The sulfide ion, S^2-, reacts with water as a weak base:
S^2- + H₂O SH^- + OH^-K = 1.0
If sodium sulfide were dissolved in water to make a solution of 0.50 M, what would be the resulting concentration of OH^-?

Question

The sulfide ion, S^2-, reacts with water as a weak base:
S^2- + H₂O The sulfide ion, S2-, reacts with water as a weak base: S2- + H2O SH- + OH- K = 1.0 If sodium sulfide were dissolved in water to make a solution of 0.50 M, what would be the resulting concentration of OH-? A) 0.50 M B) 0.37 M C) 0.63 M D) 1.0 M E) 0.19 M

The sulfide ion, S<sup>2-</sup>, reacts with water as a weak base: S<sup>2-</sup> + H<sub>2</sub>O SH<sup>-</sup> + OH<sup>-</sup><sup> </sup>K = 1.0 If sodium sulfide were dissolved in water to make a solution of 0.50 M, what would be the resulting concentration of OH<sup>-</sup>? A) 0.50 M B) 0.37 M C) 0.63 M D) 1.0 M E) 0.19 M

SH^- + OH^-K = 1.0
If sodium sulfide were dissolved in water to make a solution of 0.50 M, what would be the resulting concentration of OH^-?

Quizplus · Accepted Answer

When sodium sulfide is dissolved in water, it will dissociate into sulfide ions and sodium ions. However, since sulfide ions are a weak base, they will react with water to form hydroxide ions and hydrogen sulfide: S²F1F1F1S1^F1F1F10 F1F1F1S1F1F1F10 H₂O 11eaa952_9b5e_6ab6_ae90_f1a6386e916d_TB3835_11 SHF1F1F1S1^F1F1F10 F1F1F1S1F1F1F10 OHF1F1F1S1^F1F1F10K F1F1F1S1F1F1F10. The reaction shows that for every sulfide ion that reacts, one hydroxide ion is produced. Therefore, the concentration of hydroxide ion will be equal to the initial concentration of sulfide ion after the reaction with water. Since the initial concentration of sulfide ion is 0.50 M, the resulting concentration of hydroxide ion will also be 0.50 M.

The Sulfide Ion, S2-, Reacts with Water as a Weak

The Sulfide Ion, S^2-, Reacts with Water as a Weak