A 100.0 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