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.
