Question

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 1

Answer:

The concentration of sulfate ions is 0.0100 M.

Explanation:

Concentration of sulfuric acid =0.0100 M

$H_2SO_4\rightarrow 2H^++SO_4^{2-}$

1 mol of sulfuric acid gives 2 mole of $H^+$ ion and 1mole of $SO_4^{2-}$ion.

$[SO_4^{2-}]=1\times 0.010 M=0.0100 M$

$[H^+]=2\times 0.010 M=0.0200 m$

The concentration of sulfate ions is 0.0100 M.

Answer 2

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