Problem 51 - Chemistry and Chemical Reactivity

Question

Balance the following equations, and then classify each as a precipitation, acid-base, or gas-forming reaction. (a) \(\mathrm{Ba}(\mathrm{OH})_{2}(\mathrm{aq})+\mathrm{HCl}(\mathrm{aq}) \rightarrow \mathrm{BaCl}_{2}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\ell)\) (b) \(\mathrm{HNO}_{3}(\mathrm{aq})+\mathrm{CoCO}_{3}(\mathrm{s}) \rightarrow \mathrm{Co}\left(\mathrm{NO}_{3}\right)_{2}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\ell)+\mathrm{CO}_{2}(\mathrm{g})\) (c) \(\mathrm{Na}_{3} \mathrm{PO}_{4}(\mathrm{aq})+\mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}(\mathrm{aq}) \rightarrow \mathrm{Cu}_{3}\left(\mathrm{PO}_{4}\right)_{2}(\mathrm{s})+\mathrm{NaNO}_{3}(\mathrm{aq})\)

Step-by-Step Solution

Answer

(a) Acid-base reaction, balanced as \( \mathrm{Ba}(\mathrm{OH})_{2}+2\mathrm{HCl}\rightarrow\mathrm{BaCl}_{2}+2\mathrm{H}_{2}\mathrm{O} \); (b) Gas-forming reaction, balanced as \( 2\mathrm{HNO}_{3}+\mathrm{CoCO}_{3}\rightarrow\mathrm{Co(NO}_{3})_{2}+\mathrm{H}_{2}\mathrm{O}+\mathrm{CO}_{2} \); (c) Precipitation reaction, balanced as \( 2\mathrm{Na}_{3}\mathrm{PO}_{4}+3\mathrm{Cu(NO}_{3})_{2}\rightarrow\mathrm{Cu}_{3}(\mathrm{PO}_{4})_{2}+6\mathrm{NaNO}_{3} \).

Step 1: Analyzing Reaction (a)

The reaction is \( \mathrm{Ba}(\mathrm{OH})_{2}(\mathrm{aq})+\mathrm{HCl}(\mathrm{aq}) \rightarrow \mathrm{BaCl}_{2}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\ell) \). We need to balance it, and since it involves a base and an acid, it is an acid-base reaction.

Step 2: Balancing Reaction (a)

We balance the reaction equation by ensuring there are equal numbers of each type of atom on both sides: \[ \mathrm{Ba}(\mathrm{OH})_{2}(\mathrm{aq}) + 2\mathrm{HCl}(\mathrm{aq}) \rightarrow \mathrm{BaCl}_{2}(\mathrm{aq}) + 2\mathrm{H}_{2} \mathrm{O}(\ell) \]. Now, the equation is balanced.

Step 3: Analyzing Reaction (b)

The reaction is \( \mathrm{HNO}_{3}(\mathrm{aq})+\mathrm{CoCO}_{3}(\mathrm{s}) \rightarrow \mathrm{Co}\left(\mathrm{NO}_{3}\right)_{2}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\ell)+\mathrm{CO}_{2}(\mathrm{g}) \). This reaction involves the formation of a gas (\( \mathrm{CO}_{2} \)) and is classified as a gas-forming reaction.

Step 4: Balancing Reaction (b)

To balance the reaction equation, ensure there are equal numbers of each type of atom on both sides: \[ 2\mathrm{HNO}_{3}(\mathrm{aq})+\mathrm{CoCO}_{3}(\mathrm{s}) \rightarrow \mathrm{Co}\left(\mathrm{NO}_{3}\right)_{2}(\mathrm{aq}) + \mathrm{H}_{2} \mathrm{O}(\ell) + \mathrm{CO}_{2}(\mathrm{g}) \]. This equation is now balanced.

Step 5: Analyzing Reaction (c)

The reaction is \( \mathrm{Na}_{3}\mathrm{PO}_{4}(\mathrm{aq})+\mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}(\mathrm{aq}) \rightarrow \mathrm{Cu}_{3}\left(\mathrm{PO}_{4}\right)_{2}(\mathrm{s})+\mathrm{NaNO}_{3}(\mathrm{aq}) \). A solid precipitate (\( \mathrm{Cu}_{3}\left(\mathrm{PO}_{4}\right)_{2} \)) forms, indicating this is a precipitation reaction.

Step 6: Balancing Reaction (c)

To balance the equation, make sure the number of atoms of each element is the same on both sides: \[ 2\mathrm{Na}_{3}\mathrm{PO}_{4}(\mathrm{aq}) + 3\mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}(\mathrm{aq}) \rightarrow \mathrm{Cu}_{3}\left(\mathrm{PO}_{4}\right)_{2}(\mathrm{s}) + 6\mathrm{NaNO}_{3}(\mathrm{aq}) \]. This equation is now balanced.

Key Concepts

Acid-Base Reactions

Acid-base reactions are a class of chemical reactions that involve an acid and a base reacting to form water and a salt. A common feature of these reactions is the transfer of protons, or hydrogen ions, between the reacting species. When an acid donates a proton, the base accepts it. An example of an acid-base reaction is when hydrochloric acid (HCl), a strong acid, reacts with barium hydroxide (Ba(OH)2), a strong base.
In this reaction:

• The acid, HCl, donates a proton to the hydroxide ion (OH⁻) from Ba(OH)2.
• This results in the formation of water (H₂O) and barium chloride (BaCl₂).

The reaction can be balanced by ensuring equal numbers of each type of atom on both sides of the equation. The balanced equation is:\[ \mathrm{Ba(OH)_2(aq)} + 2\mathrm{HCl(aq)} \rightarrow \mathrm{BaCl_2(aq)} + 2\mathrm{H_2O(\ell)} \]Through this equation, you can see how the hydroxide ions from the base neutralize the hydrogen ions from the acid, forming water. Ideally, the properties of both the acid and base are neutralized upon completion of the reaction.

Gas-Forming Reactions

Gas-forming reactions are chemical processes in which the products include a gas. These reactions occur frequently when acids react with carbonates or sulfites, where carbon dioxide (CO₂) or sulfur dioxide (SO₂) gases are released. For example, when nitric acid (HNO₃) reacts with cobalt(II) carbonate (CoCO₃):

• The acid reacts with the solid carbonate, producing a soluble salt, cobalt nitrate (Co(NO₃)₂).
• During this process, water and carbon dioxide gas are also formed.

The balanced chemical equation for this reaction is:\[2\mathrm{HNO_3(aq)} + \mathrm{CoCO_3(s)} \rightarrow \mathrm{Co(NO_3)_2(aq)} + \mathrm{H_2O(\ell)} + \mathrm{CO_2(g)}\]This reaction shows how the combination of an acid and a carbonate decomposes the carbonate to release CO₂ gas, a signature trait of gas-forming reactions. Observing the bubbles of escaping gas is a clear indication of such a reaction taking place.

Precipitation Reactions

Precipitation reactions are a type of chemical reaction where two aqueous solutions react to form an insoluble solid called a precipitate. These reactions are governed by the solubility rules that predict if certain ions will form an insoluble compound when mixed.
For instance, when sodium phosphate (Na₃PO₄) reacts with copper(II) nitrate (Cu(NO₃)₂):

• The phosphate ions combine with copper ions to form copper(II) phosphate, Cu₃(PO₄)₂, which is insoluble in water.
• This insoluble compound precipitates out of the solution as a solid.

The corresponding balanced equation is:\[2\mathrm{Na_3PO_4(aq)} + 3\mathrm{Cu(NO_3)_2(aq)} \rightarrow \mathrm{Cu_3(PO_4)_2(s)} + 6\mathrm{NaNO_3(aq)}\]This reaction showcases how two soluble ionic compounds can result in the formation of an insoluble precipitate, evident by the appearance of a solid in the reaction mixture. Recognizing precipitation reactions helps in understanding how new compounds can be selectively separated from a solution through the formation of a solid.