Answer:
2OH-(aq) + 2H+(aq) → 2H2O(l)
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)
