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

c is having 4 valence electrons and is bonded to three hydrogens and one sulphur atom, therefore, the number of bond pair is equal to four. as all the valence electrons are used in forming the bonds so, there is no lone pair left on carbon.

s is having 6 valence electrons and is bonded to one other sulfur atom and one carbon atom, therefore, the number of bond pair is equal to 2. as only two valence electrons are used in forming the bonds so, four electrons are left. therefore, two lone pairs are present on s.