# What is the equation to find pH and pOH?

Jun 2, 2016

$p H + p O H = 14$

#### Explanation:

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

The auto-ionization constant of water is given by:

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

Apply the $\log$ function to both sides of the equation:

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

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

Multiply both sides of the equation by a negative sign

$- \left\{\log \left({K}_{W}\right)\right\} = - \left\{\setminus \log \left[{H}_{3} {O}^{+}\right] + \log \left[O {H}^{-}\right] \setminus\right\}$

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

$- \log \left(1.0 \times {10}^{-} 14\right) = - \log \left[{H}_{3} {O}^{+}\right] - \log \left[O {H}^{-}\right]$

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

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

$14 = p H + p O H \implies$

$p H + p O H = 14$