Posted by C on .
in a 0.40 M solution of diprotic acid H2A(Ka=7.4x10^5, Ka2=5.0x10^10 at 25 celsius), what is the equilibrium concentration of A^2? A)0.40m b)0.80M C)5.4x10^3m d)1.4x10^5M E)5.0x10^10M

Chem 
Devron,
H2A+ H2OH3O+ HA^2
HA^2+ H2O> H3O + A
Let x=HA^2
Let y=A
So, for the following reaction
H2C2O4 + H2OH3O+ HC2O4
H2C2O4...........H3O....HC2O4
I0.40M..............0 ........ 0
Cx.....................x............ x
E0.40x.............x.............x
Solve for x,
ka1=[x][x]/[0.40x], which turns into,
ka1=[x][x]/[0.40]
sqrt*(ka1*0.40)=x=5.4 x 10^3
HC2O4 + H2O> H3O + C2O4
HC2O4...........H3O....C2O4
I.......Y..............0 ............ 0
C....z...............z.............z
E....yz..............z.............z
But since it is a diprotic acid and x=H3O+ y= H3O+, the charts become
H2C2O4 + H2OH3O+ HC2O4
H2C2O4...........H3O....HC2O4
I0.4M..............0 ........ ..0
Cx..............,,..x............ .x
E0.4x.........x+Y..........xy
HC2O4 + H2O> H3O + C2O4
HC2O4..................H3O...............C2O4
I...5.4 x 10^3.........x.....................0
C..y......................x+y.....................y
E...5.4 x 10^3y...x+y.....................y
So, ka1=[5.4 x 10^3+y][5.4 x 10^3y]/[0.370]
and
ka2=[5.4 x 10^3+y][y]/[5.4 x 10^3]
We solved for x, so solve for y.
I believe this is how you tackle this problem. 
Chem 
DrBob222,
See the above.

Chem 
Devron,
Change H2C2O4 to H2A, HC2O4 to HA^2, and C2O4 to A. Everything else is okay.