A 0.010M NH3 solution was prepared and it was determined that the NH3 had undergone 4.2% ionization. Calculate Kb for NH3 under these conditions

Answer 1

....................NH3 + HOH ==> NH4^+ + OH^-
I....................0.01.....................0.................0
C...................-x.........................x...................x
E..................0.01-x....................x..................x
The problem tells you that a 0.01 M solution is ionized 4.2%; therefore, you know (NH4^+) = (OH^-) = x = 0.01*0.042 = 4.2E-4 and (NH3) = 0.01-4.2E-4= 9.88E-3
Plug those numbers into the Kb expression for NH3 and solve for Kb. Post your work if you get stuck.