We do not present here a general study of this situation as too many parameters are involved for a simple visualisation, however it is still possible to optimize solutions "by hand" as we exemplify below. In these asymmetrical situations, it becomes handy to work also with r ∆pKa = ∆pKa h ∆pKa l , where ∆pK ah = pK a3 -pK a2 and ∆pK al = pK a2 -pK a1 .