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 =

in the analysis of systems that involve fluid flow, we frequently encounter the combination of properties u and pv.

1. a)c-14 has two more protons

2. d)they have different numbers of protons, neutrons, and valence electrons; they are all gases at room temperature.

3. c) iron and nickel.