What is the effect of buffer concentration on pH?

1 Answer
Sep 20, 2017

Answer:

Well, it depends on how you compose your buffer....

Explanation:

See this old answer.

#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]]}#

When we have EQUAL CONCENTRATIONS of #A^-# and #HA#, clearly #pH=pK_a#. Why? Because #[A^-]=[HA]#, and thus #log_10([[A^-]]/[[HA]])=log_(10)1=0#.

Most of the time, the #pH# remains tolerably close to the #pK_a# of the weak acid used in the buffer.