Question

Oxalic acid (H2C2O4) is a diprotic acid that occurs in plants such as rhubarb and spinach.

Part A

Calculate the pH and the concentration of C2O2−4 ions in 0.27 M H2C2O4 (Ka1=5.9×10−2; Ka2=6.4×10−5).

Part B

Express your answer using two significant figure

[C2O2−4] =	________M

Answer 1

H2C2O4 ionizes in water as

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

preparing the ICE table

compoent                                     initial                              change                  equilibrium

H2C2O4                                      0.27                                   -x                            0.27-x

H+                                                 0                                       x                                  x

HC2O4-                                        0                                       x                                   x

Ka1= x²/(0.27-x)= 5.9*10^-2, when solved using excel, x=0.1 [H+] =0.1

hence HC2O4- = 0.1

This undergoes further ionization as HC2O4- ---------->H+ + C2O4-

Ka2= [H+] [C2O4-]/[HC2O4-]

preparing one more ICE Table

component                                 initial                        change                 equilibrium

HC2O4-                                   0.1 -x 0.1-x

H+    0 x    x

C2O4- 0    x    x

hence Ka2= x²/(0.1-x)= 6.4*10^-5, when solved using excel, x=0.0025, [H+] =0.0025, [C2O4-]=0.0025

hence total of H+= H+ produced during dissociation of H2C2O4+ H+ produced during dissociation of HC2O4-

= 0.1+0.0025= 0.1025, pH= -log (pH)= 0.989