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

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.

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

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

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