# 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?

Mar 8, 2016

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 $N a O H$ $=$ $25.0 \times {10}^{-} 3 \cdot L \times 0.600 \cdot m o l \cdot {L}^{-} 1$ $=$ ??mol

Moles of $H N {O}_{3}$ $=$ $15.0 \times {10}^{-} 3 \cdot L \times 0.400 \cdot m o l \cdot {L}^{-} 1$ $=$ ??mol

The balanced equation is:

$N a O H \left(a q\right) + H N {O}_{3} \left(a q\right) \rightarrow N a N {O}_{3} \left(a q\right) + {H}_{2} O \left(l\right)$.

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 $40.0$ $m L$.