Answer:
To find the concentration of HPO3 2- in a 0.500 molar solution of H3PO3, we can set up an ICE table for the dissociation of H3PO3 into H2PO3- and HPO3 2-:
H3PO3 ⇌ H2PO3- + HPO3 2-
Given that Ka1 = 1.010^-2 and Ka2 = 2.6x10^-7, we can assume the initial concentration of H3PO3 is 0.500 M and the initial concentrations of H2PO3- and HPO3 2- are both 0.
Let x be the concentration of HPO3 2- formed. Therefore, the concentrations at equilibrium will be:
[H3PO3] = 0.500 - x
[H2PO3-] = x
[HPO3 2-] = x
Using the equilibrium constant expression for the first dissociation:
Ka1 = [H2PO3-][HPO3 2-] / [H3PO3]
1.010^-2 = x * x / (0.500 - x)
Solving for x, we get:
x = 1.006x10^-3 M
Therefore, the concentration of HPO3 2- in a 0.500 M solution of H3PO3 is 1.006x10^-3 M.
Explanation:
