What is the concentration of CO3^-2 in a 0.028M 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.
A slight mix up here as this is H2CO3 and not H2C2O4; however, the problem is worked exactly the same way as H2C2O4 was solved.
As noted in another post, I thought one person kept posting the same question over and over again.
Change 0.370M to 0.028M and solve.
I think that happens in more cases than we know about.
To find the concentration of CO3^2- in the solution of carbonic acid (H2CO3), we need to consider the dissociation reactions of carbonic acid and use the equilibrium expressions.
The dissociation reactions of carbonic acid are as follows:
H2CO3 ⇌ H+ + HCO3^-
HCO3^- ⇌ H+ + CO3^2-
We are given the equilibrium constants, Ka1 and Ka2, which are the expression of the equilibrium of the two dissociation reactions.
An important assumption in this calculation is that the concentration of H2CO3 remains almost unchanged upon dissociation. This is because it is a weak acid, and the degree of dissociation is typically small.
Let's denote [H2CO3] as the initial concentration of carbonic acid, and let [H+] and [CO3^2-] represent the concentrations of the respective ions at equilibrium.
For the first dissociation reaction: H2CO3 ⇌ H+ + HCO3^-
The equilibrium expression is: Ka1 = [H+][HCO3^-] / [H2CO3]
For the second dissociation reaction: HCO3^- ⇌ H+ + CO3^2-
The equilibrium expression is: Ka2 = [H+][CO3^2-] / [HCO3^-]
Since we are given the value of Ka1 and Ka2, we can solve for [H+], and from there, determine [CO3^2-].
1. Determining the concentration of H+ from Ka1:
Ka1 = [H+][HCO3^-] / [H2CO3]
[H+][HCO3^-] = Ka1 * [H2CO3]
[H+][HCO3^-] = (4.2 × 10^-7) * (0.028M)
[H+] = (4.2 × 10^-7) * (0.028M) / [HCO3^-]
2. Determining the concentration of H+ from Ka2:
Ka2 = [H+][CO3^2-] / [HCO3^-]
[H+][CO3^2-] = Ka2 * [HCO3^-]
[H+][CO3^2-] = (4.8 × 10^-11) * (0.028M)
[H+] = (4.8 × 10^-11) * (0.028M) / [CO3^2-]
Now we have two expressions for [H+], equate them together:
(4.2 × 10^-7) * (0.028M) / [HCO3^-] = (4.8 × 10^-11) * (0.028M) / [CO3^2-]
Simplifying the equation further:
(4.2 × 10^-7) / [HCO3^-] = (4.8 × 10^-11) / [CO3^2-]
Cross-multiplying:
(4.2 × 10^-7) * [CO3^2-] = (4.8 × 10^-11) * [HCO3^-]
Now, we know that for every HCO3^- ion that dissociates, one CO3^2- ion is formed. Therefore, [HCO3^-] = [CO3^2-].
Substituting the values:
(4.2 × 10^-7) * [CO3^2-] = (4.8 × 10^-11) * [CO3^2-]
Simplifying the equation:
[(4.2 × 10^-7) - (4.8 × 10^-11)] * [CO3^2-] = 0
The equation simplifies to [CO3^2-] = 0.
Therefore, the concentration of CO3^2- in a 0.028M solution of carbonic acid is essentially zero.