1- First we will get the value of POH from:
PH + POH = 14
and we have PH = 12.01
∴ POH = 14 - 12.01 = 1.99
2- Then we need to get the concentration of OH:
when POH = -㏒[OH]
1.99 = -㏒[OH]
∴[OH] = 0.01 M
now we have the concentration at Equ by subtracting from the initial concentration of OH = 0.24 M
∴ [OH] = 0.24 - 0.01 = 0.23 M
∴ Kb = (0.01)^2 / 0.22977 M
∴ Kb value = 4.4 x 10^-4