Henderson-Hasselbalch Equation
Assuming that x << [HA] and x << [A-] for good buffer action, the equilibrium constant expression can be rearranged to give simplified calculations:
pH = pKa + log([A-]/[HA])
Consider 1.00 M HF, 0.500 M NaF
Ka = 7 x 10- 4, pKa = 3.155
log ([F-]/[HF]) = log (0.500/1.00) = -0.301
pH = 3.155 - 0.301 = 2.854 (same result as before)