# What is the net ionic equation for the double replacement reaction between sodium sulfate and barium nitrate?

Aug 26, 2016

See explanation

#### Explanation:

$N {a}_{2} S {O}_{4} \left(a q\right) + B a {\left(N {O}_{3}\right)}_{2} \left(a q\right) \to B a S {O}_{4} \left(s\right) + 2 N a N {O}_{3} \left(a q\right)$

Break up the compounds into ions from which they came and cross out the ones that appear on both sides; those are your $\text{spectator ions}$.

$2 \cancel{N {a}^{+} \left(a q\right)} + S {O}_{4}^{2 -} \left(a q\right) + B {a}^{2 +} \left(a q\right) + 2 \cancel{N {O}_{3}^{-} \left(a q\right)} \to B a S {O}_{4} \left(s\right) \downarrow + 2 \cancel{N {a}^{+} \left(a q\right)} + 2 \cancel{N {O}_{3}^{-} \left(a q\right)}$

$\text{Barium sulfate}$ is insoluble so it doesn't ionize in solution

Write net ionic equation

$S {O}_{4}^{2 -} \left(a q\right) + B {a}^{2 +} \left(a q\right) \to B a S {O}_{4} \left(s\right) \downarrow$