Question

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 30.0 ml of lioh.

Answer 1

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 

Answer 2

From the molarity and volume of HClO4, we can determine how many moles of H+ we initially have: 
     0.18 M HClO4 * 0.100 L HClO4 = 0.018 moles H+

We can determine how many moles of OH- we have from the molarity and volume of LiOH:
     0.27 M LiOH * 0.030 L LiOH = 0.0081 moles OH-

When the HClO4 and LiOH neutralize each other, the remaining will be
     0.018 moles H+ - 0.0081 moles OH- = 0.0099 moles of excess H+

This means that the molarity [H+] will be
     [H+] = 0.0099 moles H+ / (0.100 L + 0.030 L) = 0.07615 M

The pH of the solution will therefore be
     pH = -log [H+] = -log 0.07615 = 1.12