Question

For the diprotic weak acid H2A, Ka1 = 2.2 × 10-6 and Ka2 = 8.3 × 10-9. What is the pH of a 0.0700 M solution of H2A? What are the equilibrium concentrations of H2A and A2– in this solution?

Answer 1

Let's represent diprotic acid as H2A

it will dissociate as follows:

           Ka1

Ka1 = [HA-][H+]/[H2A]

2. Then HA- will dissociate as follows:

        Ka2

HA^- -----> A^2- + H⁺

^{Here Ka2 =[A2-][H+]/[HA-]}

^{COnstruct ICE table for First Equilibrium Reaction:}

	H2A ------->	HA-    +	H+
Initial	0.07M	0	0
Concentration	0.07 -x	+x	+x
Equilibrium	0.07 -x	x	x

As we know Ka1 = [HA-][H+]/[H2A]

Given Ka1 = 2.2 x 10^-6

2.2 x 10^-6 = [x][x]/[0.07-x]

2.2 X 10^-6 = [x^2] /0.07 ..........neglecting value of x

2.2 X 10^-6 X 0.07 = x^2

x = 3.9 X 10^-4

pH = -log [H+] = - log [3.9 X10^-4 ] = 3.4

Now Construct ICE table for second equilibrium as follows;

	HA- ------>	A2-    +	H+
Initial	3.9X 10^-4	0	3.9X10^-4
Conc.	-x	+x	+x
Equi.	3.9X10^-4 - x	x	3.9X10^-4 +x

Given Ka2 = 8.3 X 10^-9

Ka2 = [A2-][H+]/[HA-]

8.3 X 10^-9 = [x][3.9X10^-4 + x]/[3.9X10^-4 -x]

x = 8.3 X 10^-9