equilibrium reaction involved is;
NH3 + H2O <--->NH4+ + OH-
so at equilibrium we will be having [NH4+]=[OH-]
expressions for Kb = [NH4+][OH-]/[NH3]
so [NH3]=[NH4+][OH-]/Kb
given pH=11.5
so pOH=14-pH= 2.5
[OH-]=10-2.5=[NH4+]
on solving we will get [NH3]=0.55M
so no. of moles of ammonia in 1L= 0.55 moles
mass of ammonia = 0.55*17 = 9.44g
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