Question

A solution is prepared by adding 0.10 mol of sodium sulfide, Na2S , to 1.00 L of water. Which statement about the solution is correct? a. The solution is basic. b. The solution is neutral, c. The solution is acidic. d. The concentration of sodium ions and sulfide ions will be identical. e. The concentration of sulfide ions will be greater than the concentration of sodium ions.

Answer 1

The solution prepared by adding sodium sulfide to water will be basic.

Step-by-step explanation:

The solution prepared by adding 0.10 mol of sodium sulfide, Na2S, to 1.00 L of water will be basic.

Sodium sulfide, Na2S, dissociates in water to form sodium ions (Na+) and sulfide ions (S2-). The concentration of Na+ ions will be equal to the concentration of S2- ions.

Since sodium ions are cations and sulfide ions are anions, their presence in the solution will increase the concentration of hydroxide ions (OH-) and make the solution basic.

Answer 2

A solution prepared by dissolving sodium sulfide in water is basic due to the formation of hydroxide ions. The concentration of sodium ions will be twice the concentration of sulfide ions because each Na2S unit dissociates into two Na+ ions and one S2- ion.

Step-by-step explanation:

When sodium sulfide (Na2S) is added to water, it dissociates into two sodium ions (Na+) and one sulfide ion (S2-). The sulfide ion is a strong base because, when in water, it undergoes hydrolysis to produce hydroxide ions (OH-). The reaction is S2- (aq) + H2O (l) → HS- (aq) + OH- (aq). This increases the hydroxide ion concentration in the solution, making it basic.

The concentration of sodium ions will be twice that of sulfide ions because each formula unit of Na2S produces two Na+ ions for every one S2- ion that is formed. Therefore, statement a is correct, the solution is basic, and statement d is correct, the concentration of sodium ions will be twice the concentration of sulfide ions, making statement e incorrect.