What is the correct net ionic equation, including all coefficients, charges, and phases, for the following set of reactants? Assume that the contribution of protons from H2SO4 is near 100 %.Ba(OH)2(aq)+H2SO4(aq)→Express your answer as a chemical equation.

Ba^2+(aq) + 2OH-(aq) + 2H+(aq) + SO4^2-(aq) → BaSO4(s) + 2H2O(l)

Explanation:

Step 1: Data given

the contribution of protons from H2SO4 is near 100 %.

Step 2: The unbalanced equation

.Ba(OH)2(aq) + H2SO4(aq) → BaSO4(s) + H2O(l)

Step 3: Balancing the equation

On the left side we have 4x H (2x in Ba(OH)2 and 2x in H2SO4). On the right side, we have 2x H (in H2O).

To balance the amount of H on both sides, we have to muliply H2O (on the right side) by 2.

Ba(OH)2(aq) + H2SO4(aq) → BaSO4(s) + 2H2O(l)

Step 4: The net ionic equation

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.

Ba^2+(aq) + 2OH-(aq) + 2H+(aq) + SO4^2-(aq) → BaSO4(s) + 2H2O(l)

After canceling those spectator ions in both side, look like this:

Ba^2+(aq) + 2OH-(aq) + 2H+(aq) + SO4^2-(aq) → BaSO4(s) + 2H2O(l)

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