# How would you write the chemical equation for how the acetate buffer neutralizes excess acid, H_3O^+? How do you write the chemical equation for how the acetate buffer neutralizes excess base, OH^-?

Jul 6, 2017

Well, it is a buffer, the which moderates GROSS changes in $p H$.

#### Explanation:

The parent acid undergoes the equilibrium........

HOAc(aq)+H_2O(l)rightleftharpoonsH_3O^+ + ""^(-)OAc,

i.e. ${\text{HOAc = HO(O=)CCH}}_{3}$

And thus if there are significant concentrations of acetic acid, AND acetate ions.......then if there is excess acid......

H_3O^+ + ""^(-)OAc rarrHOAc(aq) + H_2O(l)

..........and if there is excess base......

HO^(-) + HOAc(aq) rarr H_2O(l) + ""^(-)OAc

As to the background. The weak acid $H O A c$ undergoes an acid base equilibrium in water according to the equation:

HOAc(aq) + H_2O(l) rightleftharpoons H_3O^+ + ""^(-)OAc

As with any equilibrium, we can write the equilibrium expression:

${K}_{a}$ $=$ ([H_3O^+][""^(-)OAc])/([HOAc])

This is a mathematical expression, which we can divide, multiply, or otherwise manipulate PROVIDED that we do it to both sides of the expression. Something we can do is to take ${\log}_{10}$ of BOTH sides.

log_10K_a=log_10[H_3O^+] + log_10{([""^(-)OAc]]/[[HOAc]]}

(Why? Because ${\log}_{10} A B = {\log}_{10} A + {\log}_{10} B$.)

Rearranging,

-log_10[H_3O^+] - log_10{[[""^(-)OAc]]/[[HOAc]]}=-log_10K_a

But BY DEFINITION, $- {\log}_{10} \left[{H}_{3} {O}^{+}\right] = p H$
, and $- {\log}_{10} {K}_{a} = p {K}_{a}$.

Thus pH=pK_a+log_10{[[""^(-)OAc]]/[[HOAc]]}

For $\text{acetic acid}$, $p {K}_{a} = 4.76$. And thus if there are significant quantities of acetic acid and acetate anion, the $p H$ of the solution should be tolerably close to $4.76$, i.e. close to $p {K}_{a}$. If there are EQUAL concentrations of acetic acid and acetate, $p H = p {K}_{a}$. Why?