-log[H+]=10.64
[H+]=2.29*10-11
Kb=[OH-][CH3NH3+]/[]CH3NH2]
[CH3NH2]=4.33*10-4=mol/litre
Wt.=Molarity*vol*M=4.33*10-4*(100/1000)*31
=1.34 mg
An aqueous solution of methyl amine has pH value 10.64.How many grams of methyl amine are present in 100mL of the solution.Kb=4.4*10-4
-log[H+]=10.64
[H+]=2.29*10-11
Kb=[OH-][CH3NH3+]/[]CH3NH2]
[CH3NH2]=4.33*10-4=mol/litre
Wt.=Molarity*vol*M=4.33*10-4*(100/1000)*31
=1.34 mg
Is the question about initial weight of methyl amine or at equilibrium
initially Wt=2.697 mg (not g)
At equilibrium Wt.=1.34 mg