How do buffers maintain pH?

2 Answers
May 27, 2016

Answer:

Buffers moderate both #[H_3O^+]# and #[HO^-]#.

Explanation:

The weak acid #HA# undergoes an acid base equilibrium in water according to the equation:

#HA(aq) + H_2O(l) rightleftharpoons H_3O^+ + A^-#

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

#K_a# #=# #([H_3O^+][A^-])/([HA])#

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{([A^-]]/[[HA]]}#

(Why? Because #log_10AB=log_10A +log_10B#.)

Rearranging,

#-log_10[H_3O^+] - log_10{[[A^-]]/[[HA]]}=-log_10K_a#

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

Thus #pH=pK_a+log_10{[[A^-]]/[[HA]]}#

Do not be intimidated by the #log# function. When I write #log_ab=c#, I ask to what power I raise the base #a# to get #c#. Here, #a^c=b#. And thus #log_(10)10=1, #, #log_(10)100=2, ##log_(10)10^(-1)=-1 #. And #log_(10)1=0#

Given our equation, it tells us that when #[A^-]=[HA]#, then #pH=pK_a#. Why? Because #log_(10)1=0#, because #10^0=1#. When acid is added to the solution, #pH# decreases only moderately, because the acid protonates the #A^-# already present in solution.

Buffers thus moderate #pH#, and keep #pH# close to the #pK_a# of the weak acid initially used.

I acknowledge that I have hit you with a lot of facts. But back in the day A level students routinely used log tables before the advent of electronic calculators. If you can get your head round the logarithmic function you will get it.

Answer:

Buffers react with added acid or base to resist a change in pH.

Explanation:

Buffers are special solutions that react with added acid or base to limit the change in pH levels.

For instance, carbonic acid is a weak acid that does not dissociate completely while in water - a small amount is dissociated into #"H"^"+"# ions and hydrogen carbonate anions (conjugate base).

#"H"_2"CO"_3 ⇌ "H"^"+" + "HCO"_3^"-"#

If you add #"H"^"+"# ions to the solution, the conjugate base will react with them - reforming the weak acid - keeping the concentration of #"H"^"+"# ions (and thus the pH) nearly constant.

If you add #"OH"^"-"# ions to the solution, they will react with the #"H"^"+"# ions to form water. However, the position of equilibrium will shift to the right — again keeping the concentration of #"H"^"+"# ions and the pH nearly constant.