# Which of the following processes will likely result in a

Which of the following processes will likely result in a precipitation reaction?
(a) Mixing an
$NaN{O}_{3}$ solution
with a
$CuS{O}_{4}$, solution.
(b) Mixing a
$BaC{l}_{2}$ solution
with a
${K}_{2}S{O}_{4}$ solution.
Write a net ionic equation for the precipitation reaction.
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Fasaniu

PART A:
1. Write the balanced chemical equation for this exchange reaction. Determine the correct states based on solubility rules.
2. Rewrite all aqueous molecules as ions in solution and leave solids and/or liquids alone. This gives us a complete ionic equation
3. Cancel the spectator ions to arrive at the net ionic equation
$2NaN{O}_{3}\left(aq\right)+CuS{O}_{4}\left(aq\right)⇒N{a}_{2}S{O}_{4}\left(aq\right)+Cu{\left(N{O}_{3}\right)}_{2}\left(aq\right)$
$2N{a}^{+}\left(ag\right)+2N{O}_{3}^{-}\left(ag\right)+C{u}^{2+}\left(aq\right)+S{O}_{4}^{2-}\left(aq\right)⇒2N{a}^{+}\left(aq\right)+S{O}_{4}^{2-}\left(aq\right)+C{u}^{2+}\left(aq\right)+2N{O}_{3}^{-}\left(aq\right)$
$⇒$
Since all species are soluble, there is no reaction and therefore, no net ionic equation
PART B:
1. Write the balanced chemical equation for this double-displacement reaction. Determine the correct states based on solubility rules.
2. Rewrite all aqueous molecules as ions in solution and leave solids and/or liquids alone. This gives us a complete ionic equation
3. Cancel the spectator ions to arrive at the net ionic equation.
$BaC{l}_{2}\left(ag\right)+ZnS{O}_{4}\left(ag\right)⇒BaS{O}_{4}\left(s\right)+ZnC{l}_{2}\left(ag\right)$
$B{a}^{2+}\left(ag\right)+2C{l}^{-\left(ag\right)}+Z{n}^{2+}\left(ag\right)+S{O}_{4}^{2-}\left(ag\right)⇒BaS{O}_{4}\left(s\right)+Z{n}^{2+}\left(ag\right)+2C{l}^{-\left(ag\right)}$
$B{a}^{2+}\left(ag\right)S{O}_{4}^{2-}\left(ag\right)⇒BaS{O}_{4}\left(s\right)$

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Archie Jones
Step 1
A precipitation reaction is a type of double displacement reaction in which a solid product is formed upon reaction of two aqueous solutions, such that the solid product formed is insoluble in aqueous solutions,
Step 2
a) Mixing of $NaN{O}_{3}$ and $CuS{O}_{4}$, will not lead to a precipitation reaction as the products formed are soluble in aqueous solutions,
b) Mixing of $BaC{l}_{2}$, and ${K}_{2}S{O}_{4}$, will lead to a precipitate and the equation is shown below,
$BaC{l}_{2}\left(ag\right)+{K}_{2}S{O}_{4}\left(ag\right)⇒BaS{O}_{4}\left(s\right)+2KCl\left(ag\right)$
Net ionic equation :
$B{a}^{2+}\left(ag\right)+S{O}_{4}^{2-}\left(ag\right)⇒BaS{O}_{4}\left(s\right)$