Why does pH = pK when pH = 7, according to the Henderson-Hasselbalch equation?

1 Answer
Nov 11, 2015

Answer:

In fact #pH# #=# #pK_a# when #[HA]# #=# #[A^-]#; i.e. #pH# #=# #pK_a# at the point of half-equivalence.

Explanation:

We may represent the behaviour of weak acid in water by the following equation:

#HA(aq) rightleftharpoons H^+ + A^-#. And we write the equilibrium reaction as follows,

#K_a = [[H^+][A^-]]/[[HA]]#

We take #log_10# of both sides:

#log_10(K_a) = log_10[H^+] + log_10{[[A^-]]/[[HA]]}#

On rearrangement,

#-log_10[H^+] = -log_10(K_a) + log_10{[[A^-]]/[[HA]]}#

OR

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

(because by definition #-log_10[H^+] = pH#, and #-log_10(K_a)=pK_a#).

I have no doubt you've seen this equation before, but what does it mean? What, precisely does the log function mean? When I write #log_ab = c#, it means to get #c# I have to raise the base (of the log) #a# to the #c^(th)# power to get #b#, i.e. #a^c =b#. Before the advent of electronic calculators, student were routinely issued log books and used to do complex mulitplications and divisions by taking logs and antilogs. Note that #log_(10)1# #=# #0#. So in the equation above, at the point of half-equivalence #[A^-]=[HA]#, and the log of this quotient #=0#. Thus at half-equivalence, #pH =pK_a#.

See if you can get your head round this. In the meantime, can you tell me values for #log_(10)10#, #log_(10)100#, #log_(10)1000#, and #log_(10)1000000#? Use a calculator if you are stumped; why are these values simple whole numbers?