Calculate the equilibrium concentration of sulfurous acid in a solution labeled "0.100 M H₂SO₃(aq)" if pK_a1 = 1.81, and pK_a2 = 6.91. A) 0.068 M B) 0.015 M C) 0.100 M D) 0.050 M E) 0.032 M

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The correct answer is A because the equilibrium concentration of sulfurous acid in a solution labeled "0.100 M H₂SO₃(aq)" can be calculated using the following equation:$$H_2SO_3(aq) + H_2O(l) ightleftharpoons HSO_3^-(aq) + H_3O^+(aq)$$$$K_{a1} = \frac{[HSO_3^-][H_3O^+]}{[H_2SO_3]}$$$$pK_{a1} = -\log K_{a1} = 1.81$$$$K_{a1} = 10^{-pK_{a1}} = 10^{-1.81} = 0.0154$$$$HSO_3^-(aq) + H_2O(l) ightleftharpoons SO_3^{2-}(aq) + H_3O^+(aq)$$$$K_{a2} = \frac{[SO_3^{2-}][H_3O^+]}{[HSO_3^-]}$$$$pK_{a2} = -\log K_{a2} = 6.91$$$$K_{a2} = 10^{-pK_{a2}} = 10^{-6.91} = 1.22 imes 10^{-7}$$The equilibrium concentrations of the species in the solution can be calculated using the following ICE table:$$\begin{array}{ccc} ext{Reaction}&H_2SO_3 & HSO_3^- & SO_3^{2-} & H_3O^+ \\ ext{Initial}&0.100 & 0 & 0 & 0 \\ ext{Change}& -x & +x & +x & +x \\ ext{Equilibrium}& 0.100-x & x & x & x \end{array}$$Substituting the equilibrium concentrations into the equilibrium constant expressions, we get:$$K_{a1} = \frac{x^2}{0.100-x} = 0.0154$$$$K_{a2} = \frac{x^2}{x} = 1.22 imes 10^{-7}$$Solving for x, we get:$$x = 0.068 M$$Therefore, the equilibrium concentration of sulfurous acid in the solution is 0.100 - x = 0.032 M.

Calculate the Equilibrium Concentration of Sulfurous Acid in a Solution