Question

What reaction occurs when 25.0 mL of 0.600 M NaOH are mixed with 15.0 mL of 0.400 M HNO3? What is the balanced equation and the net ionic equation?

Answer 1

The net ionic equation is simply:

`H_3O^+ + ""^(-)OH rarr 2H_2O(l)`

Explanation:

The sodium ions and the nitrate ions are simply along for the ride.

Of course, it is clear that there is an excess of base.

Moles of `NaOH` `=` `25.0xx10^-3*Lxx0.600*mol*L^-1` `=` `??mol`

Moles of `HNO_3` `=` `15.0xx10^-3*Lxx0.400*mol*L^-1` `=` `??mol`

The balanced equation is:

`NaOH(aq) + HNO_3(aq) rarr NaNO_3(aq) + H_2O(l)`.

I leave it to you to calculate the end concentration of sodium hydroxide. Remember, you have added the volumes so now the volume of the solution is `40.0` `mL`.