Calculate the pH of the following aqueous solution: 0.17 M H₂S (pK_a1 = 7.00; pK_a2 = 12.89) A) 10.12 B) 6.23 C) 3.88 D) 7.77 E) 7.00

Question

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First, we need to write out the dissociation reactions of H₂S: H₂S + H2O ⇌ H3O+ + ₂S- ₂S- + H2O ⇌ OH- + H₂S2- Using the given pKas, we can calculate the equilibrium constants for these reactions: Ka1 = 10^-pKa1 = 10^-7.00 = 1.0 x 10^-7 Ka2 = 10^-pKa2 = 10^-12.89 = 1.3 x 10^-13 Now, we need to calculate the concentrations of the species in the solution. Let x be the concentration of H₂S that dissociates, then: [H3O+] = [H₂S-] = x [₂S2-] = 0.17 - x Using the equilibrium expressions and the Ka values, we can set up the following calculations: Ka1 = [H3O+][₂S-]/[H₂S] 1.0 x 10^-7 = x^2 / (0.17 - x) Solving for x, we get: x = 8.4 x 10^-4 Ka2 = [OH-][H₂S2-]/[₂S-] 1.3 x 10^-13 = (x)(0.17 - x) / (8.4 x 10^-4) Solving for x, we get: x = 3.88 x 10^-5 Now, we can calculate the pH of the solution: pH = -log[H3O+] pH = -log(8.4 x 10^-4) pH = 3.88 Therefore, the answer is C, 3.88.

Calculate the PH of the Following Aqueous Solution: 0