Question

In a 1.01 M aqueous solution of a monoprotic acid, 4.48% of the acid is ionized. What is the value of its Ka?

Answer 1

I

HA ---> H+ + A-
is the ionization of monoprotic acid,which ionizes only 4.48 %

So let us find out concentration of all species at equilibrium using initial
concentration of acid[1.01 M] and percent ionization value.For that construct ICE table
In ICE table we do following calculations
Initially there will be only acid in the solution So concentrations of ions are zero.
Once it is ionized using percent ionization we found out change in concentration.Then
finally take the difference between initial and change in concentration of each species.This
will provide equilibrium concentration.

Now use equilibrium concentration of each species to get Ka value which is given by the
formula
Ka= [H+][A-]/[HA]

ICE table

	HA---->	H+	A-
initial concentration	1.01 M	0	0
change in concentration	-{1.01 *0.0448 =0.0452}	+0.0452	+0.0452
equilibrium concentration	1.01-0.452 =0.9648	0.0452	0.0452

Ka= [H+][A-]/[HA]

plug in equilibrium concentrations from the table

Ka= 0.0452*0.0452/0.9648 = 2.12 x10^-3

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