For the diprotic weak acid H2A, Ka1=2.3x10^-6 and Ka2=6.2x10^-9. What is the pH of a 0.0650M solution of H2A. What are the equilibrium concentrations of H2A and A^2-?

Answer 1

When the two Ka values for a diprotic acid vary by 103 or greater, we usually do not consider the 2nd dissociation as adding significantly to the pH. That is the case here, so I will show you the calculation for the 1st dissociation, and if you wish to repeat it for the 2nd, you may do so.

H₂A ==> H⁺ + HA^- Ka₁ = 2.3x10^-5

HA^- ==> H⁺ + A^2- Ka₂ = 6.2x10^-9

2.3x10^-5 = (x)(x)/0.0650 - x (if we assume x is negligible relative to 0.0650, we can neglect it)

x² = 1.49510^-6

x = 1.22x10^-3 = [H⁺] and this is less than 2% of 0.065 so assumption above is valid and we avoid using the quadratic equation. If you wish to do so, go back and solve using the quadratic to obtain a more precise estimate of [H⁺]

pH = -log H+ = -log 1.22x10-3

pH = 2.91

If you wish to consider the contribution of the 2nd Ka, then set up an ICE table and solve for H⁺ and add it to the 1.22x10^-3 obtained above. You can also use the ICE table to answer the second part of the question as to the equilibrium concentration of A^2-.

HA^- ==> H⁺ + A^2- Ka₂ = 6.2x10^-9

0.00122...0.00122...0...Initial

-x..............+x.........+x...Change

.00122-x...0.00122+x..x..Equilibrium

Ka2 = 6.2x10^-9 = (1.22x10^-3 + x)(x) / 1.22x10^-3 - x)

Solve for x and add it to [H+] obtained above and take -log to get final pH

Equilibrium [A^2-] = x