Chapter 3, Problem 107QRT

(a)

Interpretation Introduction

Interpretation:

The ${\text{NaOH(aq) + H}}_{\text{3}}{\text{PO}}_{\text{4}}\text{(aq) }\to {\text{ NaH}}_{\text{2}}{\text{PO}}_{\text{4}}{\text{(aq) + H}}_{\text{2}}\text{O(l)}$ given reaction is redox reactions, acid-base reactions or gas-forming reaction has to be identified. If the given reaction is redox reactions, the oxidizing and reducing agents has to be identified. If the given reaction is the acid and base reaction, acid and base has to be identified.

Concept introduction:

Acids and Bases:

Acid release hydrogen ion in water (or) proton donor.

Base release hydroxide ions in water (or) proton acceptor.

Acid - base reaction: Formation of the salt from the cation from the base and anion from the acid and formation of water is also the product.

Gas forming reaction: The reaction of acid and metal carbonates which produce carbonic acid. The carbonic acid decomposes which gives water and carbon dioxide.

Oxidation - reduction reaction (Redox reaction): The electrons are transferred to one to other is called oxidation reduction reaction.

Oxidation number:

The oxidation number of an element is zero. The oxidation number of a monoatomic ion equals its charge.

The oxidation number is zero for the summation of the oxidation numbers complete atoms in a complete formula.

The charge on the ion is equal to the summation of the oxidation numbers of complete atoms in poly atomic ion.

The oxidation state of alkali metal $(\text{Li, Na, K, Rb, Cs})$ is $+1$, and alkaline earth metal is $\text{(Be, Mg, Ca, Sr, Ba)}$ is $+2$.

The oxidation state of hydrogen is $+1$ (without bonding with metals), the oxidation state of oxygen is $-2$.

An oxidizing agent gains the electrons and is reduced in a chemical reaction and it is electron acceptor.

A reducing agent loses electrons and is oxidized in a chemical reaction and it is electron donor.

The oxidation state is called as oxidation number, which describes degree of oxidation (loss of electrons) of an atom in a chemical compound. Theoretically, the oxidation state is positive, negative or zero.

Explanation of Solution

The given equation is shown below,

  ${\text{NaOH(aq) + H}}_{\text{3}}{\text{PO}}_{\text{4}}\text{(aq) }\to {\text{ NaH}}_{\text{2}}{\text{PO}}_{\text{4}}{\text{(aq) + H}}_{\text{2}}\text{O(l)}$

Oxidation number of sodium is $+1$ in the reactant, the oxidation number of phosphorous in the reactant is given below,

  $\begin{array}{l}{\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}\text{:}\\ \text{3(1)}\text{+}\text{x}\text{+4(-2)=}\text{0}\\ \text{3 + x - 8 =}\text{0}\\ \text{x}\text{=}\text{+5}\end{array}$

The oxidation number of phosphorous is $\text{+}\text{5}$.

Oxidation number of phosphorous in the product is given below,

  $\begin{array}{l}{\text{NaH}}_{\text{2}}{\text{PO}}_{\text{4}}\text{:}\\ \text{+1+ 2(1)+ x + 4(-2)}\text{=}0\\ \text{+3 + x - 8=}0\\ \text{x =}\text{+ 5}\end{array}$

The oxidation number of phosphorous is $\text{+}\text{5}$.

Oxidation number of phosphorous is $\text{+}\text{5}$ in the reactant and the product therefore, the given reaction is not redox reaction.

${\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}$ release hydrogen ion (or) proton donor in the reaction. Therefore, ${\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}$ is an acid. $\text{NaOH}$ release hydroxide ion. Therefore, $\text{NaOH}$ is a base. Hence it is acid base reaction.

(b)

Interpretation Introduction

Interpretation:

The ${\text{NH}}_{\text{3}}{\text{(g) + CO}}_{\text{2}}{\text{(g) + H}}_{\text{2}}\text{O(l) }\to {\text{ NH}}_{\text{4}}{\text{HCO}}_{\text{3}}\text{(aq)}$ given reaction is redox reactions, acid-base reactions or gas-forming reaction has to be identified. If the given reaction is redox reactions, the oxidizing and reducing agents has to be identified. If the given reaction is the acid and base reaction, acid and base has to be identified.

Concept introduction:

Refer to part (a)

Explanation of Solution

The given equation is shown below,

  ${\text{NH}}_{\text{3}}{\text{(g) + CO}}_{\text{2}}{\text{(g) + H}}_{\text{2}}\text{O(l) }\to {\text{ NH}}_{\text{4}}{\text{HCO}}_{\text{3}}\text{(aq)}$

The oxidation number of nitrogen in the reactant is given below,

  $\begin{array}{l}{\text{NH}}_{\text{3}}\text{:}\\ \text{3(1)}\text{+}\text{x}\text{=}\text{0}\\ \text{3 + x =}\text{0}\\ \text{x}\text{=}\text{-3}\end{array}$

  $\begin{array}{l}{\text{CO}}_{\text{2}}\text{:}\\ \text{x}\text{+}\text{2(-2)}\text{=}\text{0}\\ \text{x}\text{- 4 =}\text{0}\\ \text{x}\text{=}\text{+ 4}\end{array}$

The oxidation number of nitrogen is $-3$ and the oxidation state of carbon is $\text{+}4$ in the reactant.

The oxidation number of nitrogen in the product is given below,

${\text{NH}}_{\text{4}}{\text{HCO}}_{\text{3}}\text{(aq)}$ dissociates gives ${\text{NH}}_4^{+}{\text{ and HCO}}_3^{-}$.

  $\begin{array}{l}{\text{NH}}_4^{+}\text{:}\\ \text{4(1)}\text{+}\text{x}\text{=}\text{+1}\\ \text{4 + x =}+1\\ \text{x}\text{=}\text{-3}\end{array}$

    $\begin{array}{l}{\text{HCO}}_3^{-}\text{:}\\ \text{1(1)}\text{+ }\text{x + 3(-2)}\text{=}-\text{1}\\ \text{-5 + x =}+1\\ \text{x}\text{=}+4\end{array}$

The oxidation number of nitrogen is $-3$ and the oxidation state of carbon is $\text{+}4$ in the reactant as well as in the product, therefore, the given reaction is not redox reaction.

The combination of ${\text{H}}_{\text{2}}{\text{O and CO}}_{\text{2}}$ behaves as an acid. ${\text{NH}}_{\text{3}}$ is base. Hence it is acid base reaction.

(c)

Interpretation Introduction

Interpretation:

The ${\text{TiCl}}_{\text{4}}\text{(g) + 2 Mg(l) }\to {\text{ Ti(s) + 2 MgCl}}_{\text{2}}\text{(l)}$ given reaction is redox reactions, acid-base reactions or gas-forming reaction has to be identified. If the given reaction is redox reactions, the oxidizing and reducing agents has to be identified. If the given reaction is the acid and base reaction, acid and base has to be identified.

Concept introduction:

Refer to part (a)

Explanation of Solution

The given equation is shown below,

  ${\text{TiCl}}_{\text{4}}\text{(g) + 2 Mg(l) }\to {\text{ Ti(s) + 2 MgCl}}_{\text{2}}\text{(l)}$

The oxidation number of titanium in the reactant is given below,

  $\begin{array}{l}{\text{TiCl}}_4\text{:}\\ \text{x}\text{+ 4(-1)}\text{=}\text{0}\\ \text{x}\text{=}+4\end{array}$

The oxidation number of an element is zero. Therefore, the oxidation number of magnesium is zero in the reactant.

The oxidation number of titanium is $+4$ and the oxidation state of magnesium is zero in the reactant.

The oxidation number of magnesium in the product is given below,

  $\begin{array}{l}{\text{MgCl}}_2\text{:}\\ \text{x}\text{+ 2(-1)}\text{=}\text{0}\\ \text{x}\text{=}+2\end{array}$

The oxidation number of an element is zero. Therefore, the oxidation number of titanium is zero in the reactant.

The oxidation number of magnesium is $+2$ and the oxidation state of titanium is zero in the reactant.

The electrons get transferred from the reactant to product therefore the given reaction is redox reaction.

(d)

Interpretation Introduction

Interpretation:

The ${\text{NaCl(s) + NaHSO}}_{\text{4}}\text{(aq)}\to {\text{ HCl(g) + Na}}_{\text{2}}{\text{SO}}_{\text{4}}\text{(aq)}$ given reaction is redox reactions, acid-base reactions or gas-forming reaction has to be identified. If the given reaction is redox reactions, the oxidizing and reducing agents has to be identified. If the given reaction is the acid and base reaction, acid and base has to be identified.

Concept introduction:

Refer to part (a)

Explanation of Solution

The given equation is shown below,

  ${\text{NaCl(s) + NaHSO}}_{\text{4}}\text{(aq)}\to {\text{ HCl(g) + Na}}_{\text{2}}{\text{SO}}_{\text{4}}\text{(aq)}$

The oxidation number of sulfur in the reactant is given below,

${\text{NaHSO}}_{\text{4}}\text{(aq)}$ dissociates gives ${\text{Na}}^{+}{\text{ and HSO}}_4^{-}$.

    $\begin{array}{l}{\text{HSO}}_4^{-}\text{:}\\ \text{1(1)}\text{+ }\text{x + 4(-2)}\text{=}-\text{1}\\ \text{-7 + x =}+1\\ \text{x}\text{=}+6\end{array}$

The oxidation number of sodium is $\text{+}1$, chlorine is $-1$ and sulfur is $\text{+}6$ in the reactant.

The oxidation number of sulfur in the product is given below,

${\text{Na}}_2{\text{SO}}_4\text{(aq)}$ dissociates gives ${\text{2Na}}^{+}{\text{ and SO}}_4^{2-}$.

    $\begin{array}{l}{\text{SO}}_4^{2-}\text{:}\\ \text{x + 4(-2)}\text{=}-2\\ \text{-8 + x =}-2\\ \text{x}\text{=}+6\end{array}$

The oxidation number of sodium is $\text{+}1$, chlorine is $-1$ and sulfur is $\text{+}6$ in the product.

T The oxidation number of sodium is $\text{+}1$, chlorine is $-1$ and sulfur is $\text{+}6$ in the reactant as well as in the product, therefore, the given reaction is not redox reaction, and it not an acid base reaction

The reaction of metal sulfates and produces acids as a gas $\left(\text{HCl}\right)$. Hence, it is a gas forming reaction.