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^-#?

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^-#?

Owen Ambrose · Accepted Answer

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

The balance is experienced by the parent acid. #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 in the event that there is an excess of 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# We can write the equilibrium expression as we would with any other equilibrium: #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#.) Organizing, #-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?

Jeremiah Adams · Answer

undefined For neutralizing excess acid (H₃O⁺):

CH₃COOH + H₃O⁺ → CH₃COOH₂⁺ + H₂O

For neutralizing excess base (OH⁻):

CH₃COOH + OH⁻ → CH₃COO⁻ + H₂O