Chapter 14, Problem 11E

Interpretation Introduction

(a)

Interpretation:

The reactants, acid and base, of salt sodium fluoride, $\text{NaF}\left(aq\right)$ are to be identified.

Concept introduction:

According to Arrhenius theory, an acid is defined as a species which donates a proton. A base is defined as a species which can donate a hydroxide ion $\left({\text{OH}}^{-}\right)$. Neutralization reaction is the reaction in which an acid and a base react to give water and salt of an acid and base.

Answer to Problem 11E

In the formation of salt sodium fluoride, $\text{NaF}\left(aq\right)$, acid is $\text{HF}$ and base is $\text{NaOH}$.

Explanation of Solution

Acids loses their ${\text{H}}^{\text{+}}$ ion and give their anionic part to the salt. In the given salt, $\text{NaF}\left(aq\right)$, sodium is the cation and fluoride is the anion. The $\text{HF}$ molecule always loses its proton and gives fluoride as the anion. Therefore, $\text{HF}$ acts as an acid. Base is a species which can release a ${\text{OH}}^{-}$ ion and give its cationic part to the salt. In the given reaction, $\text{NaOH}$ releases ${\text{OH}}^{-}$ ion and sodium acts as a cation. Therefore, $\text{HF}$ acts as an acid and $\text{NaOH}$ acts as a base.

The chemical equation is given below.

$\text{HF}\left(\text{aq}\right)+\text{NaOH(aq)}\to \text{NaF}\left(\text{aq}\right)+{\text{H}}_2\text{O}\left(l\right)$.

Conclusion

In the formation of salt sodium fluoride, $\text{NaF}\left(aq\right)$, acid is $\text{HF}$ and base is $\text{NaOH}$.

Interpretation Introduction

(b)

Interpretation:

The reactant, acid and base, of salt sodium fluoride, $\text{NaF}\left(aq\right)$ are to be identified.

Concept introduction:

According to Arrhenius theory, an acid is defined as a species which donates a proton. A base is defined as a species which can donate a hydroxide ion $\left({\text{OH}}^{-}\right)$. Neutralization reaction is the reaction in which an acid and a base react to give water and salt of an acid and base.

Answer to Problem 11E

In the formation of salt magnesium iodide, ${\text{MgI}}_{\text{2}}\left(aq\right)$, acid is $\text{HI}$ and base is $\text{Mg}{\left(\text{OH}\right)}_2$.

Explanation of Solution

Acids loses their ${\text{H}}^{\text{+}}$ ion and give their anionic part to the salt. In the given salt, ${\text{MgI}}_{\text{2}}\left(aq\right)$, magnesium is the cation and iodide is the anion. The $\text{HI}$ molecule always loses its proton and gives iodide as the anion. Therefore, $\text{HI}$ acts as an acid. Base is a species which can release a ${\text{OH}}^{-}$ ion and give its cationic part to the salt. In the given reaction, $\text{Mg}{\left(\text{OH}\right)}_2$ releases ${\text{OH}}^{-}$ ion and magnesium acts as a cation. Therefore, $\text{HI}$ acts as an acid and $\text{Mg}{\left(\text{OH}\right)}_2$ acts as a base.

$\text{2HI}\left(\text{aq}\right)+\text{Mg}{\left(\text{OH}\right)}_2\left(aq\right)\to {\text{MgI}}_2\left(aq\right)+2{\text{H}}_2\text{O}\left(l\right)$.

Conclusion

In the formation of salt magnesium iodide, ${\text{MgI}}_{\text{2}}\left(aq\right)$, acid is $\text{HI}$ and base is $\text{Mg}{\left(\text{OH}\right)}_2$.

Interpretation Introduction

(c)

Interpretation:

The reactant, acid and base, of salt calcium nitrate, $\text{Ca}{\left({\text{NO}}_{\text{3}}\right)}_2\left(aq\right)$ are to be identified.

Concept introduction:

According to Arrhenius theory, an acid is defined as a species which donates a proton. A base is defined as a species which can donate a hydroxide ion $\left({\text{OH}}^{-}\right)$. Neutralization reaction is the reaction in which an acid and a base react to give water and salt of an acid and base.

Answer to Problem 11E

In the formation of salt calcium nitrate, $\text{Ca}{\left({\text{NO}}_{\text{3}}\right)}_2\left(aq\right)$, acid is ${\text{HNO}}_3$ and base is $\text{Ca}{\left(\text{OH}\right)}_2$.

Explanation of Solution

Acids loses their ${\text{H}}^{\text{+}}$ ion and give their anionic part to the salt. In the given salt, $\text{Ca}{\left({\text{NO}}_{\text{3}}\right)}_2\left(aq\right)$, calcium is the cation and nitrate is the anion. The ${\text{HNO}}_3$ molecule always loses its proton and gives nitrate as the anion. Therefore, ${\text{HNO}}_3$ acts as an acid. Base is a species which can release a ${\text{OH}}^{-}$ ion and give its cationic part to the salt. In the given reaction, $\text{Ca}{\left(\text{OH}\right)}_2$ releases ${\text{OH}}^{-}$ ion and calcium acts as a cation. Therefore, ${\text{HNO}}_3$ acts as an acid and $\text{Ca}{\left(\text{OH}\right)}_2$ acts as a base.

${\text{2HNO}}_3\left(\text{aq}\right)+\text{Ca}{\left(\text{OH}\right)}_2\left(aq\right)\to \text{Ca}{\left({\text{NO}}_{\text{3}}\right)}_2\left(aq\right)+2{\text{H}}_2\text{O}\left(l\right)$.

Conclusion

In the formation of salt calcium nitrate, $\text{Ca}{\left({\text{NO}}_{\text{3}}\right)}_2\left(aq\right)$, acid is ${\text{HNO}}_3$ and base is $\text{Ca}{\left(\text{OH}\right)}_2$.

Interpretation Introduction

(d)

Interpretation:

The reactant, acid and base, of salt lithium carbonate, ${\text{Li}}_{\text{2}}{\text{CO}}_{\text{3}}\left(aq\right)$ are to be identified.

Concept introduction:

According to Arrhenius theory, an acid is defined as a species which donates a proton. A base is defined as a species which can donate a hydroxide ion $\left({\text{OH}}^{-}\right)$. Neutralization reaction is the reaction in which an acid and a base react to give water and salt of an acid and base.

Answer to Problem 11E

In the formation of salt lithium carbonate, ${\text{Li}}_{\text{2}}{\text{CO}}_{\text{3}}\left(aq\right)$, acid is ${\text{H}}_{\text{2}}{\text{CO}}_{\text{3}}$ and base is $\text{LiOH}$.

Explanation of Solution

Acids loses their ${\text{H}}^{\text{+}}$ ion and give their anionic part to the salt. In the given salt, ${\text{Li}}_{\text{2}}{\text{CO}}_{\text{3}}\left(aq\right)$, lithium is the cation and carbonate is the anion. The ${\text{H}}_{\text{2}}{\text{CO}}_{\text{3}}$ molecule always loses its proton and gives carbonate as the anion. Therefore, ${\text{H}}_{\text{2}}{\text{CO}}_{\text{3}}$ acts as an acid. Base is a species which can release a ${\text{OH}}^{-}$ ion and give its cationic part to the salt. In the given reaction, $\text{LiOH}$ releases ${\text{OH}}^{-}$ ion and lithium acts as a cation. Therefore, ${\text{H}}_{\text{2}}{\text{CO}}_{\text{3}}$ acts as an acid and $\text{LiOH}$ acts as a base.

The chemical equation is given below.

${\text{H}}_{\text{2}}{\text{CO}}_{\text{3}}\left(\text{aq}\right)+2\text{LiOH}\left(aq\right)\to {\text{Li}}_{\text{2}}{\text{CO}}_{\text{3}}\left(aq\right)+2{\text{H}}_2\text{O}\left(l\right)$.

Conclusion

In the formation of salt lithium carbonate, ${\text{Li}}_{\text{2}}{\text{CO}}_{\text{3}}\left(aq\right)$, acid is ${\text{H}}_{\text{2}}{\text{CO}}_{\text{3}}$ and base is $\text{LiOH}$.