Question

Determine the pH of 0.055 M oxalic acid (H₂C₂O₄; K_a1=5.6x10^-2 and K_a2=5.4x10^-5). Note: both steps are important here, therefore we have to set up two "ICE" boxes. Show work.

Answer 1

      H2C2O4 -----------> H^+ + HC2O4^-

I     0.055                    0             0

C    -x                        +x             +x

E    0.055-x                 +x              +x

   ka1 = [H^+][HC2O4^-]/[H2C2O4]

5.6*10^-2 = x*x/0.055-x

5.6*10^-2(0.055-x) = x^2

      x = 0.034

[H+]   =x = 0.034M

[HC2O4^-] = x= 0.034M

[H2C2O4] = 0.055-x = 0.055-0.034 = 0.021M

PH = -log[H+]

        = -log0.034

         = 1.4685 >>>>>answer