what is the hydronium ion concentration of a 0.35M oxalic acid, H2C2O4. solution? for oxalic acid, ka1= 5.6x10^-2 and ka2=5.1x10^-5. Identify and calculate the concentrations for the representative species in solution.
.....H2C2O4 ==> H^+ + HC2O4^-
I....0.35M......0......0
C.....-x........x......x
E....0.35-x.....x......x
Substitute the E line into k1 and solve for x = (H^+) = (HC2O4^-)
.......HC2O4^- ==> H^+ + C2O4^=
Write the k2 expression. You've just calculated that H^+ = HC2O4^- so cancel H^+ in the numerator with HC2O4^- in the denominator which leave k2 = (C2O4^=).
I got [H+]= 0.14M and for [HC2O4]=0.14M and [C2O4^=] 5.1x10^-5
Is this correct?
You didn't list k1 and k2 and the numbers in my text probably don't agree with the numbers in you text.
To find the hydronium ion (H3O+) concentration of a 0.35 M oxalic acid (H2C2O4) solution, we need to consider the dissociation of oxalic acid and the associated equilibrium expressions.
First, let's write the dissociation reactions for H2C2O4:
H2C2O4 ⇌ H+ + HC2O4- (Reaction 1)
HC2O4- ⇌ H+ + C2O4^2- (Reaction 2)
Next, let's interpret the given equilibrium constants (ka1 and ka2) and their relationship to the concentrations of the different species in solution.
For Reaction 1:
ka1 = [H+][HC2O4-] / [H2C2O4]
For Reaction 2:
ka2 = [H+][C2O4^2-] / [HC2O4-]
Now, let's identify and calculate the concentrations for the representative species in the solution.
Let's assume the initial concentration of H2C2O4 is [H2C2O4] = 0.35 M. Since it dissociates into H+ and HC2O4-, their initial concentrations are both zero [H+] = [HC2O4-] = 0 M.
Using the equilibrium constant (ka1) and the initial concentrations, we have:
ka1 = [H+][HC2O4-] / [H2C2O4]
5.6x10^-2 = (x)(x) / (0.35 - x)
x represents the equilibrium concentration of [H+] and [HC2O4-]. Since the initial concentrations of H+ and HC2O4- were both zero, the change in concentration is the same for both, so [H+] = [HC2O4-] = x.
Simplifying the equation, we get:
5.6x10^-2 = x^2 / (0.35 - x)
Similarly, using ka2 and the initial concentrations, we have:
ka2 = [H+][C2O4^2-] / [HC2O4-]
5.1x10^-5 = (x)(C2O4^2-) / x
Since the initial concentration of C2O4^2- is zero, we can write:
5.1x10^-5 = (x)(C2O4^2-) / x
5.1x10^-5 = C2O4^2-
Thus, the equilibrium concentration of C2O4^2- is equal to 5.1x10^-5 M.
Now, let's solve for x in the equation derived from ka1:
5.6x10^-2 = x^2 / (0.35 - x)
This is a quadratic equation, so rearrange it:
x^2 = (5.6x10^-2)(0.35 - x)
x^2 = 0.196x - 5.6x10^-2x^2
Simplifying further:
x^2 + 5.6x10^-2x - 0.196x = 0
x^2 - 0.756x + 0 = 0
Solve this quadratic equation using the quadratic formula to find x, which is the equilibrium concentration of H+ and HC2O4-.
Once you find the value of x, that will give you the equilibrium concentration of H+, which is the hydronium ion concentration in the oxalic acid solution.