Final answer:
The pH of a 0.99 M H₂S solution can be estimated by considering only the first dissociation due to the significantly higher pKa1 compared to pKa2.
Step-by-step explanation:
To calculate the pH of a 0.99 M H₂S solution, we primarily consider the first dissociation, as the second dissociation has a much weaker pKa value, making its contribution to the H+ ion concentration relatively small. Using the pKa1 value of 7.00, we can assume that the hydrogen sulfide partially dissociates to form HS- and H+ in water. Since it's a weak acid, we can set up an ICE table (Initial, Change, Equilibrium) to determine the equilibrium concentration of H+ ions.
The initial concentration of H₂S is 0.99 M, and the initial concentrations of H+ and HS- are assumed to be 0 M. After dissociation, the concentration of H+ will be the same as that of HS- due to the 1:1 stoichiometry of the reaction H₂S → HS- + H+. If x is the change in concentration for the dissociation, the equilibrium concentrations would be (0.99 - x) for H₂S and x for both HS- and H+.
Since the pKa1 is 7.00 (Ka = 10-7), the assumption in weak acid calculations is that x (the dissociation of H+ ions) is relatively small compared to 0.99 M; hence we can neglect x in the denominator when setting up the equation for Ka:
Ka = [H+][HS-]/[H₂S] ≈ x²/(0.99 - x) ≈ x²/0.99
Solving for x, we ignore the x in the denominator, which yields x² = Ka * 0.99 = 10-7 * 0.99. Solving for x gives us approximately 10-4. The pH is then -log[H+] = -log(10-4) = 4.0. However, this answer is not listed among the options. Given the options provided, since none of them are correct, the correct answer should be (e) none of these.