Question

What is the hydronium ion concentration in a 0.010M solution of carbonic acid, H2CO3? For carbonic acid, Ka1=4.2x10^-7 and Ka2=4.8x10^-11.

Answer 1

H2C2O4 + H2O----H3O+ HC2O4

HC2O4 + H2O-----> H3O + C2O4


Let x=HC2O4
Let y=C2O4

So, for the following reaction

H2C2O4 + H2O----H3O+ HC2O4

H2C2O4...........H3O....HC2O4
I0.370M..............0 ........ 0
C-x.....................x............ x
E0.370-x.............x.............x


Solve for x,

ka1=[x][x]/[0.370-x], which turns into,


ka1=[x][x]/[0.370]

sqrt*(ka1*0.370)=x

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
I0.370M..............0 ........ ..0
C-x..............,,..x............ .x
E0.370-x.........x+Y..........x-y


HC2O4 + H2O-----> H3O + C2O4
HC2O4...........H3O...............C2O4
I...x....................x.....................0
C..-y..............x+y.....................y
E...x-y...........x+y.....................y

So, ka1=[x+y][x-y]/[0.370]

and

ka2=[x+y][y]/[x-y]



We solved for x, so solve for y.


I believe this is how you tackle this problem.

Answer 2

See comments above.

Answer 3

Change the initial concentrations; I thought this was the same person posting the same problem over and over again.

Answer 4

Change 0.370M to 0.010M