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^-?

Question

The sulfide ion, S²^-, 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-? 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><sup>-</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>? 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

The reaction between sulfide ion and water can be written as:

SS1U1P12S1S1P0S1U1P1-S1S1P0 + HS1U1B12S1U1B0O → SHS1U1P1-S1S1P0 + OHS1U1P1-S1S1P0S1U1P1

Let x be the concentration of OHS1U1P1-S1S1P0 produced. Then initially the concentration of SS1U1P12S1S1P0S1U1P1-S1S1P0 is 0.50 M. Because of the 1:1 stoichiometry between SS1U1P12S1S1P0S1U1P1-S1S1P0 and OHS1U1P1-S1S1P0, the concentration of the sulfide ion will decrease by x M. And because water is present in excess, the concentration of the HS1U1B12S1U1B0O does not change significantly, so its concentration can be considered to be 0.50 M throughout the reaction.

Using the equilibrium constant expression:

K = [SHS1U1P1-S1S1P0][OHS1U1P1-S1S1P0S1U1P1] / [SS1U1P12S1S1P0S1U1P1-S1S1P0][HS1U1B12S1U1B0O]

K = 1.0 (given)

Substituting the concentrations and solving for x:

1.0 = (x)(0.50) / (0.50 - x)

0.50 - x = x/1.0

1.5x = 0.50

x = 0.33 M

Therefore, the resulting concentration of OHS1U1P1-S1S1P0 is 0.33 M. The closest answer choice is B) 0.37 M, which is the best answer choice.

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

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