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.

Answer: 0.00649M

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