Question

A 50.0 mL sample of a 0.517 M aqueous hypochlorous acid solution is titrated with a 0.206 M aqueous barium hydroxide solution. What is the pH after 40.7 mL of base have been added? pH =

Answer 1

no of moles of HClO   = molarity*volume in L

                                  = 0.517*0.05

                                  = 0.02585moles

no of moles of Ba(OH)2 = molarity*volume in L

                                       = 0.206*0.0407   = 0.0083842moles

-------- 2HClO(aq) + Ba(OH)2(aq) ---------> Ba(ClO)2(aq)   + 2H2O(l)

I ------- 0.02585 ------- 0.0083842 -----------   0

C------- -2*0.0083842 --- -0.0083842 ---------   0.0083842

E----- 0.0090816    ----      0   -------------------- 0.0083842

    PH   = PKa + log[Ba(ClO)2]/[HClO]

             = 7.53 + log0.0083842/0.0090816

              = 7.53 -0.0347

               = 7.5 >>>>answer