oxalic acid, h2c2o4, is a weak acid capable of providing two H3O+ ions.
ka=.059
k2=6.4E-5
the [h3o]+ in a .38 M solution of h2c2o4 is .12M and can be calculated by the first ionization step only. what is the equilibrium concentration of the oxalate ion c2o4^-2, in the .38 M solution of h2c2o4?
H2C2O4 ==> H^+ + HC2O4^-
HC2O4^- ==> H^+ + C2O4^-2
k2 = (H^+)(C2O4^-2)/(HC2O4^-)
You know (H^+) = 0.12M
For all practical purposes, HC2O4^- is the same; therefore, (C2O4^-2) = k2 = 6.4E-5M.
To find the equilibrium concentration of the oxalate ion (C2O4^-2) in the solution of H2C2O4, we need to use the information provided and apply the concept of equilibrium and acid dissociation.
Firstly, let's write down the balanced equation for the ionization step of H2C2O4:
H2C2O4 ⇌ H^+ + HC2O4^-
From the information given, we know the initial concentration of the H3O+ ion ([H3O+]) is 0.12 M. We also know the total concentration of H2C2O4 is 0.38 M.
Using the equilibrium expression for the ionization step of H2C2O4:
Ka = [H^+][HC2O4^-] / [H2C2O4]
We can assume that the concentration of H^+ will be twice the concentration of HC2O4^- due to the stoichiometry of the equation.
So, let's assume the concentration of HC2O4^- is x M. Then, the concentration of H^+ will be 2x M. And the concentration of H2C2O4 is 0.38 M.
Therefore, the equilibrium expression becomes:
Ka = (2x)(x) / 0.38
Now, substitute the given value for Ka (= 0.059) into the equation and solve for x:
0.059 = 2x^2 / 0.38
Cross-multiply:
0.059 * 0.38 = 2x^2
0.02242 = 2x^2
Divide both sides by 2:
x^2 = 0.01121
Take the square root of both sides:
x ≈ 0.1059
So, the equilibrium concentration of HC2O4^- (which is equal to the concentration of C2O4^-2) in the 0.38 M solution of H2C2O4 is approximately 0.1059 M.