# In a reaction of an acid with a base, the pH changes to a value that is closer to what?

May 11, 2017

If the reaction is an aqueous, i.e. water, medium, the $p H$ at equivalence will be close to $7$.

#### Explanation:

Acid-base behaviour in water is predicated on the basis of the following equilibrium, known as the $\text{autoprotolysis reaction}$:

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

${H}_{3} {O}^{+}$ is a conceptual species, and one way to look at it is to imagine a cluster of 3 or 4 water molecules with an EXTRA proton, i.e. to give ${H}_{7} {O}_{3}^{+}$, or ${H}_{9} {O}_{4}^{+}$. The proton exchanges rapidly from cluster to cluster. If you have ever played rugby, think of a maul where the forwards bind and can pass the ball from hand to hand while bound in the cluster. In water clusters, the proton tunnels, so such transfer is very rapid.

An alternative depiction that is not so favoured these days is:

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

Both representations are EQUIVALENT.

In water the $\text{autoprotolysis reaction}$ has been carefully measured, and for standard conditions, the ion product is.........

$\left[{H}_{3} {O}^{+}\right] \left[H {O}^{-}\right] = {10}^{- 14}$. And given standard definitions, $p H = - {\log}_{10} \left[{H}_{3} {O}^{+}\right]$ so that....

$p H + p O H = 14$.

And to answer your question (finally), addition of a base to an acid should result in an increase in $p H$.