Asked by Anonymous
what is the hydronium ion concentration of a 0.35M oxalic acid, H2C2O4. solution? for oxalic acid, ka1= 5.6x10^-2 and ka2=5.1x10^-5. Identify and calculate the concentrations for the representative species in solution.
Answers
Answered by
DrBob222
.....H2C2O4 ==> H^+ + HC2O4^-
I....0.35M......0......0
C.....-x........x......x
E....0.35-x.....x......x
Substitute the E line into k1 and solve for x = (H^+) = (HC2O4^-)
.......HC2O4^- ==> H^+ + C2O4^=
Write the k2 expression. You've just calculated that H^+ = HC2O4^- so cancel H^+ in the numerator with HC2O4^- in the denominator which leave k2 = (C2O4^=).
I....0.35M......0......0
C.....-x........x......x
E....0.35-x.....x......x
Substitute the E line into k1 and solve for x = (H^+) = (HC2O4^-)
.......HC2O4^- ==> H^+ + C2O4^=
Write the k2 expression. You've just calculated that H^+ = HC2O4^- so cancel H^+ in the numerator with HC2O4^- in the denominator which leave k2 = (C2O4^=).
Answered by
Anonymous
I got [H+]= 0.14M and for [HC2O4]=0.14M and [C2O4^=] 5.1x10^-5
Is this correct?
Is this correct?
Answered by
DrBob222
You didn't list k1 and k2 and the numbers in my text probably don't agree with the numbers in you text.
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