightleftharpoons H_3O^+ + ""^(-)OH

Under standard conditions, $\left[{H}_{3} {O}^{+}\right] \left[H {O}^{-}\right] = {10}^{-} 14$.

And if we take ${\log}_{10}$ of both sides:

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

Om rearrangement:

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

But by definition, $- {\log}_{10} \left[{H}_{3} {O}^{+}\right] = p H$, and ${\log}_{10} \left[H {O}^{-}\right] = p O H$.

And thus our definining relationship: $p H + p O H = 14$.

Since (finally!) $p O H = - {\log}_{10} \left[H {O}^{-}\right] = - {\log}_{10} \left(1 \times {10}^{-} 3\right) = - \left(- 3\right) = 3$.

And if $p O H = 3$, pH=14-3=??

In A level, you do have to remember the result:

$p H + p O H = 14$. With this is mind, these problems become trivial.