Chapter 4, Problem 51A

(a)

Interpretation Introduction

Interpretation: Correct formula of the ionic compound (a) needs to be determined.

Concept introduction: Ionic compounds are made up of positively charged ions (cations) and negatively charged ions (anions) in such a ratio that the overall charge on the molecule is zero, that is, that molecule is neutral.

Compounds are named by writing the name of the cation first followed by the name of the anion.

Generally, the names of the polyatomic anion ends with −ate with a few exceptions.

Answer to Problem 51A

The correct formula for “a” is ${\text{NaH}}_{\text{2}}{\text{PO}}_{\text{4}}.$

Explanation of Solution

The cation is ${\text{Na}}^{\text{+}}$ (sodium) and anion is ${\text{H}}_{\text{2}}{\text{PO}}_{\text{4}}{}^{2-}$ (dihydrogen phosphate). To make the compound neutral, we need two Na+ and one ${\text{H}}_{\text{2}}{\text{PO}}_{\text{4}}{}^{2-}.$

(b)

Interpretation Introduction

Interpretation: Correct formula of the ionic compound (b) needs to be determined.

Concept introduction: Ionic compounds are made up of positively charged ions (cations) and negatively charged ions (anions) in such a ratio that the overall charge on the molecule is zero, that is, that molecule is neutral.

Compounds are named by writing the name of the cation first followed by the name of the anion.

Generally, the names of the polyatomic anion ends with −ate with a few exceptions.

Answer to Problem 51A

The correct formula for “b” is ${\text{LiClO}}_{\text{4}}.$

Explanation of Solution

The cation is ${\text{Li}}^{\text{+}}$ (lithium) and anion is ${\text{ClO}}_{\text{4}}{}^{-}$ (perchlorate). To make the compound neutral, we need one ${\text{Li}}^{\text{+}}$ and one ${\text{ClO}}_{\text{4}}{}^{-}.$

(c)

Interpretation Introduction

Interpretation: Correct formula of the ionic compound (c) needs to be determined.

Concept introduction: Ionic compounds are made up of positively charged ions (cations) and negatively charged ions (anions) in such a ratio that the overall charge on the molecule is zero, that is, that molecule is neutral.

Compounds are named by writing the name of the cation first followed by the name of the anion.

Generally, the names of the polyatomic anion ends with −ate with a few exceptions.

Answer to Problem 51A

The correct formula for “c” is ${\text{Cu(HCO}}_{\text{3}}{\text{)}}_{\text{2}}.$

Explanation of Solution

The cation is ${\text{Cu}}^{\text{2+}}$ (copper) and anion is ${\text{HCO}}_{\text{3}}{}^{-}$ (hydrogen carbonate). To make the compound neutral, we need one Cu²+ and two ${\text{HCO}}_{\text{3}}{}^{-}.$

(d)

Interpretation Introduction

Interpretation: Correct formula of the ionic compound (d) needs to be determined.

Concept introduction: Ionic compounds are made up of positively charged ions (cations) and negatively charged ions (anions) in such a ratio that the overall charge on the molecule is zero, that is, that molecule is neutral.

Compounds are named by writing the name of the cation first followed by the name of the anion.

Generally, the names of the polyatomic anion ends with −ate with a few exceptions.

Answer to Problem 51A

The correct formula for “d” is ${\text{CH}}_{\text{3}}\text{COOK}$

Explanation of Solution

The cation is ${\text{K}}^{+}$ (potassium) and anion is ${\text{CH}}_3{\text{COO}}^{-}$ (acetate). To make the compound neutral, we need one ${\text{K}}^{+}$ and one ${\text{CH}}_3{\text{COO}}^{-}.$

(e)

Interpretation Introduction

Interpretation: Correct formula of the ionic compound (e) needs to be determined.

Concept introduction: Ionic compounds are made up of positively charged ions (cations) and negatively charged ions (anions) in such a ratio that the overall charge on the molecule is zero, that is, that molecule is neutral.

Compounds are named by writing the name of the cation first followed by the name of the anion.

Generally, the names of the polyatomic anion ends with −ate with a few exceptions.

Answer to Problem 51A

The correct formula for “e” is ${\text{BaO}}_{\text{2}}.$

Explanation of Solution

The cation is ${\text{Ba}}^{2+}$ (barium) and anion is ${\text{O}}_2{}^{2-}$ (peroxide). To make the compound neutral, we need one ${\text{Ba}}^{2+}$ and ${\text{O}}_2{}^{2-}.$

(f)

Interpretation Introduction

Interpretation: Correct formula of the ionic compound (f) needs to be determined.

Concept introduction: Ionic compounds are made up of positively charged ions (cations) and negatively charged ions (anions) in such a ratio that the overall charge on the molecule is zero, that is, that molecule is neutral.

Compounds are named by writing the name of the cation first followed by the name of the anion.

Generally, the names of the polyatomic anion ends with −ate with a few exceptions.

Answer to Problem 51A

The correct formula for “f” is ${\text{Cs}}_{\text{2}}{\text{SO}}_{\text{3}}.$

Explanation of Solution

The cation is ${\text{Cs}}^{+}$ (cesium) and anion is ${\text{SO}}_3{}^{2-}$ (sulfite). To make the compound neutral, we need two Cs+ and one ${\text{SO}}_{\text{3}}{}^{2-}$ .