how would i find the pH of this?
1.0 M NH3 (Kb = 1.8 x 10-5)
NH3 + HOH ==> NH4^+ + OH^-
Write Kb expression.
(NH4^+) = x molar
(OH^-) = x molar
(NH3) = (1 - x) molar
Solve for x which = (OH^-)
then pOH = - log(OH^-)
and pH + pOH = pKw = 14
Solve for pH.
Post your work if you need additional help.