Chapter 2, Problem 44PS

Give the formula and the number of each ion that makes up each of the following compounds:

(a) Mg(CH₃CO₂)₂

(b) Al(OH)₃

(c) CuCO₃

(d) Ti(SO₄)₂

(e) KH₂PO₄

(f)CaHPO₄

Interpretation Introduction

Interpretation:

For the given ionic compounds, the number and formula of constituted ions present are needed to be determined.

Concept introduction:

Ionic compounds are formed by electrostatic attraction between ions such as monoatomic ions and polyatomic ions; in which positively charged ions is known as cations and negatively charged ions are known as anions.

The sum of charges of ions present in an ionic compound is equals to net charge of the compound.

Answer to Problem 44PS

$Mg{(C{H}_{3}C{O}_2)}_2$ is composed of one $M{g}^{2+}$ ion and two $C{H}_{3}C{O}_2{}^{-}$ ion.

Explanation of Solution

In the given ionic compound $Mg{(C{H}_{3}C{O}_2)}_2$, there are $Mg$ ion and $C{H}_{3}C{O}_2$ ions present.  Since the $Mg$ is group $\text{IIA}$ element, then the charge will be $\text{2+}$ and the charge of $C{H}_{3}C{O}_2$ is $\text{1-}$, so two $C{H}_{3}C{O}_2$ ions are required to balance the $\text{2+}$ charge of $M{g}^{2+}$ ion.

The relative number of each ion is indicated in suffix of each ion.

That is,

One $M{g}^{2+}$ ion and two $C{H}_{3}C{O}_2{}^{-}$ ions are present in $Mg{(C{H}_{3}C{O}_2)}_2$ compound.

Interpretation Introduction

Interpretation:

For the given ionic compounds, the number and formula of constituted ions present are needed to be determined.

Concept introduction:

Ionic compounds are formed by electrostatic attraction between ions such as monoatomic ions and polyatomic ions; in which positively charged ions is known as cations and negatively charged ions are known as anions.

The sum of charges of ions present in an ionic compound is equals to net charge of the compound.

Answer to Problem 44PS

$Al{(OH)}_3$ is composed of one $A{l}^{3+}$ ion and three $O{H}^{-}$ ions.

Explanation of Solution

In the given ionic compound $Al{(OH)}_3$, there are $Al$ ion and $OH$ ion present. Since the charge of $OH$ is $\text{1-}$, then the charge of three $O{H}^{-}$ is $3+$, so the charge of $Al$ must be $\text{3+}$ to balance the charge of the three $O{H}^{-}$ ions .

The relative number of each ion is indicated in suffix of each ion.

That is,

One $A{l}^{3+}$ ion and three $O{H}^{-}$ ions ion are present in $Al{(OH)}_3$ compound.

Interpretation Introduction

Interpretation:

For the given ionic compounds, the number and formula of constituted ions present are needed to be determined.

Concept introduction:

Ionic compounds are formed by electrostatic attraction between ions such as monoatomic ions and polyatomic ions; in which positively charged ions is known as cations and negatively charged ions are known as anions.

The sum of charges of ions present in an ionic compound is equals to net charge of the compound.

Answer to Problem 44PS

$CuC{O}_3$ is composed of one $C{u}^{2+}$ ion and one $C{O}_3{}^{2-}$ ion.

Explanation of Solution

In the given ionic compound $CuC{O}_3$, there are $Cu$ ion and $C{O}_3$ ion present. Since the charge of $C{O}_3$ is $\text{2-}$, so the charge of $Cu$ must be $\text{2+}$.

The relative number of each ion is indicated in suffix of each ion.

That is,

One $C{u}^{2+}$ ion and one $C{O}_3{}^{2-}$ ion are present in $CuC{O}_3$ compound.

Interpretation Introduction

Interpretation:

For the given ionic compounds, the number and formula of constituted ions present are needed to be determined.

Concept introduction:

Ionic compounds are formed by electrostatic attraction between ions such as monoatomic ions and polyatomic ions; in which positively charged ions is known as cations and negatively charged ions are known as anions.

The sum of charges of ions present in an ionic compound is equals to net charge of the compound.

Answer to Problem 44PS

$Ti{(S{O}_4)}_2$ is composed of one $T{i}^{2+}$ ions and two $S{O}_4{}^{2-}$ ions.

Explanation of Solution

In the given ionic compound $Ti{(S{O}_4)}_2$, there are $Ti$ ion and $S{O}_4$ ion present. Since the charge of $S{O}_4$ is $\text{2-}$, then the charge of $Ti$ must be $\text{2+}$.

The relative number of each ion is indicated in suffix of each ion.

That is,

One $T{i}^{2+}$ ions and two $S{O}_4{}^{2-}$ ions are present in $Ti{(S{O}_4)}_2$ compound.

Interpretation Introduction

Interpretation:

For the given ionic compounds, the number and formula of constituted ions present are needed to be determined.

Concept introduction:

Ionic compounds are formed by electrostatic attraction between ions such as monoatomic ions and polyatomic ions; in which positively charged ions is known as cations and negatively charged ions are known as anions.

The sum of charges of ions present in an ionic compound is equals to net charge of the compound.

Answer to Problem 44PS

$K{H}_{2}P{O}_4$ is composed of one $K{H}_2{}^{3+}$ ion and one $P{O}_4{}^{3-}$ ion.

Explanation of Solution

In the given ionic compound $K{H}_{2}P{O}_4$, there are $oneKandtwoH$ ions and $P{O}_4$ ion present. Since the $K$ is group $\text{1A}$ element, then the charge will be $\text{1+}$, similarly the charge of hydrogen ions is $\text{1+}$ and the charge of $P{O}_4$ is $\text{3-}$. So there will be $oneKandtwoH$ for each $P{O}_4{}^{3-}$ ion.

The relative number of each ion is indicated in suffix of each ion.

That is,

One $one{K}^{+}andtwo{H}^{+}$ ion and one $P{O}_4{}^{3-}$ ion are present in $K{H}_{2}P{O}_4$ compound.

Interpretation Introduction

Interpretation:

For the given ionic compounds, the number and formula of constituted ions present are needed to be determined.

Concept introduction:

Ionic compounds are formed by electrostatic attraction between ions such as monoatomic ions and polyatomic ions; in which positively charged ions is known as cations and negatively charged ions are known as anions.

The sum of charges of ions present in an ionic compound is equals to net charge of the compound.

Answer to Problem 44PS

$CaHP{O}_4$ is composed of one $Ca{H}^{3+}$ ion and one $P{O}_4{}^{3-}$ ion.

Explanation of Solution

In the given ionic compound $CaHP{O}_4$, there are $CaandH$ ion and $P{O}_4$ ion present. Since the $Ca$ is group $\text{IIA}$ element, then the charge will be $\text{2+}$, similarly the charge of hydrogen ions is $\text{1+}$ and the charge $P{O}_4$ is $\text{3-}$. So there will be $oneCaandoneH$ for each $P{O}_4{}^{3-}$ ion.

The relative number of each ion is indicated in suffix of each ion.

That is,

One $oneC{a}^{2+}andone{H}^{+}$ ion and one $P{O}_4{}^{3-}$ ion are present in $CaHP{O}_4$ compound.