Answer:
Explanation:
given PH of weak base=8.88
molarity=0.15 m
ionization equation for weak base(B) is
B+H2O⇔HB+OH
kc for this equation is
kc=(HB)(OH)/(B)(H2O)
kc(H2O)=(HB)(OH)/(B)
kc(H2O)=Kb
Kb=(HB)(OH)/(B)
now we have to find the concentration of OH (OH)
we know that
PH+POH=PKw
POH=PKw-PH
POH=14.00-8.88
POH=5.12
POH=log(OH)
(OH)=10^-POH
(OH)=10^-5.12
(OH)=7.58*10^-6
remember that due to molar ratio of 1;1 (HB)=(OH)
so (HB)=(OH)=7.58*10^-6
we will use 0.15 M for (B)
Kb=(HB)(OH)/(B)
Kb=(7.58*10^-6)(7.58*10^-6)/(0.15)
Kb=3.83*10^-10
