Suppose a 0.034 M aqueous solution of sulfuric acid (H2SO4) is prepared. Calculate the equilibrium molarity of SO4 You'll find information on the properties of sulfuric acid in the ALEKS Data resource. Round your answer to 2 significant digits.

[SO4²⁻] = 0.015M

Explanation:

When H2SO4 is dissolved in water, HSO4- is produced in a direct reaction as follows:

H2SO4 → HSO4- + H+

As 1 mole of H2SO4 produce 1 mole of HSO4-, the molarity of HSO4- in this first reaction is 0.034M

Now, the HSO4- is in equilibrium with SO42- and H+ as follows:

HSO4⁻ ⇄ SO4²⁻ + H⁺

Where the equilibrium constant, K, is defined as:

K = 1.2x10⁻² = [SO4²⁻] [H⁺] / [HSO4⁻]

Where [] are the equilibrium concentrations of each species in the reaction.

The equilibrium concentrations are:

[SO4²⁻] = X

[H⁺] = X

[HSO4⁻] = 0.034M - X

Where X is reaction coordinate

Replacing:

1.2x10⁻² = [X] [X] / [0.034-X]

4.08x10⁻⁴ - 1.2x10⁻²X = X²

4.08x10⁻⁴ - 1.2x10⁻²X - X² = 0

Solving for X:

X = -0.027M. False solution, there are no negative concentrations.

X = 0.015M. Right solution.

That means the equilibrium molarity of SO4²⁻,

[SO4²⁻] = X

it is stable as it is in group 0 which have full outer shells of electrons