Chapter 4, Problem 92AE

(a)

Interpretation Introduction

Interpretation: The three reagents that would form precipitate with chloride ion using solubility rules are to be stated.

Concept introduction: The solute’s extent of dissolution into givensolvent is designated as solubility. Certain substance’s solubility remains incomplete and their solubility extent is estimated via solubility rules. Through these rules, crystallization and precipitation corresponding to solutes can be easily predicted. The solute’s solubility also depends upon its respective highestconcentrations possible.

Answer to Problem 92AE

The three reagents that would form precipitate with chloride ion are ${\text{AgNO}}_3$ , $\text{Pb}{\left({\text{NO}}_3\right)}_2$ and $\text{Hg}{\left({\text{NO}}_3\right)}_2$ .

Explanation of Solution

In the given table, it is mentioned that most of the chloride salts are generally soluble. The exceptions are $\text{AgCl}$ , ${\text{PbCl}}_2$ and ${\text{Hg}}_2{\text{Cl}}_2$ which are insoluble. The table also indicates that most nitrates $\left({\text{NO}}_3^{-}\right)$ are also soluble, so the salts of ${\text{AgNO}}_3$ , $\text{Pb}{\left({\text{NO}}_3\right)}_2$ and ${\text{Hg}}_2{\left({\text{NO}}_3\right)}_2$ can act as reagents and thereby react with chloride ion $\left({\text{Cl}}^{-}\right)$ to yield respective precipitates. The reactions are as follows.

  $\begin{array}{c}{\text{AgNO}}_3\left(aq\right)+{\text{Cl}}^{-}\left(aq\right)\to \text{AgCl}\left(s\right)+{\text{NO}}_3^{-}\left(aq\right)\\ \text{Pb}{\left({\text{NO}}_3\right)}_2\left(aq\right)+2{\text{Cl}}^{-}\left(aq\right)\to {\text{PbCl}}_2\left(s\right)+2{\text{NO}}_3^{-}\left(aq\right)\\ {\text{Hg}}_2{\left({\text{NO}}_3\right)}_2\left(aq\right)+2{\text{Cl}}^{-}\left(aq\right)\to {\text{Hg}}_2{\text{Cl}}_2\left(s\right)+2{\text{NO}}_3^{-}\left(aq\right)\end{array}$

Therefore, ${\text{AgNO}}_3$ , $\text{Pb}{\left({\text{NO}}_3\right)}_2$ and ${\text{Hg}}_2{\left({\text{NO}}_3\right)}_2$ are the reagents that form precipitate with chloride ion.

(b)

Interpretation Introduction

Interpretation: The three reagents that would form precipitate with calcium ion using solubility rules are to be stated.

Concept introduction: The solute’s extent of dissolution into given solvent is designated as solubility. Certain substance’s solubility remains incomplete and their solubility extent is estimated via solubility rules. Through these rules, crystallization and precipitation corresponding to solutes can be easily predicted. The solute’s solubility also depends upon its respective highest concentrations possible.

Answer to Problem 92AE

The three reagents that would form precipitate with calcium ion are ${\text{Na}}_2{\text{SO}}_4$ , ${\text{Na}}_2{\text{CO}}_3$ and ${\text{Na}}_3{\text{PO}}_4$ .

Explanation of Solution

In the given table, it is mentioned that most of the sulfates salts are usually soluble. The exceptions include ${\text{CaSO}}_4$ . The table also predicts that most of ${\text{CO}}_3^{2-}$ and ${\text{PO}}_4^{3-}$ also have slight solubility. It means ${\text{CaCO}}_3$ and ${\text{Ca}}_3{\left({\text{PO}}_4\right)}_2$ will also lead to precipitate formation. It is given in table that most of ${\text{Na}}^{+}$ salts are generally soluble. So, ${\text{Na}}_2{\text{SO}}_4$ , ${\text{Na}}_2{\text{CO}}_3$ and ${\text{Na}}_3{\text{PO}}_4$ are the reagents that yields precipitate with calcium ions as follows.

  $\begin{array}{c}{\text{Na}}_2{\text{SO}}_4\left(aq\right)+{\text{Ca}}^{2+}\left(aq\right)\to {\text{CaSO}}_4\left(s\right)+2{\text{Na}}^{+}\\ {\text{Na}}_2{\text{CO}}_3\left(aq\right)+{\text{Ca}}^{2+}\left(aq\right)\to {\text{CaCO}}_3\left(s\right)+2{\text{Na}}^{+}\\ {\text{2Na}}_3{\text{PO}}_4\left(aq\right)+3{\text{Ca}}^{2+}\left(aq\right)\to {\text{Ca}}_3{\left({\text{PO}}_4\right)}_2\left(s\right)+6{\text{Na}}^{+}\end{array}$

Therefore, the three reagents that would form precipitate with calcium ion are ${\text{Na}}_2{\text{SO}}_4$ , ${\text{Na}}_2{\text{CO}}_3$ and ${\text{Na}}_3{\text{PO}}_4$ .

(c)

Interpretation Introduction

Interpretation: The three reagents that would form precipitate with iron (III) ion using solubility rules are to be stated.

Concept introduction: The solute’s extent of dissolution into given solvent is designated as solubility. Certain substance’s solubility remains incomplete and their solubility extent is estimated via solubility rules. Through these rules, crystallization and precipitation corresponding to solutes can be easily predicted. The solute’s solubility also depends upon its respective highest concentrations possible.

Answer to Problem 92AE

The three reagents that would form precipitate with iron (III) ion are $\text{NaOH}$ , ${\text{Na}}_2\text{S}$ and ${\text{Na}}_3{\text{PO}}_4$ .

Explanation of Solution

In the given table, it is mentioned that hydroxides are very slightly soluble expect for alkali as well as alkaline metal hydroxides such as $\text{NaOH}$ , $\text{KOH}$ and $\text{Ca}{\left(\text{OH}\right)}_2$ . It means that $\text{Fe}{\left(\text{OH}\right)}_{\text{3}}$ is insoluble. The table also most of ${\text{S}}^{2-}$ ions and ${\text{PO}}_4^{3-}$ ions also have slight solubility. So, $\text{Fe}{\left(\text{OH}\right)}_{\text{3}}$ , ${\text{Fe}}_2{\text{S}}_3$ and ${\text{FePO}}_4$ form precipitates. It is given in table that most of ${\text{Na}}^{+}$ salts are generally soluble. So, $\text{NaOH}$ , ${\text{Na}}_2\text{S}$ and ${\text{Na}}_3{\text{PO}}_4$ are the reagents that would form precipitate with iron (III) ion as follows.

  $\begin{array}{c}\text{3NaOH}\left(aq\right)+{\text{Fe}}^{3+}\left(aq\right)\to \text{Fe}{\left(\text{OH}\right)}_{\text{3}}\left(s\right)+3{\text{Na}}^{+}\left(aq\right)\\ {\text{3Na}}_2\text{S}\left(aq\right)+2{\text{Fe}}^{3+}\left(aq\right)\to {\text{Fe}}_2{\text{S}}_3\left(s\right)+6{\text{Na}}^{+}\left(aq\right)\\ {\text{Na}}_3{\text{PO}}_4\left(aq\right)+{\text{Fe}}^{3+}\left(aq\right)\to {\text{FePO}}_4\left(s\right)+3{\text{Na}}^{+}\left(aq\right)\end{array}$

Therefore, the three reagents that would form precipitate with iron (III) ion are $\text{NaOH}$ , ${\text{Na}}_2\text{S}$ and ${\text{Na}}_3{\text{PO}}_4$ .

(d)

Interpretation Introduction

Interpretation: The three reagents that would form precipitate with sulfate ion using solubility rules are to be stated.

Concept introduction: The solute’s extent of dissolution into given solvent is designated as solubility. Certain substance’s solubility remains incomplete and their solubility extent is estimated via solubility rules. Through these rules, crystallization and precipitation corresponding to solutes can be easily predicted. The solute’s solubility also depends upon its respective highest concentrations possible.

Answer to Problem 92AE

The three reagents that would form precipitate with sulfate ion are ${\text{BaCl}}_2$ , $\text{Pb}{\left({\text{NO}}_3\right)}_2$ and $\text{Ca}{\left({\text{NO}}_3\right)}_2$ .

Explanation of Solution

The given table indicates that most of sulfate salts are usually soluble. The exceptions include ${\text{BaSO}}_4$ , ${\text{PbSO}}_4$ and ${\text{CaSO}}_4$ . The table also tells that chloride salts are soluble except ${\text{PbCl}}_2$ . So, it cannot act as reagent for sulfate ion. Since most of the nitrate salts aregenerally soluble.One of the reagent can be $\text{Pb}{\left({\text{NO}}_3\right)}_2$ . The other reagent can be ${\text{BaCl}}_2$ . Since $\text{Ca}{\left({\text{NO}}_3\right)}_2$ is also soluble nitrate, it is the third reagent. So, the three reagents that would form precipitate with sulfate ion are ${\text{BaCl}}_2$ , $\text{Pb}{\left({\text{NO}}_3\right)}_2$ and $\text{Ca}{\left({\text{NO}}_3\right)}_2$ . The reactions are as follows.

  $\begin{array}{c}{\text{BaCl}}_2\left(aq\right)+{\text{SO}}_4^{2-}\left(aq\right)\to {\text{BaSO}}_4\left(s\right)+2{\text{Cl}}^{-}\left(aq\right)\\ \text{Pb}{\left({\text{NO}}_3\right)}_2\left(aq\right)+{\text{SO}}_4^{2-}\left(aq\right)\to {\text{PbSO}}_4\left(s\right)+2{\text{NO}}_3^{-}\left(aq\right)\\ \text{Ca}{\left({\text{NO}}_3\right)}_2\left(aq\right)+{\text{SO}}_4^{2-}\left(aq\right)\to {\text{CaSO}}_4\left(s\right)+2{\text{NO}}_3^{-}\left(aq\right)\end{array}$

Therefore, the three reagents that would form precipitate with sulfate ion are ${\text{BaCl}}_2$ , $\text{Pb}{\left({\text{NO}}_3\right)}_2$ and $\text{Ca}{\left({\text{NO}}_3\right)}_2$ .

(e)

Interpretation Introduction

Interpretation: The three reagents that would form precipitate with mercury (I) ion using solubility rules are to be stated.

Concept introduction: The solute’s extent of dissolution into given solvent is designated as solubility. Certain substance’s solubility remains incomplete and their solubility extent is estimated via solubility rules. Through these rules, crystallization and precipitation corresponding to solutes can be easily predicted. The solute’s solubility also depends upon its respective highest concentrations possible.

Answer to Problem 92AE

The three reagents that would form precipitate with mercury (I) ion are $\text{NaCl}$ , ${\text{Na}}_2{\text{SO}}_4$ and ${\text{Na}}_2\text{S}$ .

Explanation of Solution

In the given table, it is mentioned that, most chloride salts except ${\text{Hg}}_2{\text{Cl}}_2$ are insoluble. Most of the sulfides and sulfate salts are also very slightly soluble. The ${\text{Na}}^{+}$ salts are generally soluble, so the reagents can be $\text{NaCl}$ , ${\text{Na}}_2{\text{SO}}_4$ and ${\text{Na}}_2\text{S}$ to form precipitates ${\text{Hg}}_2{\text{Cl}}_2$ , ${\text{Hg}}_2{\text{SO}}_4$ and ${\text{Hg}}_2\text{S}$ . The reactions are as follows.

  $\begin{array}{c}\text{2NaCl}\left(aq\right)+{\text{Hg}}_2^{2+}\left(aq\right)\to {\text{Hg}}_2{\text{Cl}}_2\left(s\right)+2{\text{Na}}^{+}\left(aq\right)\\ {\text{Na}}_2{\text{SO}}_4\left(aq\right)+{\text{Hg}}_2^{2+}\left(aq\right)\to {\text{Hg}}_2{\text{SO}}_4\left(s\right)+2{\text{Na}}^{+}\left(aq\right)\\ {\text{Na}}_2\text{S}\left(aq\right)+{\text{Hg}}_2^{2+}\left(aq\right)\to {\text{Hg}}_2\text{S}\left(s\right)+2{\text{Na}}^{+}\left(aq\right)\end{array}$

Therefore, the three reagents that would form precipitate with mercury (I) ion are $\text{NaCl}$ , ${\text{Na}}_2{\text{SO}}_4$ and ${\text{Na}}_2\text{S}$ .

(f)

Interpretation Introduction

Interpretation: The three reagents that would form precipitate with silver ion using solubility rules are to be stated.

Concept introduction: The solute’s extent of dissolution into given solvent is designated as solubility. Certain substance’s solubility remains incomplete and their solubility extent is estimated via solubility rules. Through these rules, crystallization and precipitation corresponding to solutes can be easily predicted. The solute’s solubility also depends upon its respective highest concentrations possible.

Answer to Problem 92AE

The three reagents that would form precipitate with silver ions are $\text{NaCl}$ , ${\text{Na}}_2{\text{CO}}_3$ and ${\text{Na}}_3{\text{PO}}_4$ .

Explanation of Solution

In the given table, it is mentioned that, most chloride salts are soluble except $\text{AgCl}$ .The table also most of ${\text{CO}}_3^{2-}$ ions and ${\text{PO}}_4^{3-}$ ions also have slight solubility. So, ${\text{Ag}}_2{\text{CO}}_3$ and ${\text{Ag}}_3{\text{PO}}_4$ will also form precipitates. Since ${\text{Na}}^{+}$ salts are usually soluble, so the reagents are $\text{NaCl}$ , ${\text{Na}}_2{\text{CO}}_3$ and ${\text{Na}}_3{\text{PO}}_4$ to yield $\text{AgCl}$ , ${\text{Ag}}_2{\text{CO}}_3$ and ${\text{Ag}}_3{\text{PO}}_4$ as precipitates respectively. The reactions are as follows.

  $\begin{array}{c}\text{NaCl}\left(aq\right)+{\text{Ag}}^{+}\left(aq\right)\to \text{AgCl}\left(s\right)+{\text{Na}}^{+}\left(aq\right)\\ {\text{Na}}_2{\text{CO}}_3\left(aq\right)+2{\text{Ag}}^{+}\left(aq\right)\to {\text{Ag}}_2{\text{CO}}_3\left(s\right)+2{\text{Na}}^{+}\left(aq\right)\\ {\text{Na}}_3{\text{PO}}_4\left(aq\right)+3{\text{Ag}}^{+}\left(aq\right)\to {\text{Ag}}_3{\text{PO}}_4\left(s\right)+3{\text{Na}}^{+}\left(aq\right)\end{array}$

Therefore, the three reagents that would form precipitate with silver ions are $\text{NaCl}$ , ${\text{Na}}_2{\text{CO}}_3$ and ${\text{Na}}_3{\text{PO}}_4$ .