Question

The pH of an ammonia solution is 9.00. What is the concentration in mol L^-1 of this ammonia solution given that K_b for ammonia is 1.78 x 10^-5?

Answer 1

PH   = 9

POH = 14-PH

          = 14-9

           = 5

POH = 5

-log[OH^-] = 5

     [OH^-]   = 10^-5 M

at equilibrium [NH4^+]   = [OH^-]   = 10^-5M

                  NH3(aq) + H2O(l) ----------> NH4^+ (aq) + OH^- (aq)

Kb   = [NH4^+][OH^-]/[NH3]

1.78*10^-5   = 10^-5 *10^-5/x

x                      = 10^-5 *10^-5/1.78*10^-5    = 5.6*10^-6

[NH4^+]   = x = 5.6*10^-6M >>>>answer