What is the correct net ionic equation for the following set of reactants?
Ba(OH)2(aq)+H2SO4(aq)---->2H2O(l)+BaSO4
I keep getting the answer as 2OH^- + 2H^+ ----> 2H2O(l) but it counts it wrong.
BaSO4 is insoluble; therefore, it must appear on the right as BaSO4(s) (of course that means Ba ion and SO4 ion must be on the left).
To determine the correct net ionic equation for the given set of reactants, we need to identify the spectator ions first. Spectator ions are the ions that do not participate in the reaction and remain unchanged on both sides.
Let's break down the reaction into its ionic form:
Ba(OH)2 (aq) + H2SO4 (aq) → 2H2O (l) + BaSO4
First, we need to dissociate the compounds into their respective ions:
Ba(OH)2 dissociates into Ba^2+ and 2OH^-
H2SO4 dissociates into 2H^+ and SO4^2-
Now, let's write the complete ionic equation by showing all the ions:
Ba^2+ (aq) + 2OH^- (aq) + 2H^+ (aq) + SO4^2- (aq) → 2H2O (l) + BaSO4 (s)
Next, we need to cancel out the spectator ions, which in this case are Ba^2+ and SO4^2-. These ions appear on both sides of the equation unchanged.
After canceling the spectator ions, we have:
2OH^- (aq) + 2H^+ (aq) → 2H2O (l)
This simplified equation represents the net ionic equation for the given reactants Ba(OH)2 and H2SO4. It shows only the ions that actively participate in the reaction.
Hence, the correct net ionic equation is:
2OH^- (aq) + 2H^+ (aq) → 2H2O (l)