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.
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.
See comments above.
Change the initial concentrations; I thought this was the same person posting the same problem over and over again.
Change 0.370M to 0.010M
To determine the hydronium ion concentration in a solution of carbonic acid (H2CO3), you need to consider the dissociation of carbonic acid and the equilibrium constant expressions (K).
The dissociation of carbonic acid involves two steps:
1) Dissociation of the first proton (H+):
H2CO3 ⇌ H+ + HCO3-
2) Dissociation of the second proton (H+):
HCO3- ⇌ H+ + CO32-
The equilibrium constant expression, Ka1, for the first dissociation step is given as 4.2x10^-7 and Ka2 for the second dissociation step is given as 4.8x10^-11.
To calculate the hydronium ion concentration, [H3O+], in the solution, you need to consider the concentration of the fully dissociated form of carbonic acid, which is H+.
Since carbonic acid is a weak acid, we can assume that the concentration of CO32- (fully dissociated form) is negligible. Therefore, we can assume that the concentration of HCO3- is approximately equal to the concentration of H2CO3 initially.
Let's denote the concentration of H2CO3 as [H2CO3]. Since the dissociation of H2CO3 occurs in two steps, we can assume that most of the initial H2CO3 dissociates into HCO3-, and only a small fraction dissociates into H+.
Using the equilibrium expression for Ka1:
Ka1 = [H+][HCO3-] / [H2CO3]
Since the initial concentration of HCO3- is approximately equal to [H2CO3], we can write it as:
Ka1 = [H+][HCO3-] / [HCO3-]
Simplifying:
Ka1 = [H+]
Therefore, the concentration of H+ can be considered equal to the value of Ka1.
For a 0.010M solution of H2CO3, the concentration of H+ (hydronium ions) is approximately equal to 4.2x10^-7 M.
So, the hydronium ion concentration in the 0.010M solution of carbonic acid is approximately 4.2x10^-7 M.