Oxalic acid, found in the leaves of rhubarb and other plants, is a diprotic acid.

H2C2O4 + H2O ↔ H3O+ + HC2O4-
Ka1= ?
HC2O4- + H2O ↔ H3O+ + C2O42-
Ka2 = ?

An aqueous solution that is 1.05 M H2C2O4 has pH = 0.67. The free oxalate ion concentration in this solution is [C2O42-] = 5.3 x 10-5 M. Determine Ka1 and Ka2 for oxalic acid.

ka1 = (H^+)(HC2O4^-)/(H2C2O4)

ka2 = (H^+)(C2O4^2-)/(HC2O4^-)
You get (H3O^+) from pH.
You're given C2O4^2- and H2C2O4. Plug and chug.

Ka1= 0.6381

Ka2= 5.565 x 10-5

We can use the given information to calculate Ka1 and Ka2 for oxalic acid. First, let's calculate the concentration of H3O+ in the solution:

pH = -log[H3O+]
0.67 = -log[H3O+]

Taking the inverse logarithm of both sides:

[H3O+] = 10^(-pH)
[H3O+] = 10^(-0.67)
[H3O+] = 0.210 M

Since oxalic acid is diprotic, we need to consider the ionization of both protons. Using the given concentration of H2C2O4 and the concentration of H3O+, we can calculate Ka1:

Ka1 = ([H3O+][HC2O4-]) / [H2C2O4]
= (0.210 * [HC2O4-]) / 1.05

We also know the concentration of C2O42-, so we can calculate [HC2O4-] using the stoichiometry of the reaction:

HC2O4- + H2O ↔ H3O+ + C2O42-

[H2C2O4] = [H2C2O4]initial - [HC2O4-]
= (1.05 M) - (0.210 M)
= 0.84 M

Since the reaction is in equilibrium, the concentration of HC2O4- can be assumed to be the same as the concentration of H2C2O4:

[HC2O4-] = 0.84 M

Substituting the values into the equation for Ka1:

Ka1 = (0.210 * 0.84) / 1.05
Ka1 ≈ 0.168

Now, let's calculate Ka2 using the given concentration of C2O42-:

Ka2 = ([H3O+][C2O42-]) / [HC2O4-]

Substituting the known values:

Ka2 = (0.210 * 5.3 x 10^(-5)) / 0.84
Ka2 ≈ 1.33 x 10^(-5)

Therefore, the calculated values for Ka1 and Ka2 are approximately 0.168 and 1.33 x 10^(-5) respectively.

To determine the values of Ka1 and Ka2 for oxalic acid, we need to make use of the given information about the concentration of H2C2O4, the pH of the solution, and the concentration of C2O42-.

Step 1: Find the concentration of H3O+
The pH of the solution is given as 0.67. Since pH is the negative logarithm of the H3O+ concentration, we can calculate the H3O+ concentration by taking the antilog of the negative pH value:
[H3O+] = 10^(-pH) = 10^(-0.67)

Step 2: Calculate the ratio of [HC2O4-] to [H2C2O4]
Since oxalic acid is a diprotic acid, it can donate two protons. Therefore, we can assume that at equilibrium, [H2C2O4] = [H3O+] and [HC2O4-] = [H3O+].

Step 3: Calculate [H2C2O4] and [HC2O4-] using the concentration of H3O+
[H2C2O4] = [H3O+] = 10^(-0.67)
[HC2O4-] = [H3O+] = 10^(-0.67)

Step 4: Calculate [C2O42-]
[C2O42-] is given as 5.3 x 10^(-5) M.

Step 5: Set up the expression for Ka1 based on the equilibrium equation for H2C2O4, which is:
H2C2O4 + H2O ↔ H3O+ + HC2O4-
Ka1 = ([H3O+][HC2O4-])/[H2C2O4]

Substitute the calculated values into the equation:
Ka1 = ([H3O+][H3O+]) / [H3O+]

Simplify:
Ka1 = [H3O+]^2

Step 6: Calculate Ka1
Substitute the value of [H3O+] calculated in Step 1 into the equation:
Ka1 = (10^(-0.67))^2

Step 7: Calculate Ka1 using a calculator or computer program:
Ka1 = 10^(-2 * 0.67)

Repeat the same steps for Ka2 using H2C2O4 and C2O42-. The equilibrium equation for this step is:
HC2O4- + H2O ↔ H3O+ + C2O42-
Ka2 = ([H3O+][C2O42-])/[HC2O4-]

Substitute the calculated values into the equation and solve for Ka2.

By following these steps, you can determine the values of Ka1 and Ka2 for oxalic acid based on the given information.