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 30.0 ml of lioh.

First, we can get moles of HClO4 = molarity * volume 

                                                        = 0.18 M* 0.1 L

                                                         = 0.018 mol

then, we can get moles of LiOH = molarity * volume

                                                      = 0.27 * 0.03 L

                                                      = 0.0081 mol

when moles of HCIO4 remaining = 0.018 - 0.0081

                                                         = 0.0099 mol

when total volume = 0.1 + 0.03 = 0.13 L

when HCIO4 is a strong acid so, we can assume [HC|O4] = [H+]

[H+] = moles HCIO4 / total volume

         = 0.0099 mol / 0.13 L

         = 0.076 M

∴PH = - ㏒[H+]

         = -㏒0.076 

         = 1.12