How does acid affect the pH and H concentration when added to water?

Aug 25, 2016

An acid should reduce $p H$ upon addition to water as it (the acid) increases $\left[{H}_{3} {O}^{+}\right]$.

Explanation:

When an acid is added to water it is conceived to increase concentrations of the characteristic cation (of water!):

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

This affects the equilibrium that already operates in water under standard condtions:

$2 {H}_{2} O r i g h t \le f t h a r p \infty n s {H}_{3} {O}^{+} + H {O}^{-}$, where
[H_3O^+][""^(-)OH] $=$ ${10}^{- 14}$

Note that if we take negative logs to the base 10, we get an expression that should be familiar:

-log_(10)[H_3O^+] -log_(10)[""^(-)OH] $=$ $- {\log}_{10} {10}^{-} 14$ $=$ $14$

OR, using the standard definition, $p H = - {\log}_{10} \left[{H}_{3} {O}^{+}\right]$, and $p O H = - {\log}_{10} \left[H {O}^{-}\right]$

$p H + p O H = 14$

So, at neutrality, i.e. $\left[H {O}^{-}\right] = \left[{H}_{3} {O}^{+}\right]$, $p H = 7$, and $p O H = 7$

For strong acids, e.g. hydrogen halides, perchloric acid, sulfuric acid, the first equilibrium lies strongly to the right. Whatever the strength of the acid, $\left[{H}_{3} {O}^{+}\right]$ is altered from neutrality.

And thus in an acidic solution, $p H < 7$, whereas $p H > 7$ in an alkaline solution.

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