The measured pH of a 0.100M solution of NH3(aq) at 25C is 11.12. Calculate Kb for Nh3(aq)at 25C.

Answer 1

pH = 11.12. Convert to pOH.
pH + pOH = pKw = 14 so pOH must be 14-11.12 = 2.88
Convert to OH^-
pH = -log(OH^-)
(OH^-) = about 0.0012 but that's just a close estimate. You should redo that estimate for a better answer.
.......NH3 + HOH ==> NH4^+ + OH^-
I......0.1...........0........0
C.......-x...........x........x
E.....0.1-x..........x........x

x = 0.0012 from above. Evaluate the E line, plug that into the Kb expression for Kb and solve for Kb.