# If the hydrogen ion concentration of a solution is 10^-10M is the solution acidic, alkaline, or neutral?

Jun 12, 2017

The solution is BASIC............

#### Explanation:

In water the following equilibrium applies under standard conditions:

$2 {H}_{2} O \left(l\right) r i g h t \le f t h a r p \infty n s {H}_{3} {O}^{+} + H {O}^{-}$, and also.....

${K}_{w} = \left[{H}_{3} {O}^{+}\right] \left[H {O}^{-}\right] = {10}^{-} 14$ under standard conditions at $298 \cdot K$.

We can take ${\log}_{10}$ of both sides...........

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

And thus on rearrangement,

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

And thus our defining relationship..........

$p H + p O H = 14$

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

And thus $p H = - {\log}_{10} \left\{{10}^{-} 10\right\} = - \left(- 10\right) = 10$

$p O H = 14 - p H = 14 - 10 = 4$

And thus (finally!)

$\left[H {O}^{-}\right] = {10}^{-} 4 \cdot m o l \cdot {L}^{-} 1 = 0.0001 \cdot m o l \cdot {L}^{-} 1$.

And since $\left[H {O}^{-}\right]$ is ${10}^{6}$ times greater than $\left[{H}_{3} {O}^{+}\right]$ the solution is BASIC.

Got that? (I didn't realize I could go on for so long!)