Calculating [H+] and [A2-] at Equilibrium   in a Solution of a Weak Diprotic Acid H2A
Suppose that for the diprotic acid, H2A, Ka1 = 1.0 x 10-5 and Ka2 = 1.0 x 10-9.
 
What are the values of [H+] and [A2-] at equilibrium in 0.100 mol/L H2A?
 
                 H2A  <-------->  H+  +   HA-
Initial        0.1 M                   -----    -----
Changes     -x                         +x       +x
 
Corrections
caused by
second
equilibrium   0                          +y       -y    The second conjugate of H+ at the expense of HA-

Final         0.1 -x                    x+y      x-y    Since y is very very small x+yx and x-yx.
 

Ka1 = (x)(x) = 1.0 x 10-5
          0.1 - x
 
1.0 x 10-6 - 1.0 x 10-5x = x2

x2 + 1.0 x 10-5x -1.0 x 10-6 = 0 (use the quadratic equation on this)
 

x = 9.95 x 10-4 mol/L

Concentration of [A2-] at equilibrium

Ka2 = [H+][A2-] but [H+] = [HA-]
              [HA-]

Ka2 = [A2-] = A2-

[Ka2] = Ka2 = 1.0 x 10-9 mol/L
 

          Go to the Acid-Base Polyprotic Acid Worksheet
          Go to the Unit Review