Question

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

Answer 1

H2A+ H2O----H3O+ HA^-2

HA^-2+ H2O-----> H3O + A-


Let x=HA^-2
Let y=A-

So, for the following reaction

H2C2O4 + H2O----H3O+ HC2O4

H2C2O4...........H3O....HC2O4
I0.40M..............0 ........ 0
C-x.....................x............ x
E0.40-x.............x.............x


Solve for x,

ka1=[x][x]/[0.40-x], 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....y-z..............z.............z



But since it is a diprotic acid and x=H3O+ y= H3O+, the charts become


H2C2O4 + H2O----H3O+ HC2O4

H2C2O4...........H3O....HC2O4
I0.4M..............0 ........ ..0
C-x..............,,..x............ .x
E0.4-x.........x+Y..........x-y


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^-3-y...x+y.....................y

So, ka1=[5.4 x 10^-3+y][5.4 x 10^-3-y]/[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.

Answer 2

See the above.

Answer 3

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