Answer:
2.89x10⁻⁵M = [HPO₄²⁻]
Explanation:
The equilibrium of H3PO4 in water occurs H2PO4-:
H3PO4(aq) + H2O(l) ⇄ H3O⁺(aq) + H2PO4⁻(aq)
pKa = 2.16. And as pKa = -log Ka; Ka = 10^-2.16
Ka = 6.9183x10⁻³ = [H3O⁺] [H2PO4⁻] / [H3PO4]
As both [H3O⁺] and [H2PO4⁻] comes from the same equilibrium,
[H3O⁺]=[H2PO4⁻] :
[H3O⁺] = X
[H2PO4⁻] = X
[H3PO4] = 0.042 - X
Where X is reaction coordinate
Replacing:
6.9183x10⁻³ = [X] [X] / [0.042 - X]
6.9183x10⁻³ = X² / 0.042 - X
2.905686x10⁻⁴ - 6.9183x10⁻³X - X² = 0
Solving for X:
X = -0.02M. False solution. There is no negative concentration.
X = 0.014M. Right solution
[H2PO4⁻] = 0.014M
In the second equilibrium:
H2PO4⁻(aq) + H2O(l) ⇄ HPO4-(aq) + H3O+(aq)
Based on the same principles of the last equilibrium:
pKa2 = 7.21
Ka2 = 6.166x10⁻⁸ = [HPO4-] [H3O+] / [H2PO4⁻]
[HPO4-] = X
[H3O+] = X
[H2PO4⁻] = 0.014M - X
6.166x10⁻⁸ = X² / [0.014M - X]
8.3623x10⁻¹⁰ - 6.166x10⁻⁸X - X² = 0
Solving for X:
X = -0.0000289485. False solution.
X =
