In a 0.10 M solution of sodium hydrogen sulfate,the HSO₄^- ion is 29% dissociated.What is Ka₂ for H₂SO₄?

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The dissociation of HSO₄^- can be represented by the equation: HSO₄^- ⇌ H+ + SO₄^- The equilibrium constant expression for this dissociation can be written as: Ka₂ = [H+][SO₄^-] / [HSO₄^-] We know that the concentration of HSO₄^- at equilibrium is 0.10 M multiplied by (1 - 0.29) because only 29% of the ion has dissociated. This concentration can be calculated as: [HSO₄^-] = 0.10 M × (1 - 0.29) = 0.071 M We also know that the concentration of H+ and SO₄^- are equal at equilibrium because the dissociation produces one of each ion. Let this concentration be x. Then: [H+] = x M [SO₄^-] = x M Substituting these expressions and the concentration of HSO₄^- into the equilibrium constant expression, we get: Ka₂ = (x M) × (x M) / (0.071 M) Ka₂ = x^2 / 0.071 We also know that x represents the fraction of HSO₄^- that has dissociated, which is 0.29. Therefore: x = 0.29 × 0.10 M = 0.029 M Substituting this into the expression for Ka₂, we get: Ka₂ = (0.029 M)^2 / 0.071 M Ka₂ = 1.2 × 10^-2 Therefore, the answer is B.

In a 010 M Solution of Sodium Hydrogen Sulfate,the HSO4- Ion Is

In a 010 M Solution of Sodium Hydrogen Sulfate,the HSO₄^- Ion Is