Question

Consider the titration of 40.0 mL of 0.200 M HClO4 by 0.100 M KOH. Calculate the pH of the resulting solution after the following volumes of KOH have been added. a. 0.0 mL d. 80.0 mL b. 10.0 mL e. 100.0 mL c. 40.0 mL

answers are A) 0.699 B) 0.854 C) 1.301 D) 7.00 E) 12.15 I would just like to know how to do the work please and tbank you

Answer 1

Here it is strong acid and strong base titraton.

             HClO4    +                       KOH --------------------> KClO4 + H2O

         [No. of moles of each reactant: n= Molarity * volume in L]

a)         40.0 mL * 0.200 M                                             0.100 M * 0.0 mL

    = 0.008 moles                                                                 0 moles                    [     since 40ml=0.04litres]

pH = -log[H+]

      = -log[0.008moles/0.04L]

     = 0.698

b) No.of moles of each reactant that are reacting:

     HClO4    +                       KOH -------------------->    KClO4        +       H2O

0.04L * 0.200 M                                    0.100 M * 0.01 L

= 0.008 moles                                           = 0.001 moles

after reacting the moles of acid remaining:

0.007 moles                                                         0 moles                                       0.001moles

Hence pH = -log[H+]

= -log[0.007moles/0.05litres]                   [ since molarity= no.of moles/total volume of solutions]

= 0.853      

c) No.of moles of each reactant that are reacting:

     HClO4    +                       KOH -------------------->    KClO4        +       H2O

0.04L * 0.200 M                                       0.100 M * 0.04L

= 0.008 moles                                    = 0.004 moles

after reacting the moles of reactants remaining:

0.004 moles                                              0 moles                                                     0.004moles

Hence pH = -log[H+]

= -log[0.004moles/0.08litres]                   [ since molarity= no.of moles/total volume of solutions]

                    = 1.301

d) No.of moles of each reactant that are reacting:

     HClO4    +                       KOH -------------------->    KClO4        +       H2O

0.04L * 0.200 M                                       0.100 M * 0.08L

= 0.008 moles                                    = 0.008 moles

after reacting the moles of acid remaining:

0 moles                                                    0 moles                                                     0 moles

Means complete neutralisation takes place and the pH of the neutral solution is :

pH= 7

e) No.of moles of each reactant that are reacting:

     HClO4    +                       KOH -------------------->    KClO4        +       H2O

0.04L * 0.200 M                                       0.100 M * 0.1L

= 0.008 moles                                    = 0.01moles

after reacting the moles of acid remaining:

      0 moles                                                 0.002 moles                                        0.008moles

Hence acid is consumed completely during the reaction:

pOH = -log[0.002moles/0.14L]

         = 1.847

Hence pH = 14- 1.847

                   = 12.153