Question

Hydrazine, NH2NH2, is a weak base. The following is the equilibrium equation for its reaction with water:

NH2NH2(aq) + H2O(l) <----------> NH2NH3+(aq) + OH-(aq)     Kb = 8.5 x 10-7

What is the hydronium ion concentration, [H3O+], in a 2.03 M NH2NH2 solution?

Answer 1

Here we are given with value of Kb = 8.5 x 10^-7

Now we will find the ICE for the given equation in order to find out the concentraion of [OH-] using which we will find the concentration of [H3O+] or [H+]

NH2NH2(aq) + H2O(l) <----------> NH2NH3+(aq) + OH-(aq)     Kb = 8.5 x 10-7

I   2.03 M 0 0

C - x +x +x

E 2.03 - x x x

Now Kb for the above equation can be written as

Kb = [NH2NH3+] [OH-]/[NH2NH2]

  8.5 x 10-7 = x^2 /  (2.03 - x)

Since NH2NH2 is a weak base (K_b is small), it dissociates only slightly, x will be very small compared to 2.03

So, at equilibrium, [NH2NH2] ? 2.03 mol L^-1

     x^2 = 8.5 x 10-7 x 2.03

x = 0.001313583 mol L^-1 = [NH2NH3+] = [OH-]

Now using ionic product of water to find [H3O+]

K_w = [H3O+] [OH+]

   10^-14 = [H3O+] x 0.001313583

  [H3O+] = 10^-14 / 0.001313583

[H3O+] = 7.61 x 10^-12 mol L^-1