Question

In a titration of 47.63 mL of 0.3786 M nitrous acid with 0.3786 M aqueous sodium hydroxide, what is the pH of the solution when 47.63 mL of the base have been added?

Answer 1

no of moles of HNO2 = molarity * volume in L

                                   = 0.3786*0.04763 = 0.018 moles

no of moles of NaOH = molarity * volume in L

                                  = 0.3786*0.04763 = 0.018moles

           molarity of NO2^-1   = no of moles/total volume In L

                                        = 0.018/0.04763 + 0.04763

                                       = 0.018/0.09526 = 0.188M

           NO2^- + H2O --------> HNO2 + OH^-

I                      0.188                          0            0

C                    -x                             +x              +x

E                 0.188-x                       +x              +x

       Kb   = Kw/Ka

             = 10^-14 /4.5*10^-4     = 2.2*10^-11

Kb   = [HNO2][OH-]/[NO2-]

2.2*10^-11   = x*x/0.188-x

2.2*10^-11 *(0.188-x) = x²

x = 2.034*10^-6

[OH-] = x= 2.034*10^-6 M

POH = -log[OH-]

         = -log2.034*10^-6

         = -log0.000002034

         = 5.6916

PH   = 14-POH

       = 14-5.6916 = 8.3084 >>>>answer