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

Answer:

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?