The pH of a solution is 10. What is its [H^+] concentration?

Aug 15, 2017

$\left[{H}^{+}\right] \text{/} \left[{H}_{3} {O}^{+}\right] = {10}^{- 10} \cdot m o l \cdot {L}^{-} 1$

Explanation:

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

In water, under standard conditions, $p H$ characterizes the autoprotolysis reaction.......

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

${K}_{w} = \left[H {O}^{-}\right] \left[{H}_{3} {O}^{+}\right] = {10}^{-} 14$, we take ${\log}_{10}$ of BOTH SIDES...

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

And on rearrangement,

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

And thus $14 = p O H + p H$.