Calculate the concentration of oxalate ion (C2O42−) in a 0.175 M solution of oxalic acid (C2H2O4).
[For oxalic acid, Ka1 = 6.5 × 10−2, Ka2 = 6.1 × 10-5.]
To calculate the concentration of oxalate ion (C2O42-) in a 0.175 M solution of oxalic acid (C2H2O4), we need to consider the dissociation of oxalic acid and use the given values of Ka1 and Ka2.
Oxalic acid (C2H2O4) is a diprotic acid, which means it can dissociate in two steps. The dissociation equations for oxalic acid are as follows:
C2H2O4 ⇌ H+ + HC2O4- (step 1)
HC2O4- ⇌ H+ + C2O42- (step 2)
Let's assign the initial concentration of oxalic acid as [HA], and the concentration of oxalate ion as [A2-].
Step 1: The equilibrium constant for the first dissociation (Ka1) is given as 6.5 × 10−2. This means that:
Ka1 = [H+][HC2O4-] / [HA]
Since the concentration of oxalic acid is given as 0.175 M, we can assume that the concentration of HC2O4- is negligible compared to [HA] due to its weak acidity. Therefore, we can write:
Ka1 = [H+][HC2O4-] / [HA] ≈ [H+][HC2O4-] / (0.175 M)
Now, let's consider step 2:
Step 2: The equilibrium constant for the second dissociation (Ka2) is given as 6.1 × 10-5. This means that:
Ka2 = [H+][C2O42-] / [HC2O4-]
Since we just found that [HC2O4-] ≈ [H+], we can substitute this value in the equation:
Ka2 ≈ [H+][C2O42-] / [H+]
Ka2 ≈ [C2O42-]
Therefore, we can approximate the concentration of oxalate ion as the value of Ka2, which is 6.1 × 10-5.
So, the concentration of oxalate ion (C2O42-) in a 0.175 M solution of oxalic acid (C2H2O4) is approximately 6.1 × 10-5 M.
To calculate the concentration of oxalate ion (C2O42−) in a 0.175 M solution of oxalic acid (C2H2O4), we need to consider the dissociation of oxalic acid.
Oxalic acid (C2H2O4) is a diprotic acid, which means it can donate two protons (H+) in solution. The dissociation reactions can be represented as follows:
(1) C2H2O4 ⇌ H+ + HC2O4−
(2) HC2O4− ⇌ H+ + C2O42−
The dissociation constant (Ka) for each reaction can be determined from the given values:
Ka1 = 6.5 × 10−2
Ka2 = 6.1 × 10-5
Let's assume x is the concentration of oxalate ions produced by the dissociation of oxalic acid. Since each oxalate ion is formed from the dissociation of one oxalic acid molecule, the concentration of oxalate ion is equal to x.
Using the first dissociation reaction, we can write the expression for Ka1:
Ka1 = [H+][HC2O4−] / [C2H2O4]
Since oxalic acid is diprotic, initially, the concentration of oxalic acid (C2H2O4) is 0.175 M.
Using the second dissociation reaction, we can write the expression for Ka2:
Ka2 = [H+][C2O42−] / [HC2O4−]
Since oxalic acid is a weak acid, we can assume that the concentrations of H+ and HC2O4− are negligible compared to the initial concentration of 0.175 M oxalic acid. Therefore, we can approximate the concentration of HC2O4− as 0.175 M.
Using the information above, we can make the following approximations:
[H+] ≈ 0
[HC2O4−] ≈ 0.175 M
Substituting these values into the expression for Ka2, we get:
Ka2 = [H+][C2O42−] / 0.175
Simplifying the equation, we can solve for [C2O42−]:
[C2O42−] = (Ka2 * 0.175) / [H+]
Since the concentration of H+ is negligible, we can assume it to be approximately 0. Therefore, the concentration of oxalate ion (C2O42−) in the 0.175 M solution of oxalic acid is approximately equal to Ka2 * 0.175:
[C2O42−] ≈ (6.1 × 10-5) * 0.175
Calculating the result, we find:
[C2O42−] ≈ 1.07 × 10-5 M
Therefore, the concentration of oxalate ion in the 0.175 M solution of oxalic acid is approximately 1.07 × 10-5 M.