How do you find the Ka1 and Ka2 of oxalic acid when it is in a solution that is 1.05 M H2C2O4 and has a pH of 0.67. [C2O4^2-] = 5.3x10^-5 M. I have tried using an ICE table for both reactions, as oxalic acid is a diprotic acid, but I am having trouble knowing where to put concentrations on these tables, and where to put unknowns. I have found [H3O+] to be 0.214 M, but am not sure where to go from there. The reactions are H2C2O4 + H2O --> H3O^+ + HC2O4^- (Ka1 = ?) and HC2O4^- + H2O --> H3O^+ + C2O4^2- (Ka2 = ?)
You know the concentration of the chromate ion (above), then let the concentration of the romate ion be x, then x+concentrationroamte=1.05m
It seems to me then you can figure out the H+ ion concentration in each reaction, and thence figure Ka for each. I didn't work it out, but it seems straightforward to my simplistic mind.
To find the Ka1 and Ka2 values of oxalic acid (H2C2O4), you can use the known concentration of oxalate ion ([C2O4^2-]) and the pH of the solution. The dissociation reactions of oxalic acid are as follows:
H2C2O4 + H2O ⇌ H3O^+ + HC2O4^- (Reaction 1, with Ka1)
HC2O4^- + H2O ⇌ H3O^+ + C2O4^2- (Reaction 2, with Ka2)
Here's how you can approach the problem step by step:
Step 1: Start with the given pH value.
pH = -log[H3O^+]
0.67 = -log[H3O^+]
Step 2: Solve for [H3O^+].
[H3O^+] = 10^(-pH)
[H3O^+] = 10^(-0.67)
Step 3: Calculate the concentration of oxalate ion ([C2O4^2-]).
[C2O4^2-] = 5.3 x 10^(-5) M (Given)
Step 4: Use the known concentration of oxalate ion and the concentration of hydronium ion to create an ICE (Initial, Change, Equilibrium) table for Reaction 2:
HC2O4^- + H2O ⇌ H3O^+ + C2O4^2-
Initial: 0 0 [H3O^+] [C2O4^2-]
Change: -x +x +x +x
Equilibrium: x x [H3O^+] + x [C2O4^2-] + x
Step 5: Write the expression for Ka2:
Ka2 = ([H3O^+][C2O4^2-]) / [HC2O4^-]
Since you know [H3O^+] and [C2O4^2-], plug them into the expression and solve for [HC2O4^-], which is x in the equilibrium row of the ICE table.
Step 6: Repeat Steps 4 and 5 for Reaction 1:
H2C2O4 + H2O ⇌ H3O^+ + HC2O4^-
Step 7: Solve for Ka1, using the known concentration of hydronium ion and the concentration of the acid.
Ka1 = ([H3O^+][HC2O4^-]) / [H2C2O4]
Since you know [H3O^+] and [HC2O4^-], plug them into the expression and solve for [H2C2O4], which is x in the equilibrium row of the ICE table.
By following these steps, you should be able to determine the values of Ka1 and Ka2 for oxalic acid when given the pH and the concentration of oxalate ion.
To find the Ka1 and Ka2 of oxalic acid (H2C2O4), we can follow these steps:
Step 1: Write the balanced chemical equations for the dissociation of oxalic acid:
H2C2O4 + H2O ⇌ H3O+ + HC2O4^- (Ka1)
HC2O4^- + H2O ⇌ H3O+ + C2O4^2- (Ka2)
Step 2: Set up an ICE table for the first reaction (Ka1):
H2C2O4 + H2O ⇌ H3O+ + HC2O4^-
Initial: 1.05 M 0 M 0 M
Change: -x -x +x
Equilibrium: (1.05 - x) (0 - x) x
Step 3: Write the expression for Ka1:
Ka1 = [H3O+][HC2O4^-] / [H2C2O4]
We know [H3O+] = 0.214 M and [HC2O4^-] = x
So, Ka1 = (0.214)(x) / (1.05 - x)
Step 4: Set up an ICE table for the second reaction (Ka2):
HC2O4^- + H2O ⇌ H3O+ + C2O4^2-
Initial: x 0 M 5.3x10^-5 M
Change: -x -x +x
Equilibrium: (x - x) (0 - x) (5.3x10^-5 + x)
Step 5: Write the expression for Ka2:
Ka2 = [H3O+][C2O4^2-] / [HC2O4^-]
We know [H3O+] = 0.214 M and [C2O4^2-] = 5.3x10^-5 M
So, Ka2 = (0.214)(5.3x10^-5 + x) / x
Step 6: Now we need to solve these two equations for x, which represents x = [HC2O4^-] = [H2C2O4] in the first reaction and x = [C2O4^2-] = [HC2O4^-] in the second reaction. However, it is worth noting that the concentration of H2C2O4 is initially 1.05 M, and we assumed that it fully dissociates, resulting in (1.05 - x) M. This assumption is justified since x is expected to be small compared to 1.05.
Step 7: Solve the first equation (Ka1):
Ka1 = (0.214)(x) / (1.05 - x)
Step 8: Solve the second equation (Ka2):
Ka2 = (0.214)(5.3x10^-5 + x) / x
Unfortunately, we cannot further simplify these equations since two unknowns are present, x and Ka. To solve for Ka1 and Ka2, you may need additional information or better estimates for x.