. It then comes pK a1 = pK a -∆pK a and pK a4 = pK a + ∆pK a . Moreover, we have r ∆pKa = pKa 3 -pKa 2 pKa 4 -pKa 1 , which gives pK a2 = pK a -r ∆pKa ∆pK a and pK a3 = pK a + r ∆pKa ∆pK a . In that case, it comes: