# How does the pH of an aqueous solution as [H_3O^+] is INCREASED?

May 19, 2017

$\text{.....decreases}$$\text{.........decreases............increases.......}$

#### Explanation:

We assess the autoprotolysis reaction of water:

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

Equivalently we could write this is as...........

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

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

We can take ${\log}_{10}$ of BOTH sides to give........

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

${\log}_{10} \left[{H}_{3} {O}^{+}\right] + {\log}_{10} \left[H {O}^{-}\right] = - 14$, or, multiplying each side by $- 1$

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

But by definition, $- {\log}_{10} \left[{H}_{3} {O}^{+}\right] = p H$ and $- {\log}_{10} \left[H {O}^{-}\right] = p O H$. And so after all that..........

$p H + p O H = 14$, which is our defining relationship.

At LOW $p H$, $\left[{H}_{3} {O}^{+}\right]$ is HIGH and $\left[H {O}^{-}\right]$ is low; and at high $p H$, $\left[{H}_{3} {O}^{+}\right]$ is LOW and $\left[H {O}^{-}\right]$ is HIGH.

When $p H = p O H$, $p H = 7$. Why?

Remember the definition of the $\text{log function}$. If ${\log}_{a} b = c$, then ${a}^{c} = b$, and thus ${\log}_{10} 1 = 0$, ${\log}_{10} 10 = 1$, ${\log}_{10} {10}^{-} 1 = - 1$,${\log}_{10} 100 = 2$.