Answered By Jonny C

NH3 is a weak base so it will undergo an equilibrium reaction. You need to first make an ICE table.

                NH3 + H2O <=> NH4+ + OH- Initial                0.3                      0            0 Change                 -x                        +x           +x Equilibrium                 0.3-x                    x             x 

Find the Kb of NH3 using the data booklet (You have the Ka given so divide 1.0*10^-14 by the Ka given)

Kb = 1.8 x 10^-5 = x^2 / 0.3-x 

Using the "rule of 100" (meaning if you divide concentration by Kb and the answer gives you a number greater than 100, you can take out the x on the bottom fraction, different teachers have different variants of this rule) you can take out the x on the bottom fraction.

Kb = 1.8 x 10^-5 = x^2 / 0.3x = [OH-]= 0.0023 M pOH = - log 0.0023=2.6pH = 14 - 2.6 = 11.4