Question

1) Phosphorous acid, H₃PO₃, is actually a diprotic acid for which Ka₁= 5.0 × 10^-2 and Ka₂= 2.0 × 10^-7. What are the values of [H⁺], [H₂PO₃^-], and [HPO₃^2-] in a 6.0 M solution of H₃PO₃? What is the pH of the solution?

2) What is the pH of a 1.6 M solution of Na₃PO₄? In this solution, what are the concentrations of HPO₄^2-, H₂PO₄^-, and H₃PO₄?

3)Ascorbic acid (vitamin C) is a diprotic acid, H₂C₆H₆O₆. Calculate [H⁺], pH, and [C₆H₆O₆^2-] in a 0.14 M solution of ascorbic acid.

Answer 1

1)

The concentration of the hydrogen ion in a solution of phosphorous acid is mostly from the first dissociation.
From the second dissociation of phosphorous acid the H+ concentration is negligible.

The first dissociation reaction of phosphorous acid is:

               H3PO3 <==> H+ + H2PO3-
initial           6.0        0       0
change           -x            +x    +x
Equilibrium       6-x        x        x

Ka = 1.0x10^-2

Ka = [H+] [H2PO3-] / [H3PO3]
5.0x10^-2 = x² / (6.0-x) M

x² - 0.3 + 0.05 x = 0

By soloving the quadratic equation:
we get;
x = 0.523

Therefore; [H+] = 0.523 M and [H2PO3-] = 0.523 M

We can calculate the [HPO3-2] from the second dissociation reaction

                   H2PO3- <==> H+     +   HPO32-
Initial           0.523 M           0.523 M     0  
Change               -x            +x         +x
Equilibrium       0.523 M -x        0.523 M +x    x  
  
Ka2 = 2.0x10^-7

Ka2 = [H+] [HPO32-] / [H2PO3-]
2.0x10^-7 = (0.523 M +x)x / (0.523 M -x) (since, x is small,0.523 M +x ~ 0.523 and 0.523 M -x ~ 0.523 M)

2.0x10^-7 = 0.523x/0.523
x = 2.0x10^-7 M

Therefore, [HPO32-] = 2.0x10^-7 M

[H+] = 0.523 M , [H2PO3-] = 0.523 M and [HPO32-] = 2.0x10^-7 M