Use the values of Ka1 = 0.056 and Ka2 = 5.4 x 10-5 to calculate the [HC2O4-] in
ID: 746398 • Letter: U
Question
Use the values of Ka1 = 0.056 and Ka2 = 5.4 x 10-5 to calculate the [HC2O4-] in a 0.509 M solution of oxalic acid (H2C2O4)Explanation / Answer
H2C2O4 ===> HC2O4- + H+ Initially you have 0.509 M of H2C2O4 At equilibrium you have [H2C2O4] = .509 - x [HC2O4] = [H+] = +x Ka1 = x^2/(.509-x) = .056 Solve for x = 0.1431 M [HC2O4] = [H+] = 0.1431 M Now you have HC2O4 - ====> C2O42- + H+ You have at equilibrium [HC204-] = .1431 - x [C2O42-] = +x [H+] = .1431 + x Ka2 = (.1431+x)*x/(.1431-x) = 5.4 x 10^-5 Solve for x = 5.4 x 10^-5 So [HC2O4-] = .1431 - x = .1431 - 5.4 x 10^-5 = 0.143 M Hope this helped
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