Calculate the concentration of so42− ions in a 0.010 m aqueous solution of sulfuric acid. express your answer to four decimal places and include the appropriate units.

### Answers

The question is incomplete,

You are told that the first ionization of the sulfuric acid is complete and the second ionization of the sulfuric acid has a constant Ka₂ = 0.012

With that you can solve the question following these steps"

1) First ionization:

H₂SO₄(aq) --> H⁺ (aq) + HSO₄⁻ (aq)

Under the fully ionization assumption the concentration of HSO4- is the same of the acid = 0.01 M

2) Second ionization

HSO₄⁻ (aq) ⇄ H⁺ + SO₄²⁻ with a Ka₂ = 0.012

Do the mass balance:

HSO₄⁻ (aq) H⁺ SO₄²⁻

0.01 M - x x x

Ka₂ = [H⁺] [SO₄²⁻] / [HSO₄⁻]

=> Ka₂ = (x²) / (0.01 - x) = 0.012

3) Solve the equation:

x² = 0.012(0.01 - x) = 0.00012 - 0.012x

x² + 0.012x - 0.0012 = 0

Using the quadratic formula: x = 0.00649

So, the requested concentratioN is [SO₄²⁻] = 0.00649M

The concentration of sulfate ions is 0.0100 M.

Explanation:

Concentration of sulfuric acid =0.0100 M

1 mol of sulfuric acid gives 2 mole of ion and 1mole of ion.

The concentration of sulfate ions is 0.0100 M.

The concentration of sulfate ions in the solution is 0.010 M

Explanation:

The chemical formula of sulfuric acid solution is

We are given:

Concentration of sulfuric acid = 0.010 M

The chemical equation for the ionization of sulfuric acid follows:

1 mole of sulfuric acid produces 2 moles of hydrogen ions and 1 mole of sulfate ions

Concentration of sulfate ions = (1 × 0.010) = 0.010 M

Hence, the concentration of sulfate ions in the solution is 0.010 M