The leaves of the rhubarb plant contain high concentrations of diprotic oxalic acid (hooccooh) and must be removed before the stems are used to make rhubarb pie. if pka1 = 1.23 and pka2 = 4.19, what is the ph of a 0.0321 m solution of oxalic acid?

Making the ICE table for the dissociation of oxalic acid:

H₂C₂O₄ -->  H⁺ + HC₂O₄⁻

0.0321           0        0  

 -x                 + x     +x  

0.0321-x         +x     +x  

Kₐ = x²/ 0.0321 –x  

pKₐ = - logKₐ

Kₐ = e⁻pKₐ = e⁻1.23 = 0.3

0.3 = x²/ 0.0321 – x

0.3 (0.0321 – x) = x²

-x² -0.3 x + 9.63 x 10⁻³

Using the quadratic formula:

a = -1, b = - 0.3, c = 9.63 x 10⁻³  

x = - 0.33, 0.03

Using Henderson-Hasselbalch equation

pH = pKₐ + log [conjugate base]/ [acid]

pH = pKₐ₁ + log [HC₂O₄⁻]/ [H₂C₂O₄]

pH = 1.23 + log (0.03/0.0321) = 1.23 - 0.03 = 1.20

Thus the pH of a 0.0321 M solution of oxalic acid is 1.20

answer; i believe that the correct answer is (a) potassium(k)

atomic number of carbon is 6

electronic configuration of carbon is 1s²2s²2p²

orbital diagram of c is as shown here in the image.