Oxalic acid is a relatively strong diprotic acid with Ka1 = 5.6 ´ 10-2 and Ka2 = 5.1 ´ 10-5.
What is the oxalate ion, C2O42-, concentration in a 0.10 M oxalic acid solution?
a. 0.052
b. 0.073
c. 0.10
d. 5.6 ´ 10-2
e. 5.1 ´ 10-5
To find the concentration of the oxalate ion (C2O42-) in a 0.10 M oxalic acid solution, we need to consider the dissociation of oxalic acid.
Oxalic acid (H2C2O4) is a diprotic acid, meaning it can donate two hydrogen ions (H+). It dissociates in water as follows:
H2C2O4 ⇌ 2 H+ + C2O42-
The equilibrium constant (Ka1) for the first dissociation step is given as 5.6 ´ 10-2, which represents the ratio of the concentration of the products (H+ and C2O42-) to the concentration of the reactant (H2C2O4).
So we can write the expression for Ka1 as:
Ka1 = [H+]^2 / [H2C2O4]
Since the concentration of H2C2O4 is given as 0.10 M, we can substitute this value into the equation to solve for [H+]:
5.6 ´ 10-2 = [H+]^2 / 0.10
Simplifying the equation, we can cross multiply to isolate [H+]^2:
[H+]^2 = (5.6 ´ 10-2) * (0.10)
[H+]^2 = 5.6 ´ 10-3
Taking the square root of both sides gives us:
[H+] ≈ √(5.6 ´ 10-3) ≈ 0.075
Since the H+ ions are formed in a 1:1 ratio with the oxalate ions (C2O42-), the concentration of C2O42- is also 0.075 M.
Now, we can check the answer options to find the correct concentration:
a. 0.052 - Not the correct concentration
b. 0.073 - Not the correct concentration
c. 0.10 - Not the correct concentration
d. 5.6 ´ 10-2 - Not the correct concentration
e. 5.1 ´ 10-5 - Not the correct concentration
None of the provided answer options matches the calculated concentration of 0.075 M, so none of the given options are correct.