# What is the concentration of hydroxide ions in a solution that has a pH of 6.0?

Mar 11, 2017

$\left[H {O}^{-}\right] = {10}^{-} 8 \cdot m o l \cdot {L}^{-} 1$..........

#### Explanation:

We know that water undergoes self-ionization, which we may represent by the following reaction:

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

This equilibrium has been carefully measured, and at $298 \cdot K$, we may write,

$\left[{H}^{+}\right] \left[H {O}^{-}\right] = {10}^{-} 14$, and taking ${\log}_{10}$ of both sides,

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

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

Given the definition of $p H = - {\log}_{10} \left[{H}^{+}\right]$, then...........

$p H + p O H = 14$

So if $p H = 6$, then $p O H = 8$ and $\left[H {O}^{-}\right] = {10}^{-} 8 \cdot m o l \cdot {L}^{-} 1$.