<span>Answer: 0.00649M
The question is incomplete,
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<span>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
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With that you can solve the question following these steps"
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<span>1) First ionization:
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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
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<span>HSO₄⁻ (aq) ⇄ H⁺ + SO₄²⁻ with a Ka₂ = 0.012
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<span>Do the mass balance:
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<span><span> HSO₄⁻ (aq) H⁺ SO₄²⁻</span>
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<span /><span /><span> 0.01 M - x x x
</span><span>Ka₂ = [H⁺] [SO₄²⁻] / [HSO₄⁻]</span>
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=> Ka₂ = (x²) / (0.01 - x) = 0.012
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<span>3) Solve the equation:
</span><span>x² = 0.012(0.01 - x) = 0.00012 - 0.012x</span>
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x² + 0.012x - 0.0012 = 0
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<span>Using the quadratic formula: x = 0.00649
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<span>So, the requested concentratioN is [SO₄²⁻] = 0.00649M</span>
By Tri chlorination, three chlorine atoms can replace the 3 hydrogen atoms in propane molecule to give tri-chlorinated products according to the attached picture.
Hope this helps! If you dont understand balancing equations in general, say so in the comments, I’m happy to help
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