pka= log ka.

ka=[products]/[reagents] both in the equilibrium

in the reaction: CH3NH3+H2O==(CH3NH4+)+OH-

       initial        3*10^-3

equilibrium       3*10^-3 -x          x              x

So ka=x^2/3*10^-3-x

in this equation you can resolve the x value, so you know [OH-] --> pOH=log[OH]

then, pH=14-pOH

Might there not be a difference between the formulas for acids and bases? I think the formula goes for acids, but bases like NH3 increase the OH- not H3O+.

i know that rule but i don't have the product and the reagent concentration

i only have the initial concentration which is  3*10^-3

sarah, what is the initial concentration? 3*10^-3? Right?

Ph = Pka + log [ products]/[reagents].  good lucky