Chapter 35, Problem 1ASA

Each of the observations in the following list was made on a different solution. Given the observations, state which ion studied in this experiment is present. If the test is not definitive, indicate that with a question mark.

A. Addition of $6\mathrm{M\; NAOH}$ and $\mathrm{A}1$ to the solution produces a vapor that turns red litmus blue.

Ion present:

B. Addition $6\mathrm{M\; HCI}$ produces a vapor with an acrid odor.

Ion present:

C. Addition of $6\mathrm{M\; HCI}$ produces an effervescence.

Ion present:

D. Addition of $6\mathrm{M\; HN}{\mathrm{O}}_3$ plus $0.1\mathrm{M\; AgN}{\mathrm{O}}_3$ produces a precipitate.

Ion present:

E. Addition of $6\mathrm{M\; HN}{\mathrm{O}}_3$ plus $1\mathrm{M\; BaC}{\mathrm{l}}_2$ produces a precipitate.

Ion present:

F. Addition of $6\mathrm{M\; HN}{\mathrm{O}}_3$ plus $0.5\mathrm{M}{\left(\mathrm{N}{\mathrm{H}}_4\right)}_2\mathrm{Mo}{\mathrm{O}}_4$ produces a precipitate.

Ion present:

Interpretation Introduction

Interpretation:

To identify the ion present in the solution that gives the observation: The addition of 6 M NaOH and Al to the solution produces a vapor that turns red litmus blue.

Concept Introduction :

Different ions (Cations or anions) present in the solution can be detected using different tests as these ions behave differently in terms of their chemical reactions.

Answer to Problem 1ASA

Nitrate ion ( ${\text{NO}}_{\text{3}}^{-}$ ).

Explanation of Solution

This test is known as Devarda’s test and is given by nitrate ions ( ${\text{NO}}_{\text{3}}^{-}$ ).

In this test, Devarda’s alloy (Al alloy) acts as a reducing agent and is reacted with nitrate ion in the sodium hydroxide (NaOH) solution. As a result, ammonia gas is produced that turns red litmus blue.

The reaction taking place is-

  ${\text{3NO}}_{\text{3}}^{\text{-}}{\text{ (aq) + 8Al (s) + 5OH}}^{\text{-}}{\text{ (aq) + 18H}}_{\text{2}}\text{O (l)}\to {\text{3NH}}_{\text{3}}{\text{(g) + 8[Al(OH)}}_{\text{4}}{\text{]}}^{\text{-}}\text{(aq)}$

Hence, the ion present is the Nitrate ion.

Interpretation Introduction

Interpretation:

To identify the ion present in the solution that gives the observation: Adding 6 M HCl produces a vapor with an acrid odor.

Concept Introduction :

Different ions (Cations or anions) present in the solution can be detected using different tests as these ions behave differently in terms of their chemical reactions.

Answer to Problem 1ASA

Sulfite ion ( ${\text{SO}}_3^{\text{2-}}$ ).

Explanation of Solution

Acrid odor is a strong smell that may cause a burning sensation in the throat.

This acrid odor represents the release of Sulfur dioxide ( ${\text{SO}}_2$ ) gas.

And ${\text{SO}}_2$ is released when sulfite ions are treated with an acid.

  ${\text{SO}}_{\text{3}}^{\text{2-}}{\text{(aq) + 2H}}^{\text{+}}\text{(aq)}\to {\text{SO}}_{\text{2}}{\text{(g) + H}}_{\text{2}}\text{O(l)}$

Hence, the ion present is the sulfite ion.

Interpretation Introduction

Interpretation:

To identify the ion present in the solution that gives the observation: Adding 6 M HCl produces an effervescence.

Concept Introduction :

Different ions (Cations or anions) present in the solution can be detected using different tests as these ions behave differently in terms of their chemical reactions.

Answer to Problem 1ASA

Carbonate ion ( ${\text{CO}}_3^{\text{2-}}$ )

Explanation of Solution

Effervescence formation means carbon dioxide gas is released during this test and carbonate ions release ${\text{CO}}_2$ when treated with acids.

  ${\text{CO}}_{\text{3}}^{\text{2-}}{\text{(aq) + 2H}}^{\text{+}}\text{(aq)}\to {\text{CO}}_2{\text{(g) + H}}_{\text{2}}\text{O(l)}$

Hence, the ion present is the Carbonate ion.

Interpretation Introduction

Interpretation:

To identify the ion present in the solution that gives the observation: Adding 6 M ${\text{HNO}}_{\text{3}}^{}$ plus 0.1 M ${\text{AgNO}}_{\text{3}}^{}$ produces a precipitate.

Concept Introduction :

Different ions (Cations or anions) present in the solution can be detected using different tests as these ions behave differently in terms of their chemical reactions.

Answer to Problem 1ASA

Halide ions ( ${\text{Cl}}^{\text{-}}{\text{, Br}}^{\text{-}}{\text{or I}}^{\text{-}}$ )

Explanation of Solution

This test is a Silver nitrate test and is given by halide ions ( ${\text{Cl}}^{\text{-}}{\text{, Br}}^{\text{-}}{\text{or I}}^{\text{-}}$ ).

In this test, Silver ions react with halide ions to form the precipitate of silver halides. Nitric acid is added to remove interfering ions.

  $\begin{array}{l}{\text{Ag}}^{+}{\text{+Cl}}^{\text{-}}\to \text{AgCl (s)}\\ {\text{Ag}}^{+}{\text{+Br}}^{\text{-}}\to \text{AgBr (s)}\\ {\text{Ag}}^{+}{\text{+I}}^{-}\to \text{AgI (s)}\end{array}$

Hence, the ion present is either chloride ions or bromide ions, or iodide ions.

Interpretation Introduction

Interpretation:

To identify the ion present in the solution that gives the observation: Adding of 6 M ${\text{HNO}}_{\text{3}}^{}$ plus 1 M ${\text{BaCl}}_{\text{2}}$ produces a precipitate.

Concept Introduction :

Different ions (Cations or anions) present in the solution can be detected using different tests as these ions behave differently in terms of their chemical reactions.

Answer to Problem 1ASA

Sulfate ( ${\text{SO}}_4^{2-}$ ) ions

Explanation of Solution

Sulfate ( ${\text{SO}}_4^{2-}$ ) ions form a white precipitate with Barium ions in an acidic medium.

  ${\text{SO}}_{\text{4}}^{\text{2-}}{\text{(aq) + Ba}}^{\text{2+}}\text{(aq)}\to {\text{BaSO}}_{\text{4}}\text{ (s)}$

Hence, the ion present is a sulfate ion.

Interpretation Introduction

Interpretation:

To identify the ion present in the solution that gives the observation: Adding of 6 M ${\text{HNO}}_{\text{3}}^{}$ plus 0.5 M ${{\text{(NH}}_4\text{)}}_{\text{2}}\text{MoO}$ produces a precipitate.

Concept Introduction :

Different ions (Cations or anions) present in the solution can be detected using different tests as these ions behave differently in terms of their chemical reactions.

Answer to Problem 1ASA

Phosphate ( ${\text{PO}}_4^{3-}$ ) ion

Explanation of Solution

Phosphate ( ${\text{PO}}_4^{3-}$ ) ions form a yellow precipitate with ammonium molybdate in an acidic medium.

  ${\text{2PO}}_4^{\text{3-}}{\text{(aq) + 6H}}^{\text{+}}{\text{(aq) + 3(NH}}_4{\text{)}}_{\text{2}}\text{MoO (aq)}\to 2{(N{H}_4\text{)}}_3{\text{PO}}_4{\text{ (aq) + 3MoO}}_3{\text{ (s) + 3H}}_{\text{2}}\text{O(l)}$

Hence, the ion present is the phosphate ion.