Suppose a 0.042M aqueous solution of phosphoric acid (H3PO4) is prepared. Calculate the equilibrium molarity of HPO4^−2.

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 =

hey there!

boiling point is the temperature at which the vapor pressure of the liquid equals atmospheric pressure. the vapor pressure of solvent molecules is decreased when a solute is added, so a higher temperature is required to increase the number of solvent molecules in the gas phase above the liquid.

at the freezing point, the vapor pressures of the solid and liquid are equal; a lower temperature is needed to reduce the number of solvent particles above the liquid.

hope this you

tobey

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