Question

The dissociation of hydroxylamine H2NOH in water Kb = 1.5x10-8 forms what two products. (Show ICE Table).

You have a 0.340 M solution of the base above. You create an ICE table and solve for x and get x = 1.45x10^-6

1. What is the hydroxide concentration of your base is [x]

2. the pH of your base is [y]

Answer 1

The dissociation of Hydroxylamine is as follows:

NH₂OH(aq)  +  H₂O(l)  <----->  NH₃OH⁺(aq)  +  OH^-(aq)

initial         0.34                                                                     0                            0          

change       -x                                                                        +x                          +x

eqm            0.34-x                                                                  x                          x

              The dissociation constant can be expressed as:

         K_b = [Nh3OH+] [OH-] / [NH2OH]

               1.5 x 10-8 = x² / (0.34-x)

            Given the value of x=1.45x10-6

            a) Conc of OH- = x = 1.45 x 10-6

            b) pH= 14-pOH = 14-[- log (1.45 x 10-6)] = 14-5.838 = 8.162

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