Question
In order to prepare a buffer of pH 8.26, the amount of (NH4 )2SO4 required to be mixed with 1L of 0.1M NH3 (pKb of NH3 = 4.74) is
Solution by Joshi sir.
NH3 + H2O ------------> NH4+ + OH-
so Kb = [NH4+] [OH-]/[NH3][H2O]
so Kb = [NH4+]/[NH3][H+] * [H+][OH-]/[H2O]
so Kb = [NH4+]/[NH3][H+] * Kw
now take log and solve. concentration of (NH4 )2SO4 will be double of [NH4+]
if you find any problem on solving this then reply