Oxalic acid , HOOCCOOH (aq) , is a diprotic acid with Ka1= 0.0560 and Ka2=0.0145. Determine the [-OOCCOO-] in a solution with an initial concentration of 0.139 M oxalic acid.
HOOCCOOH(aq) = HOOCCOO-(aq) + H+(aq)
HOOCCOO-(aq) = -OOCCOO-(aq) + H+(aq)
I have tried a ICE diagram:
HOOCCOO^- <=> ^-OOCCOO^-+ H^+
I: 0......0.139......0
C: -X......+X.......+X
E: -X......0.139+x......X
K= 0.139+X *X / -X
0.0145 = 0.139+X *X /-X
X= 0.03823
[-OOCCOO-] = 0.139-.03823= 0.10077
Not sure where I went wrong. Please help
Amanda
Where did you get these Ka values. The literature gives values that are about 1000 between k1 and k2 but your values have a ratio of about 4 and not 1000.
the Ka values are given in the question from the homework.
Ka1 = 0.0560, for the equation: HOOCCOOH(aq) = HOOCCOO-(aq) + H+(aq)
While Ka2 = 0.0145, for the equation:
HOOCCOO-(aq) = -OOCCOO-(aq) + H+(aq)
Your calculations are almost correct, but there is a small error in the setup of your ICE table. When you write the concentration of the oxalic acid, HOOCCOOH, as 0.139 M in the initial row, it should not have a negative sign in front of it in the change row. The concentrations in the change row only change by a positive value, so it should be:
HOOCCOO^- <=> ^-OOCCOO^- + H^+
I: 0......0.139......0
C: -X.....+X.......+X
E: 0......0.139+X....X
Now, you can calculate the value of X again using your k expression:
0.0145 = (0.139+X) * X / 0
Simplifying this equation gives:
0.0145 = X^2 / (0.139+X)
Rearranging the equation gives:
X^2 = 0.0145 * (0.139+X)
X^2 = 0.0020155 + 0.0145X
Rearranging again:
X^2 - 0.0145X - 0.0020155 = 0
Using a quadratic solver or factoring, you can find that X is approximately 0.0379.
Now, you can calculate the concentration of [-OOCCOO-]:
[-OOCCOO-] = 0.139+X - X = 0.139.
So, the concentration of [-OOCCOO-] in the solution is approximately 0.139 M.
I hope this clarifies the steps for you. Let me know if you have any further questions!
To determine the concentration of [-OOCCOO-] in the solution, you need to consider both the dissociation reactions of oxalic acid:
1) HOOCCOOH(aq) ⇌ HOOCCOO-(aq) + H+(aq) (equation 1)
2) HOOCCOO-(aq) ⇌ -OOCCOO-(aq) + H+(aq) (equation 2)
You correctly set up an ICE table for equation 1, but the equilibrium constant (Ka1) you used is incorrect. The given Ka1 value is actually the Ka2 value.
So let's correct that and solve the problem:
1. Set up the ICE table for equation 1:
HOOCCOOH(aq) ⇌ HOOCCOO-(aq) + H+(aq)
Initial: 0.139 M 0 0
Change: -x +x +x
Equilibrium: 0.139 - x x x
2. Write the expression for Ka1:
Ka1 = [HOOCCOO-][H+] / [HOOCCOOH]
3. Plug in the equilibrium concentrations:
0.0560 = x^2 / (0.139 - x)
Since Ka1 represents a small dissociation constant, we can assume that x is small compared to 0.139, so we can approximate 0.139 - x ≈ 0.139.
4. Simplify the equation:
0.0560 = x^2 / 0.139
5. Solve for x:
x^2 = 0.0560 * 0.139
x^2 = 0.007784
x ≈ 0.0881
Now we have the concentration of HOOCCOO-, which is 0.0881 M.
6. Now, let's set up an ICE table for equation 2:
HOOCCOO-(aq) ⇌ -OOCCOO-(aq) + H+(aq)
Initial: 0.0881 M 0 0
Change: -x +x +x
Equilibrium: 0.0881 - x x x
7. Write the expression for Ka2:
Ka2 = [-OOCCOO-][H+] / [HOOCCOO-]
8. Plug in the equilibrium concentrations:
0.0145 = x^2 / (0.0881 - x)
Again, since Ka2 represents a small dissociation constant, we can assume that x is small compared to 0.0881, so we can approximate 0.0881 - x ≈ 0.0881.
9. Simplify the equation:
0.0145 = x^2 / 0.0881
10. Solve for x:
x^2 = 0.0145 * 0.0881
x^2 = 0.00127745
x ≈ 0.0358
Now we have the concentration of -OOCCOO-, which is 0.0358 M.
Therefore, the concentration of [-OOCCOO-] in the solution with an initial concentration of 0.139 M oxalic acid is approximately 0.0358 M.