#HA(aq) + H_2O(l) rightleftharpoons H_3O^+ + A^-#
#K_a=([H_3O^+][A^-])/([HA(aq)])#
Now #pK_a=-log_10K_a#......
And thus #K_a=10^(-pK_a)#
And thus where #pK_a# is small (negative), this specifies a strong acid where the given equilibrium lies to the RIGHT.............
Given #K_a=([H_3O^+][A^-])/([HA(aq)])# we can take #log_10# of both sides to give......
#log_10K_a=log_10[H_3O^+]+log_10{[[A^-]]/[[HA]]}#
On rearrangement.............
#underbrace(-log_10[H_3O^+])_(pH)=underbrace(-log_10K_a)_(pK_a)+log_10{[[A^-]]/[[HA]]}#
And thus......
#pH=pK_a+log_10{[[A^-]]/[[HA]]}#
This is a form of the #"buffer equation"#, which is used when a weak acid is mixed with its conjugate base in appreciable quantities. Protonation (or deprotonation) of the base or the acid moderates GROSS changes in #pH#, and solution #pH# remains tolerably close to the #pK_a# value.
