Chapter 7, Problem 1ALQ

onsider the mixing of aqueous solutions of lead(II) nitrate and sodium iodide to form a solid.

. Name the possible products, and determine the formulas of these possible products.

. What is the precipitate? How do you know?

. Must the subscript for an ion in a reactant stay the same as the subscript of that ion in a product? Explain your answer.

Interpretation Introduction

(a)

Interpretation:

To write the name of product and formula of product when lead nitrate, ${\text{Pb(NO}}_{\text{3}}{\text{)}}_{\text{2}}(aq)$ and sodium iodide, $\text{NaI }(aq)$ solutions are mixed.

Concept Introduction:

A balanced chemical equation is an equation that contains same number of atoms as well as of each element of reactants and products of reaction.

For example, the reaction between lead sulphide and oxygen is as follows:

$\begin{array}{l}2\text{ PbS }+{\text{ 3 O}}_2\to 2\text{PbO }+{\text{ 2SO}}_{\text{2}}\\ \text{ Reactants Products}\end{array}$

In balanced chemical equation the numbers of atoms of each element on each side means product as well reactants of reaction are always equal. Calculate the number of atoms of reactants and products as follows:

$\begin{array}{l}\underset{\_}{\text{Element Reactants Products }}\\ \text{Pb }22\\ \text{S }22\\ \text{O }66\end{array}$

In the given reaction, the number of all atoms, on the both sides are equal hence this is a balance reaction.

The most common driving forces which create product in chemical reactions are as follows:

1. Formation of a solid.
2. Formation of water.
3. Transfer of electrons.
4. Formation of gas.

Answer to Problem 1ALQ

Lead iodide ${\text{PbI}}_2$ and sodium nitrate ${\mathrm{NaNO}}_3$.

Explanation of Solution

Precipitation reaction means formation of solids or formation of any Precipitate; when solutions of two ionic substances are mixed and any solid will forms in the solution mixture, the reaction is known as Precipitation reaction.

The name of product and formula of product when lead nitrate, ${\text{Pb(NO}}_{\text{3}}{\text{)}}_{\text{2}}(aq)$ and sodium iodide, $\text{NaI }(aq)$ solutions is as follows:

${\text{Pb(NO}}_{\text{3}}{\text{)}}_{\text{2}}(aq)+2\text{N}\text{a}\text{I}(aq)\to {\text{PbI}}_2(s)+\text{2 }{\mathrm{NaNO}}_3(aq)$.

Interpretation Introduction

(b)

Interpretation:

The precipitate formed on mixing of lead nitrate, ${\text{Pb(NO}}_{\text{3}}{\text{)}}_{\text{2}}(aq)$ and sodium iodide, $\text{NaI }(aq)$ solutions should be determined.

Concept Introduction:

There are following solubility rules:

1. The salts formed from sodium, potassium and ammonium ions are soluble in nature.

All sulfate salts are soluble except ${\text{CaSO}}_4$, ${\text{PbSO}}_4$ and ${\text{BaSO}}_4$

All nitrate salts are soluble. The formula of nitrate polyatomic ion is ${\text{NO}}_3^{-}$.

All the salts formed from chloride, bromide and iodide are soluble except those with ${\text{Pb}}^{2+}$ and ${\text{Ag}}^{+}$.

All sulfate salts are insoluble except ${\text{Na}}_{\text{2}}{\text{SO}}_{\text{3}}$, ${\text{K}}_{\text{2}}{\text{SO}}_{\text{3}}$ and ${\left({\text{NH}}_4\right)}_{\text{2}}{\text{SO}}_{\text{3}}$.

All sulfides are insoluble except ${\text{Na}}_{\text{2}}\text{S}$, ${\text{K}}_{\text{2}}\text{S}$ and ${\left({\text{NH}}_4\right)}_{\text{2}}\text{S}$.

All oxides and hydroxides are insoluble except ${\text{Na}}_{\text{2}}\text{O}$ /NaOH, ${\text{K}}_{\text{2}}\text{O}$ /KOH and $\text{BaO}/\text{Ba}{\left(\text{OH}\right)}_{\text{2}}$.

All carbonates are insoluble except ${\text{Na}}_{\text{2}}{\text{CO}}_3$, ${\text{K}}_{\text{2}}{\text{CO}}_{\text{3}}$ and ${\left({\text{NH}}_4\right)}_{\text{2}}{\text{CO}}_{\text{3}}$.

Answer to Problem 1ALQ

${\mathrm{PbI}}_2$.

Explanation of Solution

According to rule, all the salts formed from chloride, bromide and iodide are soluble except those with ${\text{Pb}}^{2+}$ and ${\text{Ag}}^{+}$.

When lead nitrate, ${\text{Pb(NO}}_{\text{3}}{\text{)}}_{\text{2}}(aq)$ and sodium iodide, $\text{NaI }(aq)$ solutions are mixed then lead iodide, a solid is formed as precipitate.

${\text{Pb(NO}}_{\text{3}}{\text{)}}_{\text{2}}(aq)+2\text{N}\text{a}\text{I}(aq)\to {\text{PbI}}_2(s)\downarrow +\text{2 }{\mathrm{NaNO}}_3(aq)$

Therefore, precipitate formed is ${\mathrm{PbI}}_2$.

Interpretation Introduction

(c)

Interpretation:

When lead nitrate, ${\text{Pb(NO}}_{\text{3}}{\text{)}}_{\text{2}}(aq)$ and sodium iodide, $\text{NaI }(aq)$ solutions are mixed then why the subscript for an ion in both side reactant and product stay same should be explained.

Concept Introduction:

Molecular equation:

The equation which shows all of reactants and products in molecular or un-dissociated form is known as Molecular equation. For example, the Molecular equation of solutions of potassium bromide and silver nitrate is as follows:

$\text{KBr }(aq)+{\text{ AgNO}}_{\text{3}}(aq)\to {\text{ KNO}}_{\text{3}}(aq)+\text{ AgBr }(s)$

Complete ionic equation:

In the complete ionic equation, the strong electrolytes indicate by as ions. For example, the complete ionic equation of solutions of potassium bromide and silver nitrate is following

$\begin{array}{l}{\text{K}}^{\text{+ }}(aq)+{\text{Br}}^{-}(aq)+{\text{Ag}}^{+}(aq)+{\text{NO}}_3^{-}(aq)\to {\text{K}}^{+}(aq)+{\text{NO}}_3^{-}(aq)+\text{AgBr }(s)\\ \end{array}$

Net ionic equation:

In the net ionic equation there is no any spectator ion, only those species or ions which undergoes change is present called net ionic equation. For example, the net ionic equation of solutions of potassium bromide and silver nitrate is following:

${\text{ Br}}^{-}(aq)+{\text{ Ag}}^{+}(aq)\to \text{AgBr }(s)$.

Answer to Problem 1ALQ

To balance the number of atoms and charge.

Explanation of Solution

When lead nitrate, ${\text{Pb(NO}}_{\text{3}}{\text{)}}_{\text{2}}(aq)$ and sodium iodide, $\text{NaI }(aq)$ solutions are mixed then the subscript for an ion in both side reactant and product stay same to balance the charge. The reaction is as follows:

${\text{Pb(NO}}_{\text{3}}{\text{)}}_{\text{2}}(aq)+\text{ 2 NaI}(aq)\to {\text{PbI}}_2(s)+{\text{2 NaNO}}_{\text{3}}(aq)$

For the above reaction, the complete ionic equation can be written as follows:

${\text{Pb}}^{2+}(aq)+2{\text{NO}}_3^{+}(aq){\text{ + 2 Na}}^{+}(aq){\text{+ 2 I}}^{-}(aq)\to {\text{PbI}}_2(s)++2{\text{NO}}_3^{+}(aq){\text{ + 2 Na}}^{+}(aq)$

Here, subscript of each ions on left side are equal to the subscript of the same ion on right side to balance the number of atoms and charge on both side of the reaction arrow.