What is the hydroxide ion concentration of a 0.370M oxalic acid, H2C2O4, solution? For oxalic acid, Ka1=5.6x10^-2 and Ka2=5.1x10^-5
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To find the hydroxide ion concentration of a 0.370M oxalic acid solution, we first need to determine the concentration of the oxalic acid dissociated into hydrogen ions and the corresponding hydroxyl ions.
Oxalic acid (H2C2O4) has two dissociation steps, as represented by the two given Ka values: Ka1 = 5.6x10^-2 and Ka2 = 5.1x10^-5.
Step 1: Dissociation of the first hydrogen ion (H+):
H2C2O4 (aq) ⇌ H+ (aq) + HC2O4^- (aq)
Let x be the concentration of H+ ions formed. Thus, the concentration of HC2O4^- will also be x, assuming the concentration of original H2C2O4 decreases by a negligible amount compared to its initial value. The Ka1 expression is given by:
Ka1 = [H+] [HC2O4^-] / [H2C2O4]
Given the initial concentration of oxalic acid (H2C2O4) as 0.370M, we have:
5.6x10^-2 = x * x / (0.370 - x)
Since the value of x is expected to be relatively small compared to 0.370, we can assume that (0.370 - x) ≈ 0.370. Thus, we can simplify the equation to:
5.6x10^-2 = x^2 / 0.370
Cross-multiplying and rearranging the equation, we have:
0.370 * (5.6x10^-2) = x^2
0.02072 = x^2
Taking the square root of both sides, we get:
x ≈ 0.144 M
Therefore, the concentration of H+ ions formed from the dissociation of the first hydrogen ion is approximately 0.144M.
Step 2: Dissociation of the second hydrogen ion (H+):
HC2O4^- (aq) ⇌ H+ (aq) + C2O4^2- (aq)
The second dissociation occurs to a much lesser extent compared to the first dissociation. Therefore, we can assume the concentration of H+ formed is negligible compared to the concentration of HC2O4^-.
Since we already found that [H+] ≈ 0.144M, we can treat it as a value we already know. Additionally, the concentration of C2O4^2- ions formed will also be approximately equal to [H+] because they are in a 1:1 ratio with H+.
Thus, we can say that [OH^-] ≈ [H+] ≈ 0.144M.
Therefore, the hydroxide ion concentration of the 0.370M oxalic acid solution is approximately 0.144M.