The sulfide ion, S^2- 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^-?

Question

The sulfide ion, S^2- reacts with water as a weak base:
S²^-+ 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-?

The sulfide ion, S<sup>2-</sup> reacts with water as a weak base: S<sup>2</sup><sup>-</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>?

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 Answer of The sulfide ion, S^2- 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^-?

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

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