Question

If 4.25g of sodium Azide NaN3 is dissolved in pure water to produce a solution with a total volume of 500 mL, what will the pH of this solution be when equilibrium is established?

Ka for hydroazoic acid (HN3) is 1.9x10^-5

please help answer is 8.9.

Answer 1

molarity    = W*1000/G.M.Wt * volume of solution in ml

                 = 4.25*1000/(65*500)    = 0.1307 M

                         NaN3(aq) ------------> Na^+ (aq) + N3^- (aq)

                        0.1307M                                         0.1307M

                        N3^- (aq) + H2O ----------------> HN3(aq) + OH^-

    I                  0.1307                                           0               0

C                    -x                                                  +x              +x

E                     0.1307                                         +x                  +x

                       Kb      = Kw/Ka

                                   = 1*10^-14/1.9*10^-5   = 5.26*10^-10

                         Kb     =   [HN3][OH^-]/[N3^-]

                            5.26*10^-10   = x*x/(0.1307-x)

                             5.26*10^-10*(0.1307-x)   = x^2

                             x   = 8.3*10^-6

                 [OH^]    = x    = 8.3*10^-6M
                     POH    = -log[OH^-]

                                = -log8.3*10^-6

                                = 5.1

                  PH         = 14-POH

                               = 14-5.1   = 8.9    >>>>answer