Calculate the concentrations of all species present in a 0.29 M solution of ethyleneammonium chloride?

Question

Question Details:The species are [ C2H5NH2] , [ H]+, [ C2H5NH3] +, [Cl]- and [OH]- Please help me in solving this problem related to acids and bases. Thanx in advance...

Please help in solving this problem related to acids and bases

Answer 1

First of all you should have provided the equilibrium constant.
pKa=9.5

The dissociation
C2H5NH3Cl -> C2H5NH3+ + Cl- is 100%, thus
[Cl-]= C= 0.29 M
The cation formed is a weak acid with pKa=9.5, so you have

.. .. .. .. .. .. .. C2H5NH3+ <=> C2H5NH2 + H+
Initial .. .. .. .. .. .. 0.29
Dissociate .. .. .. .. x
Produce .. .. .. .. .. .. .. .. .. .. .. .. .. x .. .. .. .. x
At Equil. .. .. .. 0.29-x .. .. .. .. .. .. .. x .. .. .. .. x

Ka= [C2H5NH2][H+] / [C2H5NH3+] = x^2/(0.29-x) (1)
Let's assume that x<<0.29 and practically 0.29-x=0.29
so (1) is simplified to

Ka=x^2/0.29 => x= squareroot( 0.29*Ka)
Ka=10^-pKa= 10^-9.5 so
x= SQRT(0.29*10^-9.5) =9.6 *10^-6 M
thus our assumption was correct

[C2H5NH3+]=0.29 (you could substract 9.6*10^-6 but the result will not change-remember that this was our assumption. if you want an exact value should solve the quadratic, find the exact x and then [C2H5NH3+]=0.29-x)

[C2H5NH2]= [H+]=x = 9.6*10^-6 M

[OH-]= Kw/[H+] = (10^-14)/(9.6*10^-6) =1.04*10^-9 M