A100.0 ml sample of 0.18 m hclo4 is titrated with 0.27 m lioh. determine the ph of the solution after the addition of 100.0 ml of lioh.

(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