# If an aqueous solution has an H_3O, concentration of 5.4 xx 10^9 M, is the solution acidic, basic, or neutral?

Apr 12, 2017

I think you mean that $\left[{H}_{3} {O}^{+}\right] = 5.4 \times {10}^{-} 9 \cdot m o l \cdot {L}^{-} 1$, and such a solution is $\text{BASIC}$.

#### Explanation:

In water, the following equilibrium operates under standard conditions of temperature and pressure.......

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

And as for any equilibrium, we can define an equilibrium constant. And in this instance, we write ${K}_{w} = \left[{H}_{3} {O}^{+}\right] \left[H {O}^{-}\right] = {10}^{- 14}$.

Given $\left[{H}_{3} {O}^{+}\right] = 5.4 \times {10}^{-} 9 \cdot m o l \cdot {L}^{-} 1$, then..............

$\left[H {O}^{-}\right] = \frac{{10}^{-} 14}{5.4 \times {10}^{-} 9} = 1.85 \times {10}^{-} 5 \cdot m o l \cdot {L}^{-} 1$.

Since $\left[H {O}^{-}\right] \text{>>} \left[{H}_{3} {O}^{+}\right]$ the solution is BASIC.

Note that we can use $\text{the log function}$ to make this calculation easier.

If we take ${\log}_{10}$ of both sides of ${K}_{w} = \left[{H}_{3} {O}^{+}\right] \left[H {O}^{-}\right] = {10}^{- 14}$, we get:

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

OR..........

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

But $- {\log}_{10} {K}_{w} = p {K}_{w}$, and $- {\log}_{10} \left[{H}_{3} {O}^{+}\right] = p H$ and $- {\log}_{10} \left[H {O}^{-}\right] = p O H$ BY DEFINITION.

And also by definition of the $\text{log function}$, ${\log}_{10} {10}^{-} 14 = - 14$.

And thus $14 = p H + p O H$. And so (finally!) we can use this relationship in the given problem:

$\left[{H}_{3} {O}^{+}\right] = 5.4 \times {10}^{-} 9$, and thus $p H = 8.27$, $p O H = 5.73$. At neutrality, $p O H = p H = 7$. And here the solution is clearly basic.