Chapter 14, Problem 12E

Interpretation Introduction

(a)

Interpretation:

The reactant, acid and base, of salt potassium bromide, $\text{KBr}\left(aq\right)$ 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 12E

In the formation of salt potassium bromide, $\text{KBr}\left(aq\right)$, acid is $\text{HBr}$ and base is $\text{KOH}$.

Explanation of Solution

Acids loses their ${\text{H}}^{\text{+}}$ ion and give their anionic part to the salt. In the given salt, $\text{KBr}\left(aq\right)$, potassium is the cation and bromide is the anion. The $\text{HBr}$ molecule always loses its proton and gives bromide as the anion. Therefore, $\text{HBr}$ 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, $\text{KOH}$ releases ${\text{OH}}^{-}$ ion and potassium acts as a cation. Therefore, $\text{HBr}$ acts as an acid and $\text{KOH}$ acts as a base.

The chemical equation is given below.

$\text{HBr}\left(\text{aq}\right)+\text{KOH(aq)}\to \text{KBr}\left(\text{aq}\right)+{\text{H}}_2\text{O}\left(l\right)$.

Conclusion

In the formation of salt potassium bromide, $\text{KBr}\left(aq\right)$, acid is $\text{HBr}$ and base is $\text{KOH}$.

Interpretation Introduction

(b)

Interpretation:

The reactant, acid and base, of salt barium chloride, ${\text{BaCl}}_2\left(aq\right)$ 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 12E

In the formation of salt barium chloride, ${\text{BaCl}}_2\left(aq\right)$, acid is $\text{HCl}$ and base is $\text{Ba}{\left(\text{OH}\right)}_2$.

Explanation of Solution

Acids loses their ${\text{H}}^{\text{+}}$ ion and give their anionic part to the salt. In the given salt, ${\text{BaCl}}_2\left(aq\right)$, barium is the cation and chloride is the anion. The $\text{HCl}$ molecule always loses its proton and gives chloride as the anion. Base is a species which can release a ${\text{OH}}^{-}$ ion and gives cation to the salt. In the given reaction, $\text{Ba}{\left(\text{OH}\right)}_2$ releases the ${\text{OH}}^{-}$ ion and barium as a cation. Therefore, $\text{HCl}$ acts as an acid and $\text{Ba}{\left(\text{OH}\right)}_2$ acts as a base for salt ${\text{BaCl}}_2\left(aq\right)$.

The chemical equation is given below.

$\text{2HCl}\left(\text{aq}\right)+\text{Ba}{\left(\text{OH}\right)}_2\left(aq\right)\to {\text{BaCl}}_2\left(aq\right)+2{\text{H}}_2\text{O}\left(l\right)$.

Conclusion

In the formation of salt barium chloride, ${\text{BaCl}}_2\left(aq\right)$, acid is $\text{HCl}$ and base is $\text{Ba}{\left(\text{OH}\right)}_2$.

Interpretation Introduction

(c)

Interpretation:

The reactant, acid and base, of salt cobalt (II) sulfate, ${\text{CoSO}}_{\text{4}}\left(aq\right)$ 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 12E

In the formation of salt cobalt (II) sulfate, ${\text{CoSO}}_{\text{4}}\left(aq\right)$, acid is ${\text{H}}_2{\text{SO}}_4$ and base is $\text{CoO}$.

Explanation of Solution

Acids loses their ${\text{H}}^{\text{+}}$ ion and give their anionic part to the salt. In the given salt, ${\text{CoSO}}_{\text{4}}\left(aq\right)$, cobalt is the cation and sulphate is the anion. The ${\text{H}}_2{\text{SO}}_4$ molecule always loses its proton and gives sulphate as the anion. Oxides of metals are basic in nature. Metal of oxides reacts to give cation of the given salt. Therefore, $\text{CoO}$ acts as base and ${\text{H}}_2{\text{SO}}_4$ acts as an acid.

The chemical equation is given below.

${\text{H}}_2{\text{SO}}_4\left(\text{aq}\right)+\text{CoO}\left(aq\right)\to {\text{CoSO}}_4\left(aq\right)+{\text{H}}_2\text{O}\left(l\right)$.

Conclusion

In the formation of salt cobalt (II) sulfate, ${\text{CoSO}}_{\text{4}}\left(aq\right)$, acid is ${\text{H}}_2{\text{SO}}_4$ and base is $\text{CoO}$.

Interpretation Introduction

(d)

Interpretation:

The reactant, acid and base, of salt sodium phosphate, ${\text{Na}}_{\text{3}}{\text{PO}}_{\text{4}}\left(aq\right)$ 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 12E

In the formation of salt sodium phosphate, ${\text{Na}}_{\text{3}}{\text{PO}}_{\text{4}}\left(aq\right)$, acid is ${\text{H}}_3{\text{PO}}_4$ 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, ${\text{Na}}_{\text{3}}{\text{PO}}_{\text{4}}\left(aq\right)$, sodium is the cation and phosphate is the anion. The ${\text{H}}_3{\text{PO}}_4$ molecule always loses its proton and gives phosphate as the anion. Base is a species which can release a ${\text{OH}}^{-}$ ion and gives cation to the salt.. In the given reaction, $\text{NaOH}$ releases ${\text{OH}}^{-}$ ion and gives sodium ion as a cation to salt. Therefore, ${\text{H}}_3{\text{PO}}_4$ acts as an acid and $\text{NaOH}$ acts as a base.

The chemical equation is given below.

${\text{H}}_3{\text{PO}}_4\left(\text{aq}\right)+\text{3 NaOH}\left(aq\right)\to {\text{Na}}_{\text{3}}{\text{PO}}_{\text{4}}\left(aq\right)\left(aq\right)+{\text{3 H}}_{\text{2}}\text{O}\left(l\right)$.

Conclusion

In the formation of salt sodium phosphate, ${\text{Na}}_{\text{3}}{\text{PO}}_{\text{4}}\left(aq\right)$, acid is and base is $\text{NaOH}$.