Question

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^-`?

Answer 1

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?