Question

Calculate the concentration of each species present in a 0.050-M solution of (H₂S).

a) (0.025 , M , H₂S, 0.025 , M , HS⁻, 0.025 , M , S²-)
b) (0.050 , M , H₂S, 0.025 , M , HS⁻, 0.025 , M , S²-)
c) (0.025 , M , H₂S, 0.050 , M , HS⁻, 0.025 , M , S²-)
d) (0.025 , M , H₂S, 0.025 , M , HS⁻, 0.050 , M , S²-)

Answer 1

The correct answer is (c) (0.025 M H₂S, 0.050 M HS⁻, 0.025 M S²⁻).

Step-by-step explanation:

In a solution of hydrogen sulfide (H₂S), the compound partially ionizes into hydrosulfide (HS⁻) and sulfide (S²⁻) ions. The initial concentration of H₂S is 0.050 M. At equilibrium, let x be the change in concentration due to ionization. The balanced chemical equation for the ionization of H₂S is:

H₂S ⇌ HS⁻ + S²⁻

The equilibrium concentrations will be 0.025 M (initial concentration - x) for H₂S, 0.050 M (initial concentration + x) for HS⁻, and 0.025 M (x) for S²⁻. Therefore, the correct distribution is (0.025 M H₂S, 0.050 M HS⁻, 0.025 M S²⁻), corresponding to option (c).

This indicates that at equilibrium, H₂S has partially ionized into HS⁻ and S²⁻, with higher concentrations of HS⁻ ions than the other species in the solution