8.16: Arrhenius Acids and Bases: Reactions of Arrhenius Acids and Metal Carbonates or Bicarbonates

Learning Objectives

• Predict the products that are generated when Arrhenius acids and metal carbonates or metal bicarbonates undergo double replacement/decomposition reaction sequences.

Because the protons, H⁺¹, and hydroxide ions, OH^–1, that are formed from the dissociation of Arrhenius acids and Arrhenius bases, respectively, are analyzed as independent chemical entities, the reactivity of an Arrhenius chemical is limited, but predictable.  The previous two sections of this chapter discussed the reactions that occur between an Arrhenius acid and an elemental metal and an Arrhenius acid and an Arrhenius base, respectively.  Molecular hydrogen, H₂, is produced in the first reaction, water, H₂O, is synthesized in the second chemical change, and a salt is generated in both transformations.  The following paragraphs will describe and symbolically-represent a third Arrhenius reaction, in which an Arrhenius acid reacts with a metal carbonate or metal bicarbonate, which can be symbolized as "M₂(CO₃)_Y" and "M(HCO₃)_Y," respectively.  If the metal that is present in a metal carbonate bears a +2 charge, the subscripts that are present in the "M₂(CO₃)_Y" pattern are reduced and, subsequently, eliminated.  A specialized symbolism, "MCO₃," results upon removing the parentheses around the polyatomic ion.  The generic reactivity patterns that will be developed can be applied to predict the products that are generated during a specific chemical reaction.

Recall that, because the hydrohalogenated, "HX," and polyatomic, "H_NPoly," Arrhenius acids both contain a cationic, H⁺¹, and an anionic, X^–1 or Poly^–N, respectively, component, both of these types of acids are classified as ionic compounds.  Metal carbonates and metal bicarbonates, which are comprised of a monatomic metal cation, M^+Y, and either a carbonate, CO₃^–2, or bicarbonate, HCO₃^–1, polyatomic anion, respectively, are also ionic compounds.  Consequently, the reaction that occurs between a metal carbonate, M₂(CO₃)_Y, or metal bicarbonate, M(HCO₃)_Y, and an Arrhenius acid can be categorized as a double replacement reaction.  During this type of chemical change, either the cationic or the anionic portions of two compounds exchange their relative positions.  Therefore, in a double replacement reaction involving an metal carbonate or metal bicarbonate and an Arrhenius acid, the cationic portions of these chemicals, a metal cation, M^+Y, and a proton, H⁺¹, respectively, replace one another.  As a result, two new ionic compounds are produced by bonding the M^+Y cation with the remaining anionic portions of the Arrhenius acid, X^–1 or Poly^–N, and by bonding a proton, H⁺¹, with the polyatomic anions that are present in the metal carbonate, CO₃^–2, or metal bicarbonate, HCO₃^–1.  The reaction patterns that are shown below symbolize the double replacement reactions that occur between a metal carbonate, "M₂(CO₃)_Y," and a hydrohalogenated, "HX,"

___ \(\ce{M_2(CO_3)_{Y}}\) + ___ \(\ce{HX}\) \(\rightarrow\) ___ \(\ce{MX_{Y}} \; + \) ___ \(\ce{H_2CO_3}\)

and a polyatomic, "H_NPoly," 

___ \(\ce{M_2(CO_3)_{Y}}\) + ___ \(\ce{H_{N}Poly}\) \(\rightarrow\) ___ \(\ce{M_{N}(Poly)_{Y}} \; + \) ___ \(\ce{H_2CO_3}\)

Arrhenius acid, respectively.  The order in which the reactants are given does not impact the classification of the reaction or the identities of the products that are generated.  Similarly, the formulas for the products, which must be separated by a plus sign, can be written in any order.  Furthermore, recall that all halide ions, X^–1, bear –1 charges, that "Y" corresponds to the numerical charge of the cation that results from the ionization of a metal, "M," and that "N" represents the charge of the polyatomic ion, "Poly^–N" that is present in a polyatomic, "H_NPoly," Arrhenius acid.  Because, as stated above, a double replacement reaction generates an ionic compound by bonding a metal cation, M^+Y, with the anionic portion, X^–1 or Poly^–N, of an Arrhenius acid, the "MX_Y" and "M_N(Poly)_Y" formulas that are written in these reaction patterns are derived by applying the symbolic information in the previous sentence to the Chapter 3 rules for determining ionic chemical formulas.  Since the "MX_Y" and "M_N(Poly)_Y" chemical formulas do not contain protons, H⁺¹, or hydroxide ions, OH^–1, the corresponding ionic compounds are classified as salts, which were also produced in both of the Arrhenius reactions that were discussed in the previous sections of this chapter.

Furthermore, when a proton, H⁺¹, from the Arrhenius acid bonds with the anionic portion, CO₃^–2, of the metal carbonate, a new polyatomic Arrhenius acid is generated as a second product in this double replacement reaction.  In order to obtain the chemical formula of this acid, the "Poly" and "N" portions of the "H_NPoly" pattern can be replaced with the formula of the specific polyatomic anion and the absolute value of the charge of that ion, respectively.  Because the carbonate ion, CO₃^–2, bears a –2 charge, the corresponding polyatomic Arrhenius acid must contain 2 protons, in order to achieve charge-balance in the final compound.  Therefore, the chemical formula of the acid that is produced in this reaction is carbonic acid, H₂CO₃.  However, when generated in an aqueous solution, carbonic acid, H₂CO₃, is highly unstable and immediately decomposes to form carbon dioxide, CO₂, and water, H₂O.  Because this double replacement reaction, in which an Arrhenius acid reacts with a metal carbonate or metal bicarbonate, involves an Arrhenius chemical which must, by definition, be dissolved in water, the carbonic acid, H₂CO₃, that is produced decomposes immediately after its generation, and the patterns that are presented above must be modified to reflect this subsequent reaction.  The equation patterns that are shown below symbolize the double replacement/decomposition reaction sequences that occur between a metal carbonate, "M₂(CO₃)_Y," and a hydrohalogenated, "HX,"

___ \(\ce{M_2(CO_3)_{Y}}\) + ___ \(\ce{HX}\) \(\rightarrow\) ___ \(\ce{MX_{Y}} \; + \) ___ \(\ce{CO_2} \; + \) ___ \(\ce{H_2O}\)

and a polyatomic, "H_NPoly," 

___ \(\ce{M_2(CO_3)_{Y}}\) + ___ \(\ce{H_{N}Poly}\) \(\rightarrow\) ___ \(\ce{M_{N}(Poly)_{Y}} \; + \) ___ \(\ce{CO_2} \; + \) ___ \(\ce{H_2O}\)

Arrhenius acid, respectively.  Because a metal bicarbonate, "M(HCO₃)_Y," also contains a carbonate ion, CO₃^–2, an analogous reaction sequence occurs when a metal bicarbonate, "M(HCO₃)_Y," is exposed to a hydrohalogenated, "HX,"

___ \(\ce{M(HCO_3)_{Y}}\) + ___ \(\ce{HX}\) \(\rightarrow\) ___ \(\ce{MX_{Y}} \; + \) ___ \(\ce{CO_2} \; + \) ___ \(\ce{H_2O}\)

or a polyatomic, "H_NPoly," 

___ \(\ce{M(HCO_3)_{Y}}\) + ___ \(\ce{H_{N}Poly}\) \(\rightarrow\) ___ \(\ce{M_{N}(Poly)_{Y}} \; + \) ___ \(\ce{CO_2} \; + \) ___ \(\ce{H_2O}\)

Arrhenius acid, as shown in the equation patterns that are presented above.

Finally, after determining the chemical formula of the ionic compound that is produced, the resultant chemical equation must be balanced by incorporating coefficients, as necessary, in order to uphold the Law of Conservation of Matter.

For example, complete the following chemical equation by determining the chemical formula(s) of the product(s) that are predicted to form.  Balance the resultant equation by writing coefficients in the "blanks," as necessary.  (States of matter are not required.)

___ \(\ce{HCl}\) + ___ \(\ce{Al_2(CO_3)_3}\) \(\rightarrow \)

Because the first chemical formula in this equation contains a proton, H⁺¹, and a halide ion, Cl^–1, the chloride ion, the corresponding molecule, hydrochloric acid, HCl, can be classified as a hydrohalogenated, "HX," Arrhenius acid.  The second reactant, aluminum carbonate, Al₂(CO₃)₃, is a metal carbonate.  Consequently, these chemicals will undergo a double replacement reaction in which the cationic or the anionic portions of the given compounds exchange their relative positions.  Therefore, based on the first metal carbonate reaction pattern that is shown above, a new ionic compound, AlCl₃, forms when the metal cation from the metal carbonate, Al⁺³, bonds with the anionic portion, Cl^–1, of the Arrhenius acid.  Furthermore, this double replacement reaction generates a second compound, H₂CO₃, by bonding a proton, H⁺¹, from the Arrhenius acid with the anionic portion, CO₃^–2, of the metal carbonate.  However, because this reaction involves an Arrhenius chemical which must, by definition, be dissolved in water, the carbonic acid, H₂CO₃, that is produced is unstable and, therefore, decomposes to form carbon dioxide, CO₂, and water, H₂O, immediately after its generation.  The information that is described in this paragraph is reflected in the unbalanced reaction equation that is shown below.

___ \(\ce{HCl}\) + ___ \(\ce{Al_2(CO_3)_3}\) \(\rightarrow \) ___ \(\ce{AlCl_3} \; + \) ___ \(\ce{CO_2} \; + \) ___ \(\ce{H_2O}\)

None of the components of this reaction are balanced.  Therefore, in order to balance the aluminum ion, Al⁺³, and carbon, C, coefficients of 2 and 3, respectively, should be written in the "blanks" that correspond to these chemicals on the right side of the reaction arrow, as shown below.

___ \(\ce{HCl}\) + ___ \(\ce{Al_2(CO_3)_3}\) \(\rightarrow \) 2 \(\ce{AlCl_3} \; + \) 3 \(\ce{CO_2} \; + \) ___ \(\ce{H_2O}\)

In order to balance the chloride ion, Cl^–1, a coefficient of 6 should be written in the "blank" that corresponds to this ion on the left side of the reaction arrow, as shown below.

6 \(\ce{HCl}\) + ___ \(\ce{Al_2(CO_3)_3}\) \(\rightarrow \) 2 \(\ce{AlCl_3} \; + \) 3 \(\ce{CO_2} \; + \) ___ \(\ce{H_2O}\)

Finally, in order to balance both hydrogen, H, and oxygen, O, a coefficient of 3 should be written in the "blank" that corresponds to these elements on the right side of the reaction arrow, as shown below.  Because all of the components in the following reaction equation are balanced, this equation is the chemically-correct representation of the double replacement/decomposition reaction sequence that occurs between hydrobromic acid, HBr, an Arrhenius acid, and aluminum carbonate, Al₂(CO₃)₃, a metal carbonate.

6 \(\ce{HCl}\) + ___ \(\ce{Al_2(CO_3)_3}\) \(\rightarrow \) 2 \(\ce{AlCl_3} \; + \) 3 \(\ce{CO_2} \; + \) 3 \(\ce{H_2O}\)

Exercise \(\PageIndex{1}\)

Complete the following chemical equation by determining the chemical formula(s) of the product(s) that are predicted to form.  Balance the resultant equation by writing coefficients in the "blanks," as necessary.  (States of matter are not required.)

___ \(\ce{HC_2H_3O_2}\) + \(\ce{Sn(HCO_3)_4}\) \(\rightarrow \)

Answer

Because the first chemical formula in this equation contains a proton, H⁺¹, and a polyatomic anion, C₂H₃O₂^–1, the acetate ion, the corresponding molecule, acetic acid, HC₂H₃O₂, can be classified as a polyatomic, "H_NPoly," Arrhenius acid.  The second reactant, tin (IV) bicarbonate, Sn(HCO₃)₄, is a metal bicarbonate.  Tin, the metal that is present in this bicarbonate, is a transition metal and, therefore, is able to achieve a stable electron configuration through multiple ionization pathways.  Because tin could ionize to form either Sn⁺² or Sn⁺⁴, the specific charge of the tin cation that is contained in Sn(HCO₃)₄ must be definitively established before the correct product chemical formulas can be written.  The "Reverse Ratio Method" that was described in Section 3.10 could be utilized to determine this charge information.  Alternatively, recall that the "Y" subscript that is present in the metal bicarbonate formula pattern, "M(HCO₃)_Y," corresponds to the numerical charge of the cation that is contained in that particular bicarbonate.  Therefore, because the numerical value of "Y" in the given metal bicarbonate, Sn(HCO₃)₄, is a "4," the tin ion that is present in the this compound bears a +4 charge.

These chemicals will undergo a double replacement reaction in which the cationic or the anionic portions of the given compounds exchange their relative positions.  Therefore, based on the second metal bicarbonate reaction pattern that is shown above, a new ionic compound, Sn(C₂H₃O₂)₄, forms when the metal cation from the metal bicarbonate, Sn⁺⁴, bonds with the anionic portion, C₂H₃O₂^–1, of the Arrhenius acid.  Furthermore, this double replacement reaction generates a second compound, H₂CO₃, by bonding a proton, H⁺¹, from the Arrhenius acid with the anionic portion, HCO₃^–1, of the metal bicarbonate.  However, because this reaction involves an Arrhenius chemical which must, by definition, be dissolved in water, the carbonic acid, H₂CO₃, that is produced is unstable and, therefore, decomposes to form carbon dioxide, CO₂, and water, H₂O, immediately after its generation.  The information that is described in this paragraph is reflected in the unbalanced reaction equation that is shown below.

___ \(\ce{HC_2H_3O_2}\) + ___ \(\ce{Sn(HCO_3)_4}\) \(\rightarrow \) ___ \(\ce{Sn(C_2H_3O_2)_4} \; + \) ___ \(\ce{CO_2} \; + \) ___ \(\ce{H_2O}\)

Since the left and right sides of this equation contain equal amounts of tin (IV) ions, Sn⁺⁴, that ion is balanced.  However, hydrogen, H, the acetate ion, C₂H₃O₂^–1, carbon, C, and oxygen, O, are not balanced.  Therefore, in order to balance the acetate ion, C₂H₃O₂^–1, and carbon, C, coefficients of 4, should be written in the "blanks" that correspond to these chemicals on the left and right sides of the reaction arrow, respectively, as shown below.

4 \(\ce{HC_2H_3O_2}\) + ___ \(\ce{Sn(HCO_3)_4}\) \(\rightarrow \) ___ \(\ce{Sn(C_2H_3O_2)_4} \; + \) 4 \(\ce{CO_2} \; + \) ___ \(\ce{H_2O}\)

Finally, in order to balance both hydrogen, H, and oxygen, O, a coefficient of 4 should be written in the "blank" that corresponds to these elements on the right side of the reaction arrow, as shown below.  Because all of the components in the following reaction equation are balanced, this equation is the chemically-correct representation of the double replacement/decomposition reaction sequence that occurs between acetic acid, HC₂H₃O₂, an Arrhenius acid, and tin (IV) bicarbonate, Sn(HCO₃)₄, a metal bicarbonate.

4 \(\ce{HC_2H_3O_2}\) + ___ \(\ce{Sn(HCO_3)_4}\) \(\rightarrow \) ___ \(\ce{Sn(C_2H_3O_2)_4} \; + \) 4 \(\ce{CO_2} \; + \) 4 \(\ce{H_2O}\)