Question

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

Answer 1

An acid should reduce `pH` upon addition to water as it (the acid) increases `[H_3O^+]`.

Explanation:

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

`HX(aq) + H_2O(l) rightleftharpoons H_3O^+ + X^-`

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

`2H_2O rightleftharpoons H_3O^+ + HO^-`, 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, `pH=-log_(10)[H_3O^+]`, and `pOH=-log_(10)[HO^-]`

`pH + pOH =14`

So, at neutrality, i.e. `[HO^-]=[H_3O^+]`, `pH=7`, and `pOH=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, `[H_3O^+]` is altered from neutrality.

And thus in an acidic solution, `pH<7`, whereas `pH>7` in an alkaline solution.

I acknowledge that you have been hit with a lot of facts and definitions. Review your notes, and ask for clarification if there are queries.