Question

Polyprotic acids contain more than one dissociable proton. Each dissociation step has its own acid-dissociation constant, Ka1, Ka2, etc. For example, a diprotic acid H2A reacts as follows: H2A(aq)+H2O(l)⇌H3O+(aq)+HA−(aq) Ka1=[H3O+][HA−][H2A] HA−(aq)+H2O(l)⇌H3O+(aq)+A2−(aq) Ka2=[H3O+][A2−][HA−] In general, Ka2 = [A2−] for a solution of a weak diprotic acid because [H3O+]≈[HA−].

Many household cleaning products contain oxalic acid, H2C2O4, a diprotic acid with the following dissociation constants: Ka1=5.9×10−2, Ka2=6.4×10−5. Part A Calculate the equilibrium concentration of H3O+ in a 0.20 M solution of oxalic acid. Express your answer to two significant figures and include the appropriate units.

Answer 1

For the first dissociation

H₂C₂O₄       H⁺ + HC₂O₄^- ; K_a1 = 0.05

IC 0.2 0 0

EC 0.2-x x x

K_a1 = [H⁺] [ HC₂O₄^-] / [H₂C₂O₄]

  0.059 = x² / (0.2-x)
   x² = 0.059( 0.2 - x)
​   x² = 0.0118 - 0.059x
​   x² + 0.059x - 0.0118 = 0
Solving the quadratic for, we get

x = 0.083 and x = -0.142

Neglecting the negative value of x, we get;

x = 0.083 = [H⁺] = [ HC₂O₄^-]

For the second dissociation

HC₂O₄^- + H₂O        H₃O⁺ + C₂O₄^2-   ; K_a2 = 6.4 x 10^-5

IC 0.083 0.083 0

EC 0.083-x   0.083+x x

K_a2 = [H₃O⁺] [ C₂O₄^2-] / [HC₂O₄^-]

6.4 x 10^-5 = ( 0.083+x)(x) / 0.083-x

6.4 x 10^-5  = ( 0.083+x)(x) / 0.083 (neglecting the x in denominator since k_a2 is too small)

6.4 x 10^-5  = ( 0.083x + x²) / 0.083

​  x² + 0.083x - (5.4 x10^-6)

Solving the quadratic equation, we get

x = 0.000065 = 6.5 x 10^-5

Hence, [H₃O⁺]  =0.083 + x = 0.083 + 0.000065 = 0.083065 = 8.3 x 10^-2 M