The equilibrium molarity HPO²₄- = 6.2 x 10⁻⁸M See the explanation below.
What is Equilibrium Molarity?
The molar concentration of a certain species in a solution is known as equilibrium molarity.
What is the calculation showing the above answer?
The initial concentration H₃PO₄ = 0.31 M
Kₐ₁ = 6.92 x 10⁻³ (See Image 1 attached)
Kₐ₁ = [H⁺][tex]_{eq}[/tex][H₂PO₄-][tex]_{eq}[/tex] / [H₃PO₄][tex]_{eq}[/tex]
6.92⁻³ x 10 = [(x) * (x)] / (0.31 M - x)
Solving for x, x = 0.043 M
[H+] = x = 0.043 M
[H₂PO₄-] = x = 0.043 M 0.0429805
Kₐ₂ = 6.2 x 10⁻⁸
See the second attached table
Kₐ₂ = [H+][tex]_{eq}[/tex] [HPO₄²-]eq / [H₂PO₄-]eq
6.2 x 10⁻⁸ = [(0.043 M + x) * (x)] / (0.043 M - x)
Solving for x, x = 6.2 x 10⁻⁸ M
[H₂PO₄-]eq = x = 6.2 x 10⁻⁸ M
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