How does a pKa value relate to pH?

1 Answer
anor277
Mar 25, 2016

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

Explanation:

Consider the general acid dissociation reaction in water:

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

Now, this is an equilibrium reaction, and for a given temperature, we can write the equlibrium expression:

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

This is an equation, and like any equation we can add/subtract/multiply it etc. and maintain the equality, SO LONG AS WE DO THE SAME OPERATION TO BOTH SIDES OF THE EQUATION. We can certainly take the #log_10# of both sides give:

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

ON REARRANGEMENT, subtract #log_10[H_3O^+]# AND #log_10K_a# FROM BOTH SIDES:

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

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

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

So, upon titration of a weak acid, at the point of half-equivalence when #[HA]=[A^-]#, #pH# #=# #pK_a# because #log_10{[[A^-]]/[[HA]}}# #=# #log_10{1}# #=# #0#

Related questions
See all questions in pH, pKa, Ka, pKb, Kb
Impact of this question
2231 views around the world
You can reuse this answer
Creative Commons License