# How do you perform acid base reactions?

Jul 4, 2018

Some context is necessary to address the question...

#### Explanation:

We know that in aqueous solution under standard conditions, the following equilibrium operates...

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

And careful measurement has established the extent of this equilibrium...

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

And thus in aqueous solution, we invoke two quantities, ….

${\underbrace{{H}_{3} {O}^{+}}}_{\text{the acid}}$,

AND....

.. ${\underbrace{H {O}^{-}}}_{\text{the base}}$

Equivalently, we could take logarithms of the expression...

${\log}_{10} \left({K}_{w}\right) = {\log}_{10} \left(\left[{H}_{3} {O}^{+}\right] \left[H {O}^{-}\right]\right) = {\log}_{10} \left({10}^{- 14}\right)$...

And so ….

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

$+ 14 = {\underbrace{- {\log}_{10} \left[{H}_{3} {O}^{+}\right]}}_{\text{pH"underbrace(-log_10[HO^-])_"+ pOH}}$

And so...$14 = p H + p O H$...the defining relationship....

I acknowledge that I may not have addressed the question you wanted answered... As to a specific reaction, you may refer to a titration reaction....where a known volume of titrant of known concentration is added to an unknown quantity of acid or base in solution….