# How do pH buffer systems work?

Aug 16, 2017

When appreciable quantities of a weak acid and its conjugate base are mixed in solution......

#### Explanation:

When appreciable quantities of a weak acid and its conjugate base are mixed in solution, a buffer solution is formed that tends to resist gross changes in $p H$. How?

Because the conjugate base of a strong acid DEMONSTRABLY does not compete strongly for the proton...................whereas the conjugate base of a weak acid DOES compete to some extent for the proton.......

And thus we can write the acid-base dissociation reaction:

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

And from this equilibrium expression we can derive an expression for $p H$ of the buffer......

${K}_{a} = \frac{\left[{H}_{3} {O}^{+}\right] \left[{A}^{-}\right]}{\left[H A\right]}$ (as you know $\left[{H}_{2} O\right]$ is effectively a constant).

And if we take ${\log}_{10}$ of BOTH SIDES, we get........

${\log}_{10} {K}_{a} = {\log}_{10} \left[{H}_{3} {O}^{+}\right] + {\log}_{10} \left\{\frac{\left[{A}^{-}\right]}{\left[H A\right]}\right\}$

Which on rearrangement gives........

${\underbrace{- {\log}_{10} \left[{H}_{3} {O}^{+}\right]}}_{\text{pH}} = {\underbrace{- {\log}_{10} {K}_{a}}}_{p {K}_{a}} + {\log}_{10} \left\{\frac{\left[{A}^{-}\right]}{\left[H A\right]}\right\}$

And thus we can write the so-called buffer equation............

$p H = p {K}_{a} + {\log}_{10} \left\{\frac{\left[{A}^{-}\right]}{\left[H A\right]}\right\}$

And as we know, when $\left[{A}^{-}\right] = \left[H A\right]$ at half-equivalence, $p H = p {K}_{a}$. Do you agree? Why?

$p H$ could not be so moderated UNLESS ${A}^{-}$ competed somewhat for the proton (and if it did not it would mean that the parent acid, $H A$, was a STRONG acid).

Anyway, is this what you want?