As an example, we treat the case of the solution seen above, with C 1 = 0.1 mol.L -1 and as target pH = 3. In fig. .a are represented the roots in function of ∆pK a for C 1 = C 2 = 0.1 mol.L -1 -the starting point-and C 1 = 0.1 mol.L -1 , C 2 = 0.1104 mol.L -1 , when the convergence was satisfying. In that situation, ∆pK c a = 1.099 ≈ 1.1, value that differs significantly from 1.144 found in the absence of the H 3 O + contribution (which is the important one at pH = 3, where OH -can safely be neglected) or the one found for C 1 = C 2 = 0.1 mol.L -1 (∆pK c a = 1.249).