Problem 11
Question
Predict the products of each precipitation reaction. Balance the completed equation, and then write the net ionic equation. (a) \(\mathrm{NiCl}_{2}(\mathrm{aq})+\left(\mathrm{NH}_{1}\right)_{2} \mathrm{S}(\mathrm{aq}) \longrightarrow ?\) (b) \(\operatorname{Mn}\left(\mathrm{NO}_{3}\right)_{2}(\mathrm{aq})+\mathrm{Na}_{3} \mathrm{PO}_{4}(\mathrm{aq}) \longrightarrow ?\)
Step-by-Step Solution
Verified1Step 1: Predict the Products for Reaction (a)
The reactants are \(\mathrm{NiCl}_{2}(\mathrm{aq})\) and \((\mathrm{NH}_{4})_{2}\mathrm{S}(\mathrm{aq})\). Nickel chloride and ammonium sulfide can exchange ions to form \(\mathrm{NiS}(\mathrm{s})\) and \(\mathrm{NH}_{4}\mathrm{Cl}(\mathrm{aq})\).
Key Concepts
Chemical EquationsNet Ionic EquationBalancing Equations
Chemical Equations
Chemical equations are used to represent what happens during a chemical reaction. They show the reactants and the products formed. A reactant is a substance present before the reaction, while a product is formed as a result of the reaction.
An equation starts with the reactants on the left, an arrow pointing to the right, and the products on the right. For example, the reaction between nickel chloride and ammonium sulfide can be depicted as follows:
\[\mathrm{NiCl}_{2}(\mathrm{aq}) + (\mathrm{NH}_{4})_{2}\mathrm{S}(\mathrm{aq}) \longrightarrow \mathrm{NiS}(\mathrm{s}) + 2\mathrm{NH}_{4}\mathrm{Cl}(\mathrm{aq})\]
This equation conveys that nickel sulfide (\mathrm{NiS}(\mathrm{s})) is a solid, a precipitate formed from the reaction, and ammonium chloride (\mathrm{NH}_{4}\mathrm{Cl}(\mathrm{aq})) remains in solution. The state symbols such as \((\mathrm{aq})\) for aqueous and \((\mathrm{s})\) for solid are crucial for understanding the nature of each substance. This helps in visualizing and predicting the outcome of the reaction.
An equation starts with the reactants on the left, an arrow pointing to the right, and the products on the right. For example, the reaction between nickel chloride and ammonium sulfide can be depicted as follows:
\[\mathrm{NiCl}_{2}(\mathrm{aq}) + (\mathrm{NH}_{4})_{2}\mathrm{S}(\mathrm{aq}) \longrightarrow \mathrm{NiS}(\mathrm{s}) + 2\mathrm{NH}_{4}\mathrm{Cl}(\mathrm{aq})\]
This equation conveys that nickel sulfide (\mathrm{NiS}(\mathrm{s})) is a solid, a precipitate formed from the reaction, and ammonium chloride (\mathrm{NH}_{4}\mathrm{Cl}(\mathrm{aq})) remains in solution. The state symbols such as \((\mathrm{aq})\) for aqueous and \((\mathrm{s})\) for solid are crucial for understanding the nature of each substance. This helps in visualizing and predicting the outcome of the reaction.
Net Ionic Equation
Net ionic equations provide a simplified version of chemical reactions by including only the ions and molecules directly involved in forming the precipitate. This allows a clearer view of the actual chemical change.
To write a net ionic equation, firstly, break down the aqueous substances into their constituent ions. Consider the full ionic equation for the reaction:
Eliminating the spectator ions leaves:
To write a net ionic equation, firstly, break down the aqueous substances into their constituent ions. Consider the full ionic equation for the reaction:
- \(\mathrm{Ni}^{2+}(\mathrm{aq}) + 2\mathrm{Cl}^{-}(\mathrm{aq}) + 2\mathrm{NH}_{4}^{+}(\mathrm{aq}) + \mathrm{S}^{2-}(\mathrm{aq}) \rightarrow \mathrm{NiS}(\mathrm{s}) + 2\mathrm{NH}_{4}^{+}(\mathrm{aq}) + 2\mathrm{Cl}^{-}(\mathrm{aq})\)
Eliminating the spectator ions leaves:
- \(\mathrm{Ni}^{2+}(\mathrm{aq}) + \mathrm{S}^{2-}(\mathrm{aq}) \rightarrow \mathrm{NiS}(\mathrm{s})\)
Balancing Equations
Balancing equations ensures that the same number of each type of atom appears on both sides of a chemical equation according to the Law of Conservation of Mass. Each side of the equation must reflect the same total number of atoms for each element involved.
Take the unbalanced equation from our example again: \[\mathrm{NiCl}_{2}(\mathrm{aq}) + (\mathrm{NH}_{4})_{2}\mathrm{S}(\mathrm{aq}) \longrightarrow \mathrm{NiS}(\mathrm{s}) + \mathrm{NH}_{4}\mathrm{Cl}(\mathrm{aq})\]
Begin by identifying each element in the equation:
A nickel atom (\(\mathrm{Ni}\)) and a sulfur atom (\(\mathrm{S}\)) each appear once, and since they form a compound, they are balanced.
Now look at chlorine and ammonium ions:
Take the unbalanced equation from our example again: \[\mathrm{NiCl}_{2}(\mathrm{aq}) + (\mathrm{NH}_{4})_{2}\mathrm{S}(\mathrm{aq}) \longrightarrow \mathrm{NiS}(\mathrm{s}) + \mathrm{NH}_{4}\mathrm{Cl}(\mathrm{aq})\]
Begin by identifying each element in the equation:
A nickel atom (\(\mathrm{Ni}\)) and a sulfur atom (\(\mathrm{S}\)) each appear once, and since they form a compound, they are balanced.
Now look at chlorine and ammonium ions:
- Initially, you have 2 chlorine (\(\mathrm{Cl}^{-}\)) ions with nickel chloride and 2 ammonium (\(\mathrm{NH}_4^{+}\)) ions per unit of ammonium sulfide.
- This will yield 2 formula units of \(\mathrm{NH}_{4}\mathrm{Cl}(\mathrm{aq})\) when all ions are involved, rendering the equation balanced: \[\mathrm{NiCl}_{2}(\mathrm{aq}) + (\mathrm{NH}_{4})_{2}\mathrm{S}(\mathrm{aq}) \longrightarrow \mathrm{NiS}(\mathrm{s}) + 2\mathrm{NH}_{4}\mathrm{Cl}(\mathrm{aq})\]
Other exercises in this chapter
Problem 9
Balance the equation for the following precipitation reaction, and then write the net ionic equation. Indicate the state of each species \((s, \ell, \text { aq,
View solution Problem 10
Balance the equation for the following precipitation reaction, and then write the net ionic equation. Indicate the state of each species (s, \(\ell\), aq, or g)
View solution Problem 12
Predict the products of each precipitation reaction. Balance the completed equation, and then write the net ionic equation. (a) \(\mathrm{Pb}\left(\mathrm{NO}_{
View solution Problem 13
Write a balanced equation for the ionization of nitric acid in water.
View solution