# 8.14: Arrhenius Acids and Bases: Reactions of Arrhenius Acids and Metals


Learning Objectives
• Predict the products that are generated when Arrhenius acids and metals undergo single replacement reactions.

Because chemists are primarily interested in studying chemical changes, predicting the products of a chemical reaction is a critically-important skill.  Recall that, because the protons, H+1, and hydroxide ions, OH–1, that are formed as a result of 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 following paragraphs will describe and symbolically-represent the transformation that occurs when an Arrhenius acid reacts with a metal.  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, "HNPoly," 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.  Therefore, the reaction that occurs between a metal in its elemental, or atomic, form, and an Arrhenius acid can be categorized as a single replacement reaction.  During this type of chemical change, an elemental reactant must displace either the cationic or anionic portion of an ionic compound.  Furthermore, in order for a replacement to occur successfully, the ionized form of the elemental reactant and the component of the compound that is being displaced must have identical cationic or anionic classifications.  Therefore, in order for a single replacement reaction to occur between an elemental metal and an Arrhenius acid, the elemental reactant, a metal, ionizes to form a cation, which replaces the cationic portion, H+1, of the Arrhenius acid and forms an ionic compound by bonding with the remaining anionic portion of the acid.  The displaced proton, H+1, is reduced, as described in Section 4.19, to molecular hydrogen, H2.  The reaction patterns that are shown below symbolize the single replacement reactions that occur between a metal, "M," and a hydrohalogenated, "HX,"

___ $$\ce{M}$$ + ___ $$\ce{HX}$$ $$\rightarrow$$ ___ $$\ce{MX_{Y}} \; +$$ ___ $$\ce{H_2}$$

and a polyatomic, "HNPoly,"

___ $$\ce{M}$$ + ___ $$\ce{H_{N}Poly}$$ $$\rightarrow$$ ___ $$\ce{M_{N}(Poly)_{Y}} \; +$$ ___ $$\ce{H_2}$$

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, "HNPoly," Arrhenius acid.  Because, as stated above, a single 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 "MXY" and "MN(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 "MXY" and "MN(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 first discussed in Exercise 8.12.3.  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.)

___ $$\ce{H_2SO_4}$$ + ___ $$\ce{Li}$$ $$\rightarrow$$

Because the first chemical formula in this equation contains two protons, H+1, and a polyatomic anion, SO4–2, the sulfate ion, the corresponding molecule, sulfuric acid, H2SO4, can be classified as a polyatomic, "HNPoly," Arrhenius acid.  Since the second reactant, lithium, Li, is a metal that is present in its elemental form, these chemicals will undergo a single replacement reaction.  Therefore, based on the second reaction pattern that is shown above, the metal, lithium, Li, ionizes to form a cation, Li+1, which replaces the cationic portion, H+1, of the Arrhenius acid and forms an ionic compound, Li2SO4, by bonding with the remaining anionic portion, SO4–2, of the acid.  The displaced proton, H+1, is reduced, as described in Section 4.19, to molecular hydrogen, H2.  The information that is described in this paragraph is reflected in the unbalanced reaction equation that is shown below.

___ $$\ce{H_2SO_4}$$ + ___ $$\ce{Li}$$ $$\rightarrow$$ ___ $$\ce{Li_2SO_4}$$ + ___ $$\ce{H_2}$$

Since the left and right sides of this equation contain equal amounts of sulfate ions, SO42, and hydrogen, H, those chemicals are balanced.  However, lithium, Li, is not balanced, and, therefore, a coefficient of 2 should be written in the "blank" that corresponds to this chemical on the left side of the reaction arrow, as shown below.  Because all of the components in the following reaction equation are balanced, the equation that is shown below is the chemically-correct representation of the single replacement reaction that occurs between sulfuric acid, H2SO4, an Arrhenius acid, and elemental lithium, Li, a metal.

___ $$\ce{H_2SO_4}$$ + 2 $$\ce{Li}$$ $$\rightarrow$$ ___ $$\ce{Li_2SO_4}$$ + ___ $$\ce{H_2}$$

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.)

___ $$\ce{Be}$$ + ___ $$\ce{HI}$$ $$\rightarrow$$

___ $$\ce{Be}$$ + ___ $$\ce{HI}$$ $$\rightarrow$$ ___ $$\ce{BeI_2}$$ + ___ $$\ce{H_2}$$
___ $$\ce{Be}$$ + 2 $$\ce{HI}$$ $$\rightarrow$$ ___ $$\ce{BeI_2}$$ + ___ $$\ce{H_2}$$