#NH_4^+(aq) + H_2O(l) rightleftharpoons NH_3(aq) + H_3O^+#
Now #K_a# #=# #([NH_3][H_3O^+])/([NH_4^+])#.
We can take #log_10# OF BOTH SIDES to give:
#log_10K_a# #=# #log_10[H_3O^+]# #+# #log_10(([NH_3])/([NH_4^+]))#.
On rearrangement:
#-log_10[H_3O^+]# #=# #-log_10K_a# #+# #log_10(([NH_3])/([NH_4^+]))#
But by definition, #-log_10[H_3O^+]# #=# #pH#, and, #-log_10K_a# #=# #pK_a#.
So,
#pH# #=# #pK_a# #+# #log_10(([NH_3])/([NH_4^+]))#
So given concentrations of ammonia and ammonium chloride, the #pH# of the SOLUTION should remain TOLERABLY CLOSE to the #pK_a#, which from memory is #4.76#? When #[NH_3]# #=# #[NH_4^+]# then #pH# #=# #pK_a#, because #log_(10)1# #=# #0#.
