PLEASE HELP!!!! IMPORTANT!!!! Use the values of Ka1 = 0.056 and Ka2 = 5.4e-5 to calculate the [C2O4 2-] in ...

Question

Use the values of Ka1 = 0.056 and Ka2 = 5.4 x 10-5 to calculate the [C2O4 2-] in a 0.116 M solution of oxalic acid (H2C2O4). Use E notation and 3 significant figures.

Thanks!!!

Answer 1

Ka1 only has 2 sigfigs @ 0.056

H2 C2O4 --> H+ & HC2O4-

Ka1 = [H+] [HC2O4-] / [H2C2O4]

0.056 = [x] [x] / [0.116-x]

(0.056) [0.116-x] = x2

0.006496 - 0.056x = x2

X2  +  0.056x   - 0.006496 = 0

http://www.1728.com/quadratc.htm

x = [H+]  =    [HC2O4-] =  0.0573

=====================================

Ka2

HC2O4 -  -->  H+  &  (C2O4)-2

Ka2 = [H+] [ (C2O4)-2]  /  [HC2O4-]

5.4 e-5 = [0.0573] [ (C2O4)-2]  /  [0.0573]

5.4 e-5 =  [ (C2O4)-2]  

your answer is 5.4 e-5 molar

I know you were asked touse 3 sig figs, but C2O4)-2  equals your Ka2, and it only has 2