Question

Write a net ionic equation for the overall reaction that occurs when aqueous solutions of sodium hydroxide and hydrosulfuric acid are combined. Assume excess base.

Answer 1

The net ionic equation for the reaction between sodium hydroxide and hydrosulfuric acid with excess base is 2 OH−(aq) + H2S(aq) → S2−(aq) + 2 H2O(l).

Step-by-step explanation:

To write a net ionic equation for the overall reaction that occurs when aqueous solutions of sodium hydroxide and hydrosulfuric acid are combined, we first need to consider the full balanced chemical equation. Hydrosulfuric acid is better known as hydrogen sulfide, H2S. The balanced chemical equation, assuming excess base (sodium hydroxide, NaOH), would be:

2 NaOH(aq) + H2S(aq) → Na2S(aq) + 2 H2O(l)

To derive the net ionic equation, we exclude the spectator ions, which in this case are the sodium ions (Na+). The net ionic equation for the reaction then becomes:

2 OH−(aq) + H2S(aq) → S2−(aq) + 2 H2O(l)

This equation represents the reaction of the hydroxide ions from the sodium hydroxide with the hydrosulfuric acid to produce sulfide ions and water.

Answer 2

2OH-(aq) + 2H+(aq) → 2H2O(l)

Step-by-step explanation:

Step 1: Data given

sodium hydroxide = NaOH

hydrosulfuric acid = H2S

Step 2: The unbalanced equation

NaOH(aq) + H2S(aq) → Na2S(aq) + H2O(l)

Step 3: Balancing the equation

On the left side we have 1x Na (in NaOH), on the right side we have 2x Na (in Na2S). To balance the amount of Na on both sides, we have to multiply NaOH on the left side by 2.

2NaOH(aq) + H2S(aq) → Na2S(aq) + H2O(l)

On the left side we have 4x H (2x in NaOH and 2x in H2S), on the right side we have 2x H (in H2O). To balance the amount of H on both sides, we have to multiply H2O on the right side by 2. Now the equation is balanced.

2NaOH(aq) + H2S(aq) → Na2S(aq) + 2H2O(l)

Step 4: The net ionic equation

2Na+(aq) + 2OH-(aq) + 2H+(aq) + S^2-(aq) → 2Na+(aq) + S^2-(aq) + 2H2O(l)

The net ionic equation, for which spectator ions are omitted - remember that spectator ions are those ions located on both sides of the equation - will , after canceling those spectator ions in both side, look like this:

2OH-(aq) + 2H+(aq) → 2H2O(l)