The sulfide ion, S²^-, reacts with water as a weak base:
S²^- + H₂O $\leftrightarrows$ 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, S²^-, reacts with water as a weak base:
S²^- + H₂O

\leftrightarrows

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

The equilibrium constant K = 1.0 suggests that the reaction is balanced, so the concentration of OH⁻ will be approximately equal to the concentration of S²⁻ that reacts. Since the initial concentration of S²⁻ is 0.50 M, and the reaction is not complete, the concentration of OH⁻ will be slightly less than 0.50 M, approximately 0.37 M.

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

The Sulfide Ion, S²^-, Reacts with Water as a Weak $\leftrightarrows$