# The pH of a solution is 5.6. What are the hydrogen and hydroxide concentrations?

May 13, 2016

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

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

#### Explanation:

The pH of a given solution is determined using the following formula:

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

To find the $\left[{H}_{3} {O}^{+}\right]$ the formula must be manipulated as follows:

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

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

Plugging the value of the pH in the equation gives:

$\left[{H}_{3} {O}^{+}\right] = {10}^{-} 5.6$

$\left[{H}_{3} {O}^{+}\right] = {10}^{0.4} \cdot {10}^{- 6}$

$\left[{H}_{3} {O}^{+}\right] = 2.51 \cdot {10}^{-} 6 M$

In an aqueous solution,

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

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

$\left[O {H}^{-}\right] = \frac{{10}^{-} 14}{2.51 \cdot {10}^{-} 6} = 3.98 \cdot {10}^{-} 9 M$