Not quite sure what you're asking, but I think what you're looking for is 'Native Species'.
(I think you have a mistake in your question as the addition is 30mL, not 100mL)
when PH = - ㏒[H+]
and here we have HClO4 is the strong acid
So PH = - ㏒[HClO4]
moles of HClO4 = 0.1 L *0.18 m = 0.018 M
moles of LiOH = 0.03 L * 0.27 m = 0.0081 M
when the total volume = 0.1L + 0.03L = 0.13 L
∴ [HClO4] = (0.018-0.0081)/0.13 L
= 0.076 M
PH = -㏒ 0.076
= 1.12
