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^-?

1 Answer
Jul 6, 2017

Well, it is a buffer, the which moderates GROSS changes in pH.

Explanation:

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

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

i.e. "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 HOAc 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_10AB=log_10A +log_10B.)

Rearranging,

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

But BY DEFINITION, -log_10[H_3O^+]=pH
, and -log_10K_a=pK_a.

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

For "acetic acid", pK_a=4.76. And thus if there are significant quantities of acetic acid and acetate anion, the pH of the solution should be tolerably close to 4.76, i.e. close to pK_a. If there are EQUAL concentrations of acetic acid and acetate, pH=pK_a. Why?