What is the concentration of C2O4^-2 in a 0.370M oxalic acid solution? Ka1=5.6x10^-2 and Ka2=5.1x10^-5
H2C2O4 + H2O----H3O+ HC2O4
HC2O4 + H2O-----> H3O + C2O4
Let x=HC2O4
Let y=C2O4
So, for the following reaction
H2C2O4 + H2O----H3O+ HC2O4
H2C2O4...........H3O....HC2O4
I0.370M..............0 ........ 0
C-x.....................x............ x
E0.370-x.............x.............x
Solve for x,
ka1=[x][x]/[0.370-x], which turns into,
ka1=[x][x]/[0.370]
sqrt*(ka1*0.370)=x
HC2O4 + H2O-----> H3O + C2O4
HC2O4...........H3O....C2O4
I.......Y..............0 ............ 0
C....-z...............z.............z
E....y-z..............z.............z
But since it is a diprotic acid and x=H3O+ y= H3O+, the charts become
H2C2O4 + H2O----H3O+ HC2O4
H2C2O4...........H3O....HC2O4
I0.370M..............0 ........ ..0
C-x..............,,..x............ .x
E0.370-x.........x+Y..........x-y
HC2O4 + H2O-----> H3O + C2O4
HC2O4...........H3O...............C2O4
I...x....................x.....................0
C..-y..............x+y.....................y
E...x-y...........x+y.....................y
So, ka1=[x+y][x-y]/[0.370]
and
ka2=[x+y][y]/[x-y]
We solved for x, so solve for y.
I believe this is how you tackle this problem.
I don't have any real quarrel with Devron's solution but I think there is an easier way to do it. First I should point out the following:
1. Since k1 is so large in comparison with C, it requires the quadratic; i.e., x cannot be neglected as was done in the above solution. My solution below doesn't need to do that; in fact, we need not calculate it at all. Here is what I would do.
.......H2C2O4 ==> H^+ + HC2O4^-
I.......0.370......0.......0
C........-x.........x......x
E......0.370-x......x......x
BUT there is no need to solve the equation as was done above. We simply note that (H^+) = (HC2O4^-) and move on to k2.
k2
.......HC2O4^- ==> H^+ + C2O4^2-
I.....x(from above)..0.....0
k2 = (H^+)(C2O4^2-)/(HC2O4^-)
BUT from k1 we saw (H^+) = (HC2O4^-); therefore, H^+ in the numerator cancels with HC2O4^- in the denominator so that k2 = (C2O4^2-)
You can do this for several reasons. Note that we assumed the total H^+ = just H^+ from k1. We can do that because the H^+ supplied by k2 is 1000 times less than that supplied by k1. And we can assume HC2O4^- we used for k2 is the same as (H^+) because only 1/1000 of those HC2O4^- ions will ionize further.
Yes you can not ignore it, anything above 1 x 10^-2 exceeds the 5% rule.
So, If you are going to tackle this problem they way that I have, you will have to solve for x using the quadratic equation, and then you will have to solve for y.
I don't think you need to do that. No matter what number you obtain, whether it be by the simple route of ignoring x or by using the quadratic formula, the answer will ALWAYS be (H^+ = (HC2O4^-). And when you substitute those answers into k2, whatever they are, they will ALWAYS cancel and k2 = (C2O4^2-). What we're ignoring in this case is the contribution of the H+ from k2 and that is 1000 x less than H^+ from k1 which is so small it can be ignored.
Ok now I'm more lost and confuse with both ways if solving this problem...HELP!
Solve it the way that Dr. Bob222 recommended. He is Jiskha's expert in chemistry; I just come on here and play around with science questions from time to time.
I don't want to confuse you, so that is why I suggested using his method to solve the problem. If you still do not get it then you can try it the way that I went about doing it. If you still are lost repost the question and show what you did so far, so someone can try and help you on where you on the parts that you are lost on.
To find the concentration of C2O4^-2 in the oxalic acid solution, we need to determine the degree of ionization of the oxalic acid and apply the appropriate equilibrium expressions.
Oxalic acid (H2C2O4) is a diprotic acid, meaning it can donate two protons (H+) in separate ionization steps. The two ionization steps can be represented as follows:
H2C2O4 ⇌ H+ + HC2O4^-
HC2O4^- ⇌ H+ + C2O4^-2
Given that Ka1 = 5.6x10^-2 for the first ionization step and Ka2 = 5.1x10^-5 for the second ionization step, we can assume that the first ionization step is complete because Ka1 is relatively large compared to Ka2.
Therefore, we can conclude that [H+] ≈ [HC2O4^-] in the solution since the first ionization is complete.
If we define x as the concentration of C2O4^-2 that forms from the second ionization step, then [H+] = [C2O4^-2] = x.
We can express the equilibrium constant (Ka2) for the second ionization step as:
Ka2 = [H+][C2O4^-2] / [HC2O4^-]
Rearranging the equation, we have:
x^2 / (0.370 - x) = 5.1x10^-5
Simplifying the equation, we can assume that x is much smaller than 0.370, so we can approximate 0.370 - x ≈ 0.370.
x^2 / 0.370 = 5.1x10^-5
Multiplying both sides by 0.370, we get:
x^2 = (5.1x10^-5)(0.370)
Solving for x, we find:
x ≈ 0.000664
Therefore, the concentration of C2O4^-2 in the 0.370M oxalic acid solution is approximately 0.000664M.