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 M NAOH and A 1 to the solution produces a vapor that turns red litmus blue. Ion present: B. Addition 6 M HCI produces a vapor with an acrid odor. Ion present: C. Addition of 6 M HCI produces an effervescence. Ion present: D. Addition of 6 M HN O 3 plus 0.1 M AgN O 3 produces a precipitate. Ion present: E. Addition of 6 M HN O 3 plus 1 M BaC l 2 produces a precipitate. Ion present: F. Addition of 6 M HN O 3 plus 0.5 M N H 4 2 Mo O 4 produces a precipitate. Ion present:

Question

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Answer 1

Textbook Question

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 $6 M NAOH$ and $A 1$ to the solution produces a vapor that turns red litmus blue.

Ion present:

B. Addition $6 M HCI$ produces a vapor with an acrid odor.

Ion present:

C. Addition of $6 M HCI$ produces an effervescence.

Ion present:

D. Addition of $6 M HN O 3$ plus $0.1 M AgN O 3$ produces a precipitate.

Ion present:

E. Addition of $6 M HN O 3$ plus $1 M BaC l 2$ produces a precipitate.

Ion present:

F. Addition of $6 M HN O 3$ plus $0.5 M N H 4 2 Mo O 4$ produces a precipitate.

Ion present:

(a)

Expert Solution

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 ( $NO3−$ ).

Explanation of Solution

This test is known as Devarda’s test and is given by nitrate ions ( $NO3−$ ).

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-

$3NO3- (aq) + 8Al (s) + 5OH- (aq) + 18H2O (l)→3NH3(g) + 8[Al(OH)4]-(aq)$

Hence, the ion present is the Nitrate ion.

(b)

Expert Solution

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 ( $SO32-$ ).

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 ( $SO2$ ) gas.

And $SO2$ is released when sulfite ions are treated with an acid.

$SO32-(aq) + 2H+(aq)→SO2(g) + H2O(l)$

Hence, the ion present is the sulfite ion.

(C)

Expert Solution

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 ( $CO32-$ )

Explanation of Solution

Effervescence formation means carbon dioxide gas is released during this test and carbonate ions release $CO2$ when treated with acids.

$CO32-(aq) + 2H+(aq)→CO2(g) + H2O(l)$

Hence, the ion present is the Carbonate ion.

(D)

Expert Solution

Interpretation Introduction

Interpretation:

To identify the ion present in the solution that gives the observation: Adding 6 M $HNO3$ plus 0.1 M $AgNO3$ 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 ( $Cl-, Br-or I-$ )

Explanation of Solution

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

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

$Ag++Cl-→AgCl (s)Ag++Br-→AgBr (s)Ag++I−→AgI (s)$

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

(E)

Expert Solution

Interpretation Introduction

Interpretation:

To identify the ion present in the solution that gives the observation: Adding of 6 M $HNO3$ plus 1 M $BaCl2$ 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 ( $SO42−$ ) ions

Explanation of Solution

Sulfate ( $SO42−$ ) ions form a white precipitate with Barium ions in an acidic medium.

$SO42-(aq) + Ba2+(aq)→BaSO4 (s)$

Hence, the ion present is a sulfate ion.

(F)

Expert Solution

Interpretation Introduction

Interpretation:

To identify the ion present in the solution that gives the observation: Adding of 6 M $HNO3$ plus 0.5 M $(NH4)2MoO$ 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 ( $PO43−$ ) ion

Explanation of Solution

Phosphate ( $PO43−$ ) ions form a yellow precipitate with ammonium molybdate in an acidic medium.

$2PO43-(aq) + 6H+(aq) + 3(NH4)2MoO (aq)→2(NH4)3PO4 (aq) + 3MoO3 (s) + 3H2O(l)$

Hence, the ion present is the phosphate ion.

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Answer 2

Textbook Question

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 $6 M NAOH$ and $A 1$ to the solution produces a vapor that turns red litmus blue.

Ion present:

B. Addition $6 M HCI$ produces a vapor with an acrid odor.

Ion present:

C. Addition of $6 M HCI$ produces an effervescence.

Ion present:

D. Addition of $6 M HN O 3$ plus $0.1 M AgN O 3$ produces a precipitate.

Ion present:

E. Addition of $6 M HN O 3$ plus $1 M BaC l 2$ produces a precipitate.

Ion present:

F. Addition of $6 M HN O 3$ plus $0.5 M N H 4 2 Mo O 4$ produces a precipitate.

Ion present:

(a)

Expert Solution

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 ( $NO3−$ ).

Explanation of Solution

This test is known as Devarda’s test and is given by nitrate ions ( $NO3−$ ).

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-

$3NO3- (aq) + 8Al (s) + 5OH- (aq) + 18H2O (l)→3NH3(g) + 8[Al(OH)4]-(aq)$

Hence, the ion present is the Nitrate ion.

(b)

Expert Solution

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 ( $SO32-$ ).

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 ( $SO2$ ) gas.

And $SO2$ is released when sulfite ions are treated with an acid.

$SO32-(aq) + 2H+(aq)→SO2(g) + H2O(l)$

Hence, the ion present is the sulfite ion.

(C)

Expert Solution

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 ( $CO32-$ )

Explanation of Solution

Effervescence formation means carbon dioxide gas is released during this test and carbonate ions release $CO2$ when treated with acids.

$CO32-(aq) + 2H+(aq)→CO2(g) + H2O(l)$

Hence, the ion present is the Carbonate ion.

(D)

Expert Solution

Interpretation Introduction

Interpretation:

To identify the ion present in the solution that gives the observation: Adding 6 M $HNO3$ plus 0.1 M $AgNO3$ 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 ( $Cl-, Br-or I-$ )

Explanation of Solution

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

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

$Ag++Cl-→AgCl (s)Ag++Br-→AgBr (s)Ag++I−→AgI (s)$

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

(E)

Expert Solution

Interpretation Introduction

Interpretation:

To identify the ion present in the solution that gives the observation: Adding of 6 M $HNO3$ plus 1 M $BaCl2$ 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 ( $SO42−$ ) ions

Explanation of Solution

Sulfate ( $SO42−$ ) ions form a white precipitate with Barium ions in an acidic medium.

$SO42-(aq) + Ba2+(aq)→BaSO4 (s)$

Hence, the ion present is a sulfate ion.

(F)

Expert Solution

Interpretation Introduction

Interpretation:

To identify the ion present in the solution that gives the observation: Adding of 6 M $HNO3$ plus 0.5 M $(NH4)2MoO$ 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 ( $PO43−$ ) ion

Explanation of Solution

Phosphate ( $PO43−$ ) ions form a yellow precipitate with ammonium molybdate in an acidic medium.

$2PO43-(aq) + 6H+(aq) + 3(NH4)2MoO (aq)→2(NH4)3PO4 (aq) + 3MoO3 (s) + 3H2O(l)$

Hence, the ion present is the phosphate ion.

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Answer 3

Textbook Question

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 $6 M NAOH$ and $A 1$ to the solution produces a vapor that turns red litmus blue.

Ion present:

B. Addition $6 M HCI$ produces a vapor with an acrid odor.

Ion present:

C. Addition of $6 M HCI$ produces an effervescence.

Ion present:

D. Addition of $6 M HN O 3$ plus $0.1 M AgN O 3$ produces a precipitate.

Ion present:

E. Addition of $6 M HN O 3$ plus $1 M BaC l 2$ produces a precipitate.

Ion present:

F. Addition of $6 M HN O 3$ plus $0.5 M N H 4 2 Mo O 4$ produces a precipitate.

Ion present:

(a)

Expert Solution

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 ( $NO3−$ ).

Explanation of Solution

This test is known as Devarda’s test and is given by nitrate ions ( $NO3−$ ).

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-

$3NO3- (aq) + 8Al (s) + 5OH- (aq) + 18H2O (l)→3NH3(g) + 8[Al(OH)4]-(aq)$

Hence, the ion present is the Nitrate ion.

(b)

Expert Solution

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 ( $SO32-$ ).

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 ( $SO2$ ) gas.

And $SO2$ is released when sulfite ions are treated with an acid.

$SO32-(aq) + 2H+(aq)→SO2(g) + H2O(l)$

Hence, the ion present is the sulfite ion.

(C)

Expert Solution

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 ( $CO32-$ )

Explanation of Solution

Effervescence formation means carbon dioxide gas is released during this test and carbonate ions release $CO2$ when treated with acids.

$CO32-(aq) + 2H+(aq)→CO2(g) + H2O(l)$

Hence, the ion present is the Carbonate ion.

(D)

Expert Solution

Interpretation Introduction

Interpretation:

To identify the ion present in the solution that gives the observation: Adding 6 M $HNO3$ plus 0.1 M $AgNO3$ 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 ( $Cl-, Br-or I-$ )

Explanation of Solution

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

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

$Ag++Cl-→AgCl (s)Ag++Br-→AgBr (s)Ag++I−→AgI (s)$

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

(E)

Expert Solution

Interpretation Introduction

Interpretation:

To identify the ion present in the solution that gives the observation: Adding of 6 M $HNO3$ plus 1 M $BaCl2$ 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 ( $SO42−$ ) ions

Explanation of Solution

Sulfate ( $SO42−$ ) ions form a white precipitate with Barium ions in an acidic medium.

$SO42-(aq) + Ba2+(aq)→BaSO4 (s)$

Hence, the ion present is a sulfate ion.

(F)

Expert Solution

Interpretation Introduction

Interpretation:

To identify the ion present in the solution that gives the observation: Adding of 6 M $HNO3$ plus 0.5 M $(NH4)2MoO$ 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 ( $PO43−$ ) ion

Explanation of Solution

Phosphate ( $PO43−$ ) ions form a yellow precipitate with ammonium molybdate in an acidic medium.

$2PO43-(aq) + 6H+(aq) + 3(NH4)2MoO (aq)→2(NH4)3PO4 (aq) + 3MoO3 (s) + 3H2O(l)$

Hence, the ion present is the phosphate ion.

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Answer 4

Textbook Question

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 $6 M NAOH$ and $A 1$ to the solution produces a vapor that turns red litmus blue.

Ion present:

B. Addition $6 M HCI$ produces a vapor with an acrid odor.

Ion present:

C. Addition of $6 M HCI$ produces an effervescence.

Ion present:

D. Addition of $6 M HN O 3$ plus $0.1 M AgN O 3$ produces a precipitate.

Ion present:

E. Addition of $6 M HN O 3$ plus $1 M BaC l 2$ produces a precipitate.

Ion present:

F. Addition of $6 M HN O 3$ plus $0.5 M N H 4 2 Mo O 4$ produces a precipitate.

Ion present:

(a)

Expert Solution

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 ( $NO3−$ ).

Explanation of Solution

This test is known as Devarda’s test and is given by nitrate ions ( $NO3−$ ).

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-

$3NO3- (aq) + 8Al (s) + 5OH- (aq) + 18H2O (l)→3NH3(g) + 8[Al(OH)4]-(aq)$

Hence, the ion present is the Nitrate ion.

(b)

Expert Solution

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 ( $SO32-$ ).

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 ( $SO2$ ) gas.

And $SO2$ is released when sulfite ions are treated with an acid.

$SO32-(aq) + 2H+(aq)→SO2(g) + H2O(l)$

Hence, the ion present is the sulfite ion.

(C)

Expert Solution

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 ( $CO32-$ )

Explanation of Solution

Effervescence formation means carbon dioxide gas is released during this test and carbonate ions release $CO2$ when treated with acids.

$CO32-(aq) + 2H+(aq)→CO2(g) + H2O(l)$

Hence, the ion present is the Carbonate ion.

(D)

Expert Solution

Interpretation Introduction

Interpretation:

To identify the ion present in the solution that gives the observation: Adding 6 M $HNO3$ plus 0.1 M $AgNO3$ 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 ( $Cl-, Br-or I-$ )

Explanation of Solution

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

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

$Ag++Cl-→AgCl (s)Ag++Br-→AgBr (s)Ag++I−→AgI (s)$

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

(E)

Expert Solution

Interpretation Introduction

Interpretation:

To identify the ion present in the solution that gives the observation: Adding of 6 M $HNO3$ plus 1 M $BaCl2$ 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 ( $SO42−$ ) ions

Explanation of Solution

Sulfate ( $SO42−$ ) ions form a white precipitate with Barium ions in an acidic medium.

$SO42-(aq) + Ba2+(aq)→BaSO4 (s)$

Hence, the ion present is a sulfate ion.

(F)

Expert Solution

Interpretation Introduction

Interpretation:

To identify the ion present in the solution that gives the observation: Adding of 6 M $HNO3$ plus 0.5 M $(NH4)2MoO$ 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 ( $PO43−$ ) ion

Explanation of Solution

Phosphate ( $PO43−$ ) ions form a yellow precipitate with ammonium molybdate in an acidic medium.

$2PO43-(aq) + 6H+(aq) + 3(NH4)2MoO (aq)→2(NH4)3PO4 (aq) + 3MoO3 (s) + 3H2O(l)$

Hence, the ion present is the phosphate ion.

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Answer 5

Textbook Question

Answer 6

Textbook Question

Answer 7

(a)

Expert Solution

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 ( $NO3−$ ).

Explanation of Solution

This test is known as Devarda’s test and is given by nitrate ions ( $NO3−$ ).

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-

$3NO3- (aq) + 8Al (s) + 5OH- (aq) + 18H2O (l)→3NH3(g) + 8[Al(OH)4]-(aq)$

Hence, the ion present is the Nitrate ion.

(b)

Expert Solution

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 ( $SO32-$ ).

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 ( $SO2$ ) gas.

And $SO2$ is released when sulfite ions are treated with an acid.

$SO32-(aq) + 2H+(aq)→SO2(g) + H2O(l)$

Hence, the ion present is the sulfite ion.

(C)

Expert Solution

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 ( $CO32-$ )

Explanation of Solution

Effervescence formation means carbon dioxide gas is released during this test and carbonate ions release $CO2$ when treated with acids.

$CO32-(aq) + 2H+(aq)→CO2(g) + H2O(l)$

Hence, the ion present is the Carbonate ion.

(D)

Expert Solution

Interpretation Introduction

Interpretation:

To identify the ion present in the solution that gives the observation: Adding 6 M $HNO3$ plus 0.1 M $AgNO3$ 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 ( $Cl-, Br-or I-$ )

Explanation of Solution

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

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

$Ag++Cl-→AgCl (s)Ag++Br-→AgBr (s)Ag++I−→AgI (s)$

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

(E)

Expert Solution

Interpretation Introduction

Interpretation:

To identify the ion present in the solution that gives the observation: Adding of 6 M $HNO3$ plus 1 M $BaCl2$ 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 ( $SO42−$ ) ions

Explanation of Solution

Sulfate ( $SO42−$ ) ions form a white precipitate with Barium ions in an acidic medium.

$SO42-(aq) + Ba2+(aq)→BaSO4 (s)$

Hence, the ion present is a sulfate ion.

(F)

Expert Solution

Interpretation Introduction

Interpretation:

To identify the ion present in the solution that gives the observation: Adding of 6 M $HNO3$ plus 0.5 M $(NH4)2MoO$ 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 ( $PO43−$ ) ion

Explanation of Solution

Phosphate ( $PO43−$ ) ions form a yellow precipitate with ammonium molybdate in an acidic medium.

$2PO43-(aq) + 6H+(aq) + 3(NH4)2MoO (aq)→2(NH4)3PO4 (aq) + 3MoO3 (s) + 3H2O(l)$

Hence, the ion present is the phosphate ion.

Answer 8

(a)

Expert Solution

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 ( $NO3−$ ).

Explanation of Solution

This test is known as Devarda’s test and is given by nitrate ions ( $NO3−$ ).

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-

$3NO3- (aq) + 8Al (s) + 5OH- (aq) + 18H2O (l)→3NH3(g) + 8[Al(OH)4]-(aq)$

Hence, the ion present is the Nitrate ion.

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

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 13

Answer to Problem 1ASA

Nitrate ion ( $NO3−$ ).

Answer 14

Explanation of Solution

This test is known as Devarda’s test and is given by nitrate ions ( $NO3−$ ).

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-

$3NO3- (aq) + 8Al (s) + 5OH- (aq) + 18H2O (l)→3NH3(g) + 8[Al(OH)4]-(aq)$

Hence, the ion present is the Nitrate ion.

Answer 15

(b)

Expert Solution

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 ( $SO32-$ ).

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 ( $SO2$ ) gas.

And $SO2$ is released when sulfite ions are treated with an acid.

$SO32-(aq) + 2H+(aq)→SO2(g) + H2O(l)$

Hence, the ion present is the sulfite ion.

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

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 20

Answer to Problem 1ASA

Sulfite ion ( $SO32-$ ).

Answer 21

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 ( $SO2$ ) gas.

And $SO2$ is released when sulfite ions are treated with an acid.

$SO32-(aq) + 2H+(aq)→SO2(g) + H2O(l)$

Hence, the ion present is the sulfite ion.

Answer 22

(C)

Expert Solution

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 ( $CO32-$ )

Explanation of Solution

Effervescence formation means carbon dioxide gas is released during this test and carbonate ions release $CO2$ when treated with acids.

$CO32-(aq) + 2H+(aq)→CO2(g) + H2O(l)$

Hence, the ion present is the Carbonate ion.

Answer 23

(C)

Expert Solution

Answer 24

(C)

Expert Solution

Answer 25

Expert Solution

Answer 26

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 27

Answer to Problem 1ASA

Carbonate ion ( $CO32-$ )

Answer 28

Explanation of Solution

Effervescence formation means carbon dioxide gas is released during this test and carbonate ions release $CO2$ when treated with acids.

$CO32-(aq) + 2H+(aq)→CO2(g) + H2O(l)$

Hence, the ion present is the Carbonate ion.

Answer 29

(D)

Expert Solution

Interpretation Introduction

Interpretation:

To identify the ion present in the solution that gives the observation: Adding 6 M $HNO3$ plus 0.1 M $AgNO3$ 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 ( $Cl-, Br-or I-$ )

Explanation of Solution

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

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

$Ag++Cl-→AgCl (s)Ag++Br-→AgBr (s)Ag++I−→AgI (s)$

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

Answer 30

(D)

Expert Solution

Answer 31

(D)

Expert Solution

Answer 32

Expert Solution

Answer 33

Interpretation Introduction

Interpretation:

To identify the ion present in the solution that gives the observation: Adding 6 M $HNO3$ plus 0.1 M $AgNO3$ 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 34

Answer to Problem 1ASA

Halide ions ( $Cl-, Br-or I-$ )

Answer 35

Explanation of Solution

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

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

$Ag++Cl-→AgCl (s)Ag++Br-→AgBr (s)Ag++I−→AgI (s)$

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

Answer 36

(E)

Expert Solution

Interpretation Introduction

Interpretation:

To identify the ion present in the solution that gives the observation: Adding of 6 M $HNO3$ plus 1 M $BaCl2$ 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 ( $SO42−$ ) ions

Explanation of Solution

Sulfate ( $SO42−$ ) ions form a white precipitate with Barium ions in an acidic medium.

$SO42-(aq) + Ba2+(aq)→BaSO4 (s)$

Hence, the ion present is a sulfate ion.

Answer 37

(E)

Expert Solution

Answer 38

(E)

Expert Solution

Answer 39

Expert Solution

Answer 40

Interpretation Introduction

Interpretation:

To identify the ion present in the solution that gives the observation: Adding of 6 M $HNO3$ plus 1 M $BaCl2$ 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 41

Answer to Problem 1ASA

Sulfate ( $SO42−$ ) ions

Answer 42

Explanation of Solution

Sulfate ( $SO42−$ ) ions form a white precipitate with Barium ions in an acidic medium.

$SO42-(aq) + Ba2+(aq)→BaSO4 (s)$

Hence, the ion present is a sulfate ion.

Answer 43

(F)

Expert Solution

Interpretation Introduction

Interpretation:

To identify the ion present in the solution that gives the observation: Adding of 6 M $HNO3$ plus 0.5 M $(NH4)2MoO$ 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 ( $PO43−$ ) ion

Explanation of Solution

Phosphate ( $PO43−$ ) ions form a yellow precipitate with ammonium molybdate in an acidic medium.

$2PO43-(aq) + 6H+(aq) + 3(NH4)2MoO (aq)→2(NH4)3PO4 (aq) + 3MoO3 (s) + 3H2O(l)$

Hence, the ion present is the phosphate ion.

Answer 44

(F)

Expert Solution

Answer 45

(F)

Expert Solution

Answer 46

Expert Solution

Answer 47

Interpretation Introduction

Interpretation:

To identify the ion present in the solution that gives the observation: Adding of 6 M $HNO3$ plus 0.5 M $(NH4)2MoO$ 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 48

Answer to Problem 1ASA

Phosphate ( $PO43−$ ) ion

Answer 49

Explanation of Solution

Phosphate ( $PO43−$ ) ions form a yellow precipitate with ammonium molybdate in an acidic medium.

$2PO43-(aq) + 6H+(aq) + 3(NH4)2MoO (aq)→2(NH4)3PO4 (aq) + 3MoO3 (s) + 3H2O(l)$

Hence, the ion present is the phosphate ion.

Answer 50

Ch. 35 - Each of the observations in the following list was...

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Answer to Problem 1ASA

Explanation of Solution

Answer to Problem 1ASA

Explanation of Solution

Answer to Problem 1ASA

Explanation of Solution

Answer to Problem 1ASA

Explanation of Solution

Answer to Problem 1ASA

Explanation of Solution

Answer to Problem 1ASA

Explanation of Solution

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