# What is the H_3O^+ ion concentration of a solution whose OH^-  ion concentration is  1 xx 10^-3 M?

Jun 19, 2017

We know that $p H + p O H = 14$..........so $p H = 11$, and $\left[{H}_{3} {O}^{+}\right] = {10}^{-} 11 \cdot m o l \cdot {L}^{-} 1$.

#### Explanation:

We use the following equilibrium..........

${H}_{2} O r i g h t \le f t h a r p \infty n s {H}^{+} + H {O}^{-}$, and further that we know the equilibrium constant for this expression is......

$\left[{H}^{+}\right] \times \left[H {O}^{-}\right] = {10}^{-} 14$ under standard conditions.

Alternatively, $\left[{H}_{3} {O}^{+}\right] \left[H {O}^{-}\right] = {10}^{-} 14$. We can take ${\log}_{10}$ of both sides and get.........

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

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

or......$- {\log}_{10} \left({10}^{-} 14\right) = p H + p O H = 14$

But the expression on the left hand side is simply $14$, and this goes back to definition of the log function, the $\log$ of something is the exponent to which you raise the $\text{base}$ (normally $10$ or $e$) to get the bracketed term.

or......$- {\log}_{10} \left({10}^{-} 14\right) = - \left(- 14\right) = 14$

And so finally $p H + p O H = 14$, you must know this for A-level....

Here $\left[H {O}^{-}\right] = {10}^{-} 3 \cdot m o l \cdot {L}^{-} 1$, so $p O H = - {\log}_{10} \left({10}^{-} 3\right) = - \left(- 3\right) = 3$

$p H = 11$, and $p O H = 3$, i.e. taking antilogs.......

$\left[{H}_{3} {O}^{+}\right] = {10}^{-} 11 \cdot m o l \cdot {L}^{-} 1.$