# How do you write the total ionic equation and net ionic equation for the reaction of silver nitrate and potassium chloride?

Feb 9, 2016

Total ionic equation:
$A {g}^{+} \left(a q\right) + N {O}_{3}^{-} \left(a q\right) + {K}^{+} \left(a q\right) + C {l}^{-} \left(a q\right) \to A g C l \left(s\right) + {K}^{+} \left(a q\right) + N {O}_{3}^{-} \left(a q\right)$

Net ionic equation: $A {g}^{+} \left(a q\right) + C {l}^{-} \left(a q\right) \to A g C l \left(s\right)$

#### Explanation:

The reaction between silver nitrate $A g N {O}_{3}$ and potassium chloride $K C l$ is the following:

$A g N {O}_{3} \left(a q\right) + K C l \left(a q\right) \to A g C l \left(s\right) + K N {O}_{3} \left(a q\right)$

During this reaction, a precipitate will form which is the silver chloride $A g C l$.

The total ionic equation would be:

$A {g}^{+} \left(a q\right) + N {O}_{3}^{-} \left(a q\right) + {K}^{+} \left(a q\right) + C {l}^{-} \left(a q\right) \to A g C l \left(s\right) + {K}^{+} \left(a q\right) + N {O}_{3}^{-} \left(a q\right)$

The net ionic equation however, would be:

$A {g}^{+} \left(a q\right) + C {l}^{-} \left(a q\right) \to A g C l \left(s\right)$

Notice, the spectator ions - which are the common ions in both sides - are removed from the net ionic equation.