Practical Manual XI ( Simple Titrations, Redox Titrations, Element Detection, MP/BP)

  

Conceptual Chemistry

Practical Manual XI

 

I. Simple Titrations

Contents

 

1. You are given N/10 solution of HCl.  Find out normality &amount of NaOH in g/750 cm³.

2. You are given 0.25 N soln of KOH.  Find out normality & amount of hydrous oxalic acid in g/500 cm³

3. You are given N/20 soln of H₂SO₄. Find out normality & amount of anhydrous Na₂CO₃ in g/250 cm³.

4. You are given N/15 solution of HNO₃. Find Normality & amount of KOH solution in g/0.75 dm³

5. You are given 0.5 N soln. of KOH.  Find Normality & amount of sulphuric acid solution in g/600 cm³

6. You are given 0.05 N solution of H₂SO₄.  Find out Normality & amount of K₂CO₃ in g/900 ml

Acid base titration using phenolphthalein

Object....Given _____ solution of _____, find out the Normality and Amount of ______ in g/___

Chemicals     

Acid ----------  ................. (Strong or weak)
Base ---------- ................ (Strong)
Indicator ----- Phenolphthalein

Theory

1. Acid is a specie that donates H⁺ ions in aqueous medium, base is a specie that donates OH^- ions, while salt is neutralized product of acid and base. The reaction between acids and bases to form salt and water is known as Neutralization. It is the type of double displacement (decomposition) reaction and it is a reverse of process of hydrolysis. It is the process which involves the mutual combination of hydrogen ions (H⁺) of an acid and hydroxide ions (OH^-) of a base to form water molecule is termed as neutralization.

2. Acid and bases are classified on the basis of their dissociation abilities as strong and weak acid and bases.

3. The concentration (normality) of unknown acid or base can be determine through titration. Titration is the type of quantitative volumetric analysis in which burette and pipette are used to standardize a solution. It is the process by which the strength of unknown solution is determined by treating it with a definite volume of a standard solution of known concentration in the presence of an indicator.

4. Acidimetry-Alkalimetry Titrations or Acid-base titration is the type of titration in which an acid is allowed to react with a base in the presence of an indicator.

5. This is an Acidimetry/alkalimetry titration; the titration in which the strength of an alkali/acid is determined by neutralizing it with an acid/base solution of known strength in the presence of an indicator.

6. In this titration acid is strong or weak and base is strong so phenolphthalein indicators is used. As a rule when phenolphthalein is used, base will be Titrant (i.e. taken in burette) while acid will be titrate (taken in flask). Thus color transition from colourless to light pink is observed in titration.

Method/Procedure

1. Wash the burette, pipette, titration flask and funnel with water and rinse the burette with base (titrant) and pipette with acid (titrate).

2. Clamp the burette vertically in an iron stand and fill it with base/Titrant ( ______ )* solution up to the 0.0 cm³ mark carefully using funnel.

3. pipette out 10.0 cm³ of acid/Titrate ( ______ )* solution into the conical flask and add two drops of phenolphthalein (indicator) into it.

4. Run (Pour) base/Titrant ( ______ )* solution from burette into flask containing acid/titrate ( ______ )* drop-wise with constant shaking till a permanent pinkish tinge just appears. This is the end point. Note down the reading of the lower meniscus and then throw the contents of the flask and wash it with distilled water.

5. Now repeat the above process of titration for further observations till two concordant readings (similar readings) are obtained.

6. Calculate normality, equivalent weight & amount of required specie.

(*write formula according to object)

Neutralization Equation

*write equation according to object.

Observations

Solution in burette (Titrant) = ______

Solution in conical Flask (Titrate) = ______

Normality of acid (N₁) = ______ N

Volume of acid (V₁) = 10.0 cm³

Normality of Base (N₂) = ______N

Volume of Base (V₂) = ____cm³ (Comes from burette reading)

Amount of NaOH in g/...........  = _____  ?

Colour change = From Colourless to pink

Indicator used = Phenolphthalein

Burette Readings

Burette Reading Chart

Calculation

Result

Normality of _______ solution =  ............. N

Amount of _______ in g/......... = ............  g/_____

Acid base titration using Methyl Orange

Object .... Given _____ solution of _____, find out the Normality and Amount of ______ in g/___

Chemicals     

Acid ----------  ................. (Strong )

Base ---------- ................ (Strong or weak)

Indicator ----- Methyl Orange

Theory

1. Acid is a specie that donates H⁺ ions in aqueous medium, base is a specie that donates OH^- ions, while salt is neutralized product of acid and base. The reaction between acids and bases to form salt and water is known as Neutralization. It is the type of double displacement (decomposition) reaction and it is a reverse of process of hydrolysis. It is the process which involves the mutual combination of hydrogen ions (H⁺) of an acid and hydroxide ions (OH^-) of a base to form water molecule is termed as neutralization.

2. Acid and bases are classified on the basis of their dissociation abilities as strong and weak acid and bases.

3. The concentration (normality) of unknown acid or base can be determine through titration. Titration is the type of quantitative volumetric analysis in which burette and pipette are used to standardize a solution. It is the process by which the strength of unknown solution is determined by treating it with a definite volume of a standard solution of known concentration in the presence of an indicator.

4. Acidimetry-Alkalimetry Titrations or Acid-base titration is the type of titration in which an acid is allowed to react with a base in the presence of an indicator.

5. This is an Acidimetry/alkalimetry titration; the titration in which the strength of an alkali/acid is determined by neutralizing it with an acid/base solution of known strength in the presence of an indicator.

6. In this titration acid is strong and base is weak so methyl orange indicator is used. As a rule when methyl orange is used, acid will be Titrant (i.e. taken in burette) while base will be titrate (taken in flask).Thus color transition from yellow to light red is observed in titration.

Method/Procedure

1. Wash the burette, pipette, titration flask and funnel with water and rinse the burette with acid (titrant) and pipette with acid (titrate).

2. Clamp the burette vertically in an iron stand and fill it with acid/Titrant ( ______ )* solution up to the 0.0 cm3 mark carefully using funnel.

3. pipette out 10.0 cm3 of base/Titrate ( ______ )* solution into the conical flask and add two drops of methyl orange (indicator) into it.

4. Run (Pour) acid/Titrant ( ______ )* solution from burette into flask containing base/titrate ( ______ )* drop-wise with constant shaking till a permanent faint red colour just appears. This is the end point. Note down the reading of the lower meniscus and then throw the contents of the flask and wash it with distilled water.

5. Now repeat the above process of titration for further observations till two concordant readings (similar readings) are obtained.

6. Calculate normality, equivalent weight & amount of required specie.

(*write formula according to object)

Neutralization Equation

Observations

Burette Reading Chart

Calculation

Result

Normality of _______ solution =  ............. N

Amount of _______ in g/......... = ............  g/_____

 

Simple Titration (Experiment No. 1)

Object .... You are given N/10 (0.1 N) solution of HCl.  Find out the normality and amount of NaOH in g/750 cm³.

Chemicals     

Acid ----------   HCl (Strong )

Base ----------  NaOH(Strong )

Indicator ----- Phenolphthalein

Theory

1. Acid is a specie that donates H⁺ ions in aqueous medium, base is a specie that donates OH^- ions, while salt is neutralized product of acid and base. The reaction between acids and bases to form salt and water is known as Neutralization. It is the type of double displacement (decomposition) reaction and it is a reverse of process of hydrolysis. It is the process which involves the mutual combination of hydrogen ions (H⁺) of an acid and hydroxide ions (OH^-) of a base to form water molecule is termed as neutralization.

2. Acid and bases are classified on the basis of their dissociation abilities as strong and weak acid and bases.

3. The concentration (normality) of unknown acid or base can be determine through titration. Titration is the type of quantitative volumetric analysis in which burette and pipette are used to standardize a solution. It is the process by which the strength of unknown solution is determined by treating it with a definite volume of a standard solution of known concentration in the presence of an indicator.

4. Acidimetry-Alkalimetry Titrations or Acid-base titration is the type of titration in which an acid is allowed to react with a base in the presence of an indicator.

5. This is an Acidimetry/alkalimetry titration; the titration in which the strength of an alkali/acid is determined by neutralizing it with an acid/base solution of known strength in the presence of an indicator.

6. In this titration both acid and base are strong so any of the indicators may be used. Here phenolphthalein indicators is used. As a rule when phenolphthalein is used, base will be Titrant (i.e. taken in burette) while acid will be titrate (taken in flask). Thus color transition from colourless to light pink is observed in titration.

Method/Procedure

1. Wash the burette, pipette, titration flask and funnel with water and rinse the burette with base (titrant) and pipette with acid (titrate).

2. Clamp the burette vertically in an iron stand and fill it with base/Titrant (NaOH) solution up to the 0.0 cm³ mark carefully using funnel.

3. pipette out 10.0 cm³ of acid/Titrate (HCl) solution into the conical flask and add two drops of phenolphthalein (indicator) into it.

4. Run (Pour) base/Titrant (NaOH) solution from burette into flask containing acid/titrate (HCl) drop-wise with constant shaking till a permanent pinkish tinge just appears. This is the end point. Note down the reading of the lower meniscus and then throw the contents of the flask and wash it with distilled water.

5. Now repeat the above process of titration for further observations till two concordant readings (similar readings) are obtained.

6. Calculate normality, equivalent weight & amount of required specie.

Observations

Burette Reading Chart

Calculation

1.   Normality Determination

2.   Amount Determination

Result

Simple Titration (Experiment No. 2)

Object .... You are given 0.25 N solution of KOH.  Find out the normality and amount of hydrous oxalic acid (H₂C₂O₄.2H₂O) in g/500 cm³.

Chemicals

Acid ----------  H₂C₂O₄.2H₂O (Weak )

Base ---------- KOH(Strong )

Indicator ----- Phenolphthalein

Theory

1. Acid is a specie that donates H⁺ ions in aqueous medium, base is a specie that donates OH^- ions, while salt is neutralized product of acid and base. The reaction between acids and bases to form salt and water is known as Neutralization. It is the type of double displacement (decomposition) reaction and it is a reverse of process of hydrolysis. It is the process which involves the mutual combination of hydrogen ions (H⁺) of an acid and hydroxide ions (OH^-) of a base to form water molecule is termed as neutralization.

2. Acid and bases are classified on the basis of their dissociation abilities as strong and weak acid and bases.

3. The concentration (normality) of unknown acid or base can be determine through titration. Titration is the type of quantitative volumetric analysis in which burette and pipette are used to standardize a solution. It is the process by which the strength of unknown solution is determined by treating it with a definite volume of a standard solution of known concentration in the presence of an indicator.

4. Acidimetry-Alkalimetry Titrations or Acid-base titration is the type of titration in which an acid is allowed to react with a base in the presence of an indicator.

5. This is an Acidimetry/alkalimetry titration; the titration in which the strength of an alkali/acid is determined by neutralizing it with an acid/base solution of known strength in the presence of an indicator.

6. In this titration base is strong so phenolphthalein is used as an indicator. As a rule when phenolphthalein is used base will be Titrant (i.e. taken in burette) while acid will be titrate (taken in flask)

Method/Procedure

1. Wash the burette, pipette, titration flask and funnel with water and rinse the burette with base (titrant) and pipette with acid (titrate).

2. Clamp the burette vertically in an iron stand and fill it with base/Titrant (NaOH) solution up to the 0.0 cm³ mark carefully using funnel.

3. pipette out 10.0 cm³ of acid/Titrate (HCl) solution into the conical flask and add two drops of phenolphthalein (indicator) into it.

4. Run (Pour) base/Titrant (NaOH) solution from burette into flask containing acid/titrate (HCl) drop-wise with constant shaking till a permanent pinkish tinge just appears. This is the end point. Note down the reading of the lower meniscus and then throw the contents of the flask and wash it with distilled water.

5. Now repeat the above process of titration for further observations till two concordant readings (similar readings) are obtained.

6. Calculate normality, equivalent weight & amount of required specie.

Observations

Burette Reading Chart

Calculation

 

II. Redox Titrations

Contents

 

1. You are given N/10 solution of KMnO₄.  Find out normality & amount of FeSO₄.7H₂O in g/750 ml.

 

2. You are given 0.25 N solution of KMnO₄.  Find out normality & amount of Mohr’s salt in g/500 cm³

 

3. You are given 0.05 N soln of hydrous oxalic acid .Find normality & amount of KMnO₄ in g/950 cm³

 

4. You are given N/30 soln of FeSO₄.7H₂O.Find out normality & amount of KMnO₄ in g/850 ml

 

5. You are given N/25 soln. of KMnO₄.Find out normality & amount of hydrous oxalic acid in g/1.5 dm³

 

Redox titration General Description

Object

Given _____ solution of _____, find out the Normality and Amount of ______ in g/___

Chemicals

1.Oxidizing agent ....... Potassium permanganate (KMnO₄) solution.

2.Reducing agent ....... Hydrous Ferrous Sulphate, or Mohr’s salt or oxalic acid

3. Indicator ................. KMnO₄ itself behaves as an indicator.

4. Helper/Medium ...... Acidic (dilute H₂SO₄).

Theory

1.   Titration is the type of quantitative volumetric analysis in which burette and pipette are used to standardize a solution. It is the process by which the strength of unknown solution is determined by treating it with a definite volume of a standard solution in the presence of an indicator.

2.  Oxidation is the process during which oxidation number of an element is increased due to loss of one or more electron while reduction is the process during which oxidation number of an element is decreased due to gain of one or more electrons.

3. Oxidizing agent is a substance which gains electrons during reaction and thus they oxidize other substance and itself get reduced. e.g. KMnO₄, K₂Cr₂O₇ etc. Reducing agent is a substance which             loses electrons during reaction and thus they reduce other substance and itself get oxidized. e.g.              FeSO₄.7H₂O, FeSO₄.(NH₄)₂SO₄.6H₂O, H₂C₂O₄.2H₂O etc.

4.Both oxidation and reduction takes place simultaneously and such reactions are called Redox reactions.

5.   The titration between reducing agent and oxidizing agent in which one reactant is oxidized and other is reduced is called Redox titration. In redox titration KMnO₄ is always a Titrant (i.e. taken in the burette) while reducing agent is titrate. The titration is carried out in acidic medium so dilute H₂SO₄ is also taken along with reducing agent in flask in equal amount. No external indicator is used as KMnO₄ itself acts as an internal indicator.

Equations

 

 

 

 

 

Method

1. Wash the burette, pipette and titration flask with water and rinse the pipette with reducing agent and burette with KMnO₄.

2. Clamp the burette vertically in an iron stand and rinse the burette with Titrant (KMnO₄) solution.  Now fill it with Titrant (KMnO₄) solution carefully using funnel upto the zero mark.

3. After rinsing pipette with titrate (reducing agent) solution, pipette out 10 ml (cm³) Titrate (reducing agent) solution into the conical flask and add equal volume or half test tube of H₂SO₄ into it.

4.   Run Titrant (KMnO₄) solution from burette into flask drop-wise with constant shaking till a permanent pinkish tinge just appears by a single drop of Titrant (KMnO₄). This is the end point. Note down the reading of the upper meniscus.

 

5.   Now throw the content of the flask and wash it with distilled water and repeat the above process of titration till two similar readings (concordant readings) are obtained

6.   Calculate normality, equivalent weight & amount of required specie.

 

Observations

1.   Solution in the burette (Titrant) =  KMnO₄

2.Solution in conical flask(Titrate)  = Reducing agent +  H₂SO₄  (10 ml each)

3. Colour change                          =  From Colourless to pink

4.   Normality of Reducing agent (N₂)            =  ?

5.   Volume of Reducing agent (V₂)                =  10 ml

6.   Normality of KMnO₄          (N₁)                 =  …… N

7.   Volume of KMnO₄             (V₁)                  =  ?

8.   Amount of Reducing agent in g/750 ml    =  ?

9.   Indicator used                             = KMnO₄ as self-indicator

 

Burette Reading Chart

Calculations

1. Normality Determination

2.   Amount Determination

Result

1. Normality of _______ = _______ N

2. Amount of _______ = _______ g/……… ml

 

 

Redox Titration (Experiment No. 1)

Object

 You are given N/10 (0.1 N) solution of KMnO₄.  Find out the normality and amount of FeSO₄.7H₂O in gm/750 ml.

 

Chemicals

1.   Oxidizing agent     ––––    Potassium permanganate (KMnO₄) solution.

2.   Reducing agent     ––––    Hydrous Ferrous Sulphate (FeSO₄.7H₂O) solution.

3.   Indicator                ––––    KMnO₄ itself behaves as an indicator.

4.   Helper / Medium   ––––    Acidic (dilute H₂SO₄).

 

Theory

1.   Titration is the type of quantitative volumetric analysis in which burette and pipette are used to standardize a solution. It is the process by which the strength of unknown solution is determined by treating it with a definite volume of a standard solution in the presence of an indicator.

2.  Oxidation is the process during which oxidation number of an element is increased due to loss of        one or more electron while reduction is the process during which oxidation number of an element   is decreased due to gain of one or more electrons.

3.  Oxidizing agent is a substance which gains electrons during reaction and thus they oxidize other    substance and itself get reduced. e.g. KMnO₄, K₂Cr₂O₇ etc. Reducing agent is a substance which          loses electrons during reaction and thus they reduce other substance and itself get oxidized. e.g.   FeSO₄.7H₂O, FeSO₄.(NH₄)₂SO₄.6H₂O, H₂C₂O₄.2H₂O etc

4. Both oxidation and reduction takes place simultaneously and such reactions are called Redox  reactions.

5.  The titration between reducing agent and oxidizing agent in
which one reactant is oxidized and other is reduced is called Redox titration. In redox titration KMnO₄ is always a Titrant (i.e. taken in the burette) while reducing agent is titrate. The titration is carried out in acidic medium so dilute H₂SO₄ is also taken along with reducing agent in flask in equal amount. No external indicator is used as KMnO₄ itself acts as an internal indicator.

Molecular Equation

Method

1.   Wash the burette, pipette and titration flask with water.

 

2.   Clamp the burette vertically in an iron stand and rinse the burette with Titrant (KMnO₄) solution.  Now fill it with Titrant (KMnO₄) solution upto the zero mark.

3.   After rinsing pipette with titrate (reducing agent) solution, pipette out 10 ml Titrate (reducing agent) solution into the conical flask and add equal volume of H₂SO₄ into it.

4.   Run Titrant (KMnO₄) solution from burette into flask drop-wise with constant shaking till a permanent pinkish tinge just appears by a single drop of Titrant (KMnO₄). This is the end point. Note down the reading of the upper meniscus.

5.   Now throw the content of the flask and wash it with distilled water and repeat the above process of titration till two similar readings (concordant readings) are obtained

 

Observations

1.  Solution in the burette  (Titrant) =  KMnO₄

2. Solution in conical flask (Titrate) = FeSO₄ + H₂SO₄ (10 ml each)

3.   Colour change                            =  Colourless to pink

4.   Normality of FeSO₄.7H₂O(N₂)    =  ?

5.   Volume of FeSO₄.7H₂O    (V₂)    =  10 ml

6.   Normality of KMnO₄          (N₁)    =  0.1 N

7.   Volume of KMnO₄             (V₁)     =  ?

8.   Amount of FeSO₄.7H₂O in g/750 ml =  ?

Burette Reading Chart

Calculations

1. Normality Determination

2.  Amount Determination

Result

1.   Normality of FeSO₄.7H₂O =  _______ N

2.   Amount of FeSO₄.7H₂O   =  _______ g/750 ml

Redox Titration (Experiment No. 2)

Object

 You are given 0.25 N solution of KMnO₄.  Find out the normality and amount of Mohr’s salt (ferrous ammonium sulphate) in gm/500 cm³.

 

Chemicals

1.   Oxidizing agent     ––––    KMnO₄

2.   Reducing agent     ––––    Mohr’s salt [FeSO₄.(NH₄)₂SO₄.6H₂O].

3.   Indicator                ––––    KMnO₄ itself behaves as an indicator.

4.   Helper / Medium   ––––    Acidic (dilute H₂SO₄).

 

Theory

1.        Titration is the type of quantitative volumetric analysis in which burette and pipette are used to standardize a solution. It is the process by which the strength of unknown solution is determined by treating it with a definite volume of a standard solution in the presence of an indicator.

2.         Oxidation is the process during which oxidation number of an element is increased due to loss of  one or more electron while reduction is the process during which oxidation number of an element             is decreased due to gain of one or more electrons.

3.Oxidizing agent is a substance which gains electrons during reaction and thus they oxidize other  substance and itself get reduced. e.g. KMnO₄, K₂Cr₂O₇ etc. Reducing agent is a substance which         loses electrons during reaction and thus they reduce other substance and itself get oxidized. e.g.    FeSO₄.7H₂O, FeSO₄.(NH₄)₂SO₄.6H₂O, H₂C₂O₄.2H₂O etc

4.         Both oxidation and reduction takes place simultaneously and such reactions are called Redox             reactions.

5.         The titration between reducing agent and oxidizing agent in which one reactant is oxidized and other is reduced is called Redox titration. In redox titration KMnO₄ is always a Titrant (i.e. taken in the burette) while reducing agent is titrate. The titration is carried out in acidic medium so dilute H₂SO₄ is also taken along with reducing agent in flask in equal amount. No external indicator is used as KMnO₄ itself acts as an internal indicator.

Method

1.   Wash the burette, pipette and titration flask with water.

2.   Clamp the burette vertically in an iron stand and rinse the burette with Titrant (KMnO₄) solution.  Now fill it with Titrant (KMnO₄) solution upto the zero mark.

3.   After rinsing pipette with titrate (reducing agent) solution, pipette out 10 ml Titrate (reducing agent) solution into the conical flask and add equal volume of H₂SO₄ into it.

4.   Run Titrant (KMnO₄) solution from burette into flask drop-wise with constant shaking till a permanent pinkish tinge just appears by a single drop of Titrant (KMnO₄). This is the end point. Note down the reading of the upper meniscus.

5.   Now throw the content of the flask and wash it with distilled water and repeat the above process of titration till two similar readings (concordant readings) are obtained

Observations

 

1.   Solution in the burette       (Titrant)           =  KMnO₄

2.   Solution in conical flask     (Titrate)           =  Mohr’s salt + equal volume of dil. H₂SO₄

3.   Colour change                 =  Colourless to pink

4.   Volume of KMnO₄         (V₁)                  =  ?

5.   Normality of KMnO₄ solution       (N₁)   =  0.25 N

6.   Volume of Mohr’s salt                   (V₂)  =  10 ml

7.   Normality of Mohr’s salt                (N₂)  =  ?

8.   Amount of Mohr’s salt in g/500 cm³      =  ?

 

Burette Reading Chart

Calculations

1. Normality Determination

2.   Amount Determination

Result

1. Normality of Mohr’s salt = -------N     

2. Amount of Mohr’s salt    =--------g/500 cm³

Redox Titration (Experiment No. 3)

 

Object             You are given 0.05 N solution of hydrous oxalic acid (H₂C₂O₄.2H₂O). Find out the normality and amount of KMnO₄ in g/950 cm³.

Chemicals

1.   Oxidizing agent     ––––    KMnO₄           

2.   Reducing agent    ––––    H₂C₂O₄.2H₂O

3.   Indicator                ––––    KMnO₄ itself acts as an indicator.

4.   Helper / Medium   ––––    Acidic (dilute H₂SO₄).

 

Theory

1.         Titration is the type of quantitative volumetric analysis in which burette and pipette are used to standardize a solution. It is the process by which the strength of unknown solution is determined by treating it with a definite volume of a standard solution in the presence of an indicator.

2.         Oxidation is the process during which oxidation number of an element is increased due to loss of             one or more electron while reduction is the process during which oxidation number of an element             is decreased due to gain of one or more electrons.

3.         Oxidizing agent is a substance which gains electrons during reaction and thus they oxidize other             substance and itself get reduced. e.g. KMnO₄, K₂Cr₂O₇ etc. Reducing agent is a substance which             loses electrons during reaction and thus they reduce other substance and itself get oxidized. e.g.             FeSO₄.7H₂O, FeSO₄.(NH₄)₂SO₄.6H₂O,H₂C₂O₄.2H₂O etc

4.         Both oxidation and reduction takes place simultaneously and such reactions are called Redox             reactions.

5.         The titration between reducing agent and oxidizing agent in which one reactant is oxidized and other is reduced is called Redox titration. In redox titration KMnO₄ is always a Titrant (i.e. taken in the burette) while reducing agent is titrate. The titration is carried out in acidic medium so dilute H₂SO₄ is also taken along with reducing agent in flask in equal amount. No external indicator is used as KMnO₄ itself acts as an internal indicator.

 

Method

1.   Wash the burette, pipette and titration flask with water.

 

2.   Clamp the burette vertically in an iron stand and rinse the burette with Titrant (KMnO₄) solution.  Now fill it with Titrant (KMnO₄) solution upto the zero mark.

 

3.   After rinsing pipette with titrate (reducing agent) solution, pipette out 10 ml Titrate (reducing agent) solution into the conical flask and add equal volume of H₂SO₄ into it.

 

4.   Run Titrant (KMnO₄) solution from burette into flask drop-wise with constant shaking till a permanent pinkish tinge just appears by a single drop of Titrant (KMnO₄). This is the end point.  Note down the reading of the upper meniscus.

 

5.   Now throw the content of the flask and wash it with distilled water and repeat the above process of titration till two similar readings (concordant readings) are obtained

Observations

 

1.   Solution in the burette                   (Titrant)           =  KMnO₄

2.   Solution in conical flask                (Titrate)           =  Oxalic acid  +  Dilute H₂SO₄

3.   Colour change                                                      =  Colourless to pink

4.   Normality of KMnO₄                      (N₁)                  =  ?

5.   Volume of KMnO₄                         (V₁)                  =  ?

6.   Amount of KMnO₄ in gm/950 cm³                        =  ?

7.   Normality of Oxalic acid                (N₂)                  =  0.05 N

8.   Volume of Oxalic acid                   (V₂)                  =  10 ml

 

Burette Reading Chart

 

Calculations

1.   Normality Determination

 

2.   Amount Determination

Result

1.   Normality of KMnO₄=  _______ N

2.   Amount of KMnO₄   =  _______ g/950 cm³

 

I. Element Detection

Detection of Elements in the Given Organic Compounds

Theory

The covalent compounds of carbon with H, O, N, S, P and halogens are called organic compounds. Since organic compounds are non-ionic or covalent in nature and do not dissociate into ions when put in water, hence detection of different elements present in organic compound becomes very difficult by ordinary reactions.

To overcome above difficulty, organic compounds are fused with Na metal to form what is known as Sodium Extract or Lassaigne’s Filtrate. It is Sodium Salt containing elements as ions.

Why Sodium Extract is made

1. In order to convert covalent bonding into ionic.

2. In order to make molecular and complex reactions to ionic and simple.

Nature of Sodium Extract

It is mostly alkaline because of the following side reactions.

   4Na  +  O₂      →   2Na₂O       ( Basic oxide)

  2Na  +  2H₂O  →  2NaOH   +   H₂

Reasons for using Na metal

1. It is cheap and soft, so easy to cut into pieces.

2. It melts at low temperature.

3. It reacts readily.

4. All sodium salts are water soluble.

Reasons for not using K-metal

1. It catches fire in air at once.

2. Its melting point is high.

3. It does not react with all the elements present in O.C

Reactions of Na metal with elements of O.C

Procedure for the preparation of Na-Extract

1. A small piece of dry sodium metal is taken in an ignition tube and a small amount of the given organic compound added to it.

2. Ignition tube is then heated first slowly and then strongly till it becomes red hot.

3. It is then dropped in a china dish containing distilled water.

4. The tube is crushed into pieces with the help of glass rod and this solution of china dish is boiled for few minutes.

5. Then solution is filtered out and the filtrate is collected in a beaker.

6. This filtrate is called Sodium Extract or Lassaign’s Filtrate which is colourless and transparent.

Scheme of Element Detection

Model Experiment No 1 (for Nitrogen)

 

Result

The given compound contains nitrogen as element

Model Experiment No 2 (For nitrogen and sulphur)

 

Result

The given compound contains both nitrogen and sulphur as element

Model Experiment No 3  (For Sulphur) 

 

Result

The given compound contains sulphur as element

Model Experiment No 4 (For Chlorine)

Result

The given compound contains chlorine  as element

Model Experiment No 5 (For Bromine)

 

Result

The given compound contains bromine as element

Result

The given compound contains both sulphur as element

IV. Determination of Boiling and Melting Points

IV. Determination of Boiling and Melting Points

 

 

Determination of Boiling Point

Object                        

Determine the boiling point of given organic liquid.

Apparatus

Thermometer, ignition tube, thread, capillary tube, iron stand, tripod stand, wire gauze, Bunsen burner, glass rod, small beaker containing bath liquid (glycerin or paraffin oil

Theory           

The temperature at which the vapour pressure of the liquid becomes equal to outside pressure (atmospheric pressure) is known as Boiling Point (B.P.).The normal boiling point is the temperature at which the vapour pressure of the liquid equals to one atmosphere (standard pressure).

Boiling point is the characteristic constant of a liquid and it is the criteria of purity of liquids because pure liquids always boil at a definite temperature showing sharp B.P.

Method (Capillary Tube Method / Siwoloboff’s Method

1.   Take an ignition tube and put in it few drops of given liquid.

2.   Attach the ignition tube with the bulb of a thermometer with the help of rubber band or thread.

3.   Place a capillary tube whose upper end is sealed in the ignition tube.

4.   Now hang the thermometer along with ignition tube in a beaker containing a liquid (water) of high boiling point (water bath).

5.   Heat the beaker over wire gauze with constant stirring.

6.   When a rapid stream of bubbles of vapours rises from the lower end of the capillary tube, then the temperature in thermometer is noted which indicates b.p. of given liquid.

7.   Repeat the same process to take at least three reading and find out concordant reading.

Observations

Result

The B.P. of given liquid is found out to be _____ °C.

 

Determination of Melting Point

 

Object

Determine the melting point of given compound.

Apparatus

Thermometer, thread, capillary tube, iron stand, tripod stand, wire gauze, Bunsen burner, glass rod, small beaker containing bath liquid (glycerin or paraffin oil)

Theory           

The temperature at which there is an equilibrium between solid and liquid phases of a solid is called melting point.  OR Melting point is the temperature at which both liquid and solid phases coexist at equilibrium.

Melting point is the characteristic constant of a solid and it is the criteria of purity of solids because pure solids always melt at a definite temperature showing sharp M.P.

 

Method

1.   Take a capillary tube and seal its lower end by heating.

2.   Put a little quantity of powdered compound in a capillary tube.

3.   The capillary tube is then fixed to a bulb of thermometer with the help of rubber band or a thread.

4.   Now thermometer with capillary tube is suspended in water bath (beaker containing bath liquids like glycerin or water) by means of iron stand in such a way that upper end of capillary tube should be kept out of the liquid.

5.   Heat the beaker with constant stirring till solid given compound shows signs of melting and then remove the burner.

6.   Stir continuously till wax becomes transparent. Note this temperature at which compound just becomes transparent. This will be m.p. of given compound.

 

Observation

Result

M.P. of the given compound is found out to be _____ °C.

VI. Comprehensive Question Answer Viva-Voce

Simple and Redox Titrations

 

Q1.      What is the volume of burette?

Ans:      50.0cm³

 

Q2.      Name two organic acids which are diprotic?

Ans:     Oxalic acid, succinic acid

 

Q3.      Indicate the amounts in grams required to prepare 1N and 1M one liter of oxalic acid solution.

Ans:     63.0 g for 1N solution, 126.0 g for 1M for solution.

 

Q4.      Name any two hydrates used in simple titration.

Ans:     Oxalic acid (H₂C₂O₄.2H₂O), Washing soda (Na₂CO₃.10H₂O)

 

Q5.      What is primary standard solution?

Ans:     It is a substance which is 100 percent pure or of a known purity.

 

Q6.      Why NaOH and KOH are not primary standard?

Ans:     Because they absorb water (moisture) and CO₂ from air.

 

Q7.      Is anhydrous sodium carbonate a primary standard?

Ans:     Yes, it is.

 

Q8.      What is pH range of phenolphthalein? OR Why is phenolphthalein taken as an indicator             when weak acid is titrated against strong base?

Ans: The pH range of phenolphthalein is 8.3 to 10.0.The basic pH range of phenolphthalein makes it a suitable indicator when weak acid is titrated against strong base as it produces sharp colour change at pH round about 8 to 10.

 

Q9.      What is pH range of methyl orange? OR Why is methyl orange taken as an indicator when strong acid is titrated against weak base?

Ans:     The pH range of methyl orange is 2.9 to 4.6.The acidic pH range of methyl orange makes it a suitable indicator when strong acid is titrated against weak base as it produces sharp colour change at pH round about 3 to 5.

 

Q10.    Is KMnO₄ a primary standard solution?

Ans:No, as KMnO₄ can not be obtained in pure form and its solution in not very stable.

 

Q11.    Name two coloured salts of potassium.

Ans; i.Potassium permanganate (KMnO₄)        ii. Potassium dichromate (K₂Cr₂O₇)

 

Q12.    Name two coloured salts of sodium.

Ans:     i.Sodium chromate (Na₂CrO₄)                  ii. Sodium dichromate (Na₂Cr₂O₇)

 

Q13.    Give 3 reasons responsible for formation of brown precipitates in redox titration.

Ans;     The formation of brown ppt in KMnO₄ titration is due to:

i.              The solution may be very cold.

ii.             KMnO₄ might have been added in large amount and too quickly.

iii.            The quantity of dilute H₂SO₄ added may be insufficient.

Q14.    Why does phenolphthalein remain colourless in acidic medium and give pink colour in basic medium?

Ans:     Phenolphthalein is a weak organic acid. Its unionized molecule is colourless and its anion is pink  coloured. It is feebly ionized to give H⁺ and pink coloured anion (In^-). Under acidic conditions, its   ionization equilibrium is forced to left due to common ion effect thereby lowering the concentration of its pink coloured anion (In^-), so solution becomes colourless. Under alkaline  conditions, the added OH^¯ ions combine with its H⁺ ions to form feebly ionized water molecule   thereby increasing the ionization of HIn (phenolphthalein) giving more and more pink coloured    anion In^¯ and thus solution turns pink.

 

                                

 

Q15.    Why does methyl orange give reddish pink in acidic medium and faint yellow colour in             basic medium?

Ans:     Methyl orange is a weak organic base. Its unionized molecule is yellow and its cation is reddish  pink coloured.It is feebly ionized to give OH^¯ ions and reddish pink coloured cation (Me⁺). Under         basic conditions, its ionization equilibrium is pushed to left due to common ion effect thereby     lowering the concentration of its pink coloured cation and increasing the concentration of its     yellow coloured unionized molecule, so solution turns yellow. Under acidic conditions ,the added       H⁺ ions combine with its OH^¯ ions to form feebly ionized water molecule thereby          increasing the ionization of MeOH (methyl orange), giving more and more reddish pink coloured cation Me⁺ and           thus solution turns yellow to pink.

 

                                    

 

Q16.    What is meniscus?

Ans:     The surface of a liquid in cylindrical vessel is called meniscus means moon.

 

Q17.    Which meniscus is taken for burette reading of coloured solution?

Ans:     Upper meniscus.

 

Q18.    Which meniscus is taken for burette reading of colourless solution?

Ans:     Lower meniscus.

 

Q19.    Why do we note upper meniscus of KMnO₄ solutions?

Ans:     KMnO₄ is highly coloured compound. Its lower meniscus is not adequately observed or distinctly  visible due to its intense colour and hence upper meniscus is recorded.

 

Q20.    Why KMnO₄ solution is kept in coloured bottles?

Ans: KMnO₄ solution decomposes on exposure to light and air, hence it is kept in dark coloured             bottles.

 

Q21. How will you store a standard stock solution of KMnO₄?

Ans: It is stored in dark or amber-coloured bottles and should be protected from light

 

Q22.    Can HCl be used instead of dilute H₂SO₄ in redox titration?

Ans:     HCl cannot be used in redox titration as some of KMnO₄ may be consumed for oxidizing HCl into  chlorine.

 

Q23.    Why KMnO₄ is always acidified with H₂SO₄ and not with HCl or HNO₃?

Ans:     KMnO₄ is always acidified with H₂SO₄ because sulphuric acid is not self oxidizing and also it has             no action upon KMnO₄ in dilute solution.KMnO₄ can never be acidified with HCl or HNO₃ as         nitric acid is itself an oxidizing agent and HCl itself oxidized by KMnO₄ into chlorine gas.

 

                        2MnO₄⁻    +  16H⁺    +   10Cl⁻   ¾®   2Mn²⁺ +   8H₂O +    5Cl₂

 

Q24.    In Mohr’s salt which is oxidized and which remain unaffected?

Ans:     Ferrous sulphate is oxidized to ferric sulphate while ammonium sulphate remains unaffected             during titration.

 

Q25.    Is heating necessary for titration of KMnO₄ with ferrous sulphate?

Ans:     No, because on heating ferrous sulphate is oxidized to ferric sulphate. Hence if we heat ferrous             sulphate the titration can not be performed accurately.

 

Q26.    Why oxalic acid solution is heated before titrations?

Ans:     The reaction between KMnO₄ and cold oxalic acid solution is very slow due to slow production of             Mn²⁺ ions. In order to speed up the reaction and for rapid production of Mn²⁺ ions, to complete the             reaction, heating is done.

 

Q27.    What happens to oxalic acid in redox titration in the presence of dilute H₂SO₄?

Ans:     Oxalic acid is oxidized to CO₂ by oxidizing agent KMnO₄.

 

Q28.    Why we do not add any other indicator in KMnO₄ titration? Or                                                 Why KMnO₄ is called self-indicator?

Ans:     Because KMnO₄ is a self-indicator and highly coloured due to presence of permanganate ion             which is pink in colour .When it reacts a reducing agent solutions during titration, a very light       pink-coloured product, manganous sulphate (seem to be colourless) is obtained. When the     reaction is over, even one drop of KMnO₄ produces pink colour.

 

Q29.    What is the function of KMnO₄ in redox titration?

Ans:     KMnO₄ acts as an oxidizing agent and provides nascent oxygen or gains electrons (3e^- in basic             medium and 5e^- in acidic medium).It also acts as a self-indicator or internal indicator.

 

Q30.    What is the function of H₂SO₄ in redox titration?

Ans:     It helps to increase the oxidizing tendency of KMnO_4. In the presence of sulphuric acid, KMnO₄              becomes strong oxidizing agent and extract 5 electrons from reducing agent (which is being             oxidized) and gets reduced to manganous (Mn²⁺) ion.

 

MnO4−

+

8H+

+

5e−

¾®

Mn2+

+

4H2O

 

Q31.    Why a burette with glass tap is used in redox titrations?

Ans:     Rubber tap should not be used as KMnO₄ being an oxidant acts powerfully on rubber. Thus burette with glass tap is used in redox titrations.

 

Q32     What are the sources of following acids?

            Citric acid, Tartaric acid, Acetic acid, Lactic acid, formic acid

Ans:     (i)	Citric acid	–––––	Lemon
(ii)	Tartaric acid	–––––	Grapes
(iii)	Acetic acid	–––––	Vinegar
(iv)	Lactic acid	–––––	Fermented milk
(v)	formic acid	–––––	Stings of bees, wasps

 

 

Q33.    What is Oxonium/Hydronium/Hydroxonium Ion (H₃O⁺)?

Ans:     The solvated (hydrated) proton i.e. a proton (H⁺) which accepts a lone pair of electron from water  is known as Oxonium or Hydroxonium Ion (H₃O⁺).

               

 

Q34.    What is Degree of Dissociation?

Ans:     The ratio of the number of molecules ionized to the total number of dissolved molecules of a  substance is known as Degree of Dissociation (a) i.e.

 

           

 

Q35.    What are the precautions taken while performing simple titration?

Ans:     Following precautions must be taken while performing simple titration:

(i)            Burette must be clamped vertically.

(ii)           Burette should be filled with the help of funnel.

(iii)          Lower meniscus of liquid in the burette should be noted.

(iv)          Air bubbles must be removed from the jet of the burette before starting experiment.

(v)        Rinse the burette and the pipette with NaOH and HCl respectively but conical flask must be washed with water and never rinse with acid.

 

Q36. Why alkali is taken in burette when phenolphthalein is used as an indicator? Or Why phenolphthalein is preferred as indicator when base is taken in burette?

Ans:     Phenolphthalein is a suitable indicator when colour change is required at basic pH round about 8       to 10 which it gives by producing sudden sharp colour change when strong base is used. When       strong base is used it is always taken in burette as the appearance of pink colour at the end point       is distinctly visible

 

Q37.    What are hydrates or hydrous compounds?                   

Ans:     The compounds containing water of crystallization as an essential part of their crystals are called             hydrates or hydrous compounds.      

 

Q38.    What is meant by water of crystallization?          

Ans:     The constant number of water molecules attached to the ions (mostly cations) of ionic salts             (hydrates) is called water of crystallization.   

 

Q39.    What is meant by anhydrous salts?                      

Ans:     The hydrous salts without water of crystallization are called anhydrous salts.                    

 

Q40.    What are hygroscopic substances?                                  

Ans:     Hygroscopic substances absorb moisture on exposure to the atmosphere but not form solution   and merely get moist or sticky.

 

Q41.    Give some examples of hygroscopic substances?                    

Ans:    NaNO₃,  CuO,  CaO

 

Q42.    What is efflorescence and efflorescent?              

Ans:   The phenomenon of losing part or all water of crystallization by some crystalline salts on exposure           to the atmosphere to form a lower hydrate or anhydrous salt is called efflorescence and the salt is           said to be efflorescent.

Q43.   Give example of efflorescent.

Ans:    Na₂CO₃. 10H₂O (which loses 9 out of its 10 molecules of water of crystallization).

 

Q44.    What is meant by deliquescence and deliquescent?      

Ans:    Some compounds tend to absorb a large amount of water on exposure to the atmosphere thereby turning into solutions. This phenomenon is known as Deliquescence and the substances are said to be Deliquescent.

 

Q45.    Give examples of some deliquescent.

Ans:     NaOH,  KOH,  FeCl₃,  MgCl₂,  CaCl₂,  P₄O_10.

 

Q46.    On which factor hydration ability of ions depend?         

Ans:     Hydration ability of ion depends on its charge density. Greater the charge density of ion, greater will be its hydration ability and vice-versa.                

 

Q47.    Give name and formulae different hydrates used in theory and practical.

Ans:oxalic acid, sodium carbonate, ferrous sulphate, Mohr’s salt, blue vitriol                

 

Q48.    What is hydrated ion?                     

Ans:     The ion surrounded by water molecules is called hydrated ion.       

 

Q49.    What are strong acids and bases?

Ans:     Acids or bases which completely ionize having high %age dissociation (usually 30 % or more) in aqueous solution are considered to be strong. In terms of ionization constant strong acids and bases are those which have large value of ionization constant usually greater than 1

 

Q50.    What are weak acids and bases?                          

Ans:     Acids or bases which partially ionize having low %age dissociation (usually below 30 %) in aqueous solution are considered to be weak. In terms of ionization constant weak acid and bases are those which have small value of ionization constant usually less than 1   

           

Element Detection

 

Q51.    What are organic compounds?

Ans:     The covalent compounds of carbon with other elements such as H, O, S N, P, and             halogens are   called   organic compounds with the exception of carbonates, bicarbonates,             cyanides, sulphocyanides, thiocynates, carbides and oxides of carbon.

 

Q52.    Which element is regarded as the chief constituent of all organic compounds?

Ans:     Carbon is considered to be an essential element of all organic compounds.

 

Q53.    What are the other elements that are generally present in the organic compounds?

Ans:     Organic compound contain mostly hydrogen and oxygen. Beside these they may contain             elements like nitrogen, sulphur, chlorine, bromine, iodine and some metallic elements like             sodium, magnesium, calcium, copper, lead, silver, iron, lithium, potassium etc. Usually the atomic             number of those elements range form 1 to 20 except Br and I and some metals.

 

Q54.    Which type of bonding is present in organic compounds?

Ans      The bonding in organic compounds is essentially covalent in the form of single, double, triple,      polar or nonpolar bonds.

 

Q55.    What type of reactions take place in organic compounds?

Ans:     Organic compound undergo molecular or nonionic reactions. These reactions include substitution             reaction, addition reactions, elimination reactions, oxidation reactions etc.

Q56.    Can we detect an element directly in organic compound? Or Why are the organic    compounds fused with sodium metal?

Ans:     organic compounds are non-ionic or covalent in nature and do not dissociate into ions when put             in water. In addition, due to covalent bonding they do not undergo ionic reactions instead give             molecular reactions which are very slow. Hence detection of different elements present in             organic compounds directly becomes very difficult by ordinary reactions. Hence they are converted to water-soluble ionic sodium salts by fusing with sodium metal.

 

Q57.    Why is sodium extract prepared?

Ans:     Sodium extract is prepared to change the nature of bonding in organic compounds from covalent             to ionic.

 

Q58.    What happen when sodium extract is prepared?

Ans:     The organic compounds are decomposed and the elements present in them form ionic water             soluble salts with sodium.

 

Q59.    How does the element present in organic compounds change into sodium salt?

Ans:     When an organic compound containing nitrogen, sulphur or halogens is fuses with sodium, the             nitrogen of the organic compound changes to water-soluble sodium cyanide, sulphur changes to  sodium sulphide and halogens are converted into respective sodium halides. If both nitrogen and             sulphur are present, they are converted into sodium sulphocynides.

Q60.    How are the elements detected from the sodium extract?

Ans:     The elements are detected as radicals by means of qualitative analysis. The presence of CN^¯,    S^2-, Cl^¯, Br^- and I^¯ radicals confirm the presence of nitrogen, sulphur, chlorine, bromine and    iodine respectively in the given organic compounds.

 

Q61.    What is Lassaign’s filtrate or Sodium Extract?

Ans:     Sodium Extract or Lassaign’s Filtrate is Sodium Salt containing elements as ions. It is prepared by fusing organic compounds with sodium metal followed by dissolution in water.

 

Q62.    Why sodium metal is preferably taken to make Sodium Extract?

Ans:     It is cheap and soft, so easy to cut into pieces. It melts at low temperature (98°C).

            It is very reactive metal and due to its high chemical reactivity it reacts readily with other elements. In addition, all sodium salts are water soluble and can undergo simple ionic reactions.

 

Q63.    Why sodium metal is always kept in kerosene oil?

Ans:     Sodium is very reactive metal owing to its prior position in the reactivity series. On exposure to             atmosphere it reacts with air and moisture to form its respective oxide and then hydroxide             readily. Thus it is immersed in liquid containing no oxygen to prevent its reaction with air.

Q64.    What is the nature of sodium extract?

Ans:     It is mostly alkaline because of the oxidation of unreacted sodium to basic sodium oxide which    upon dissolving in water yield sodium hydroxide. In addition unreacted sodium directly reacts   with water to form sodium hydroxide.     

 

 

Q65.    Why is concentrated sulphuric acid added in the test for nitrogen?

Ans:     Conc. H₂SO₄ is added in N-Test because first FeSO₄ is converted into green ppt. of ferrous             hydroxide, Fe(OH)₂ by combining with NaOH formed during the preparation of Na-Extract and     these ppt will mask the green or blue colour of nitrogen test and will  mislead our result. The      green   ppt of             ferrous hydroxide is dissolved in conc. H₂SO₄ leaving Prussian blue colour which is a hallmark of nitrogen test.

 

Q66.    We only add the ferrous sulphate, from where the ferric ions come?

Ans: Although we do not provide ferric ions but they already formed by the oxidation of ferrous             sulphate upon heating.

 

Q67.    Is it necessary to add ferric chloride in the test for nitrogen?

Ans: No, it is not compulsory to add ferric chloride in the test for nitrogen. The purpose of addition of ferric chloride is to supply ferric ion, which may already be produced by the oxidation of ferrous ion upon boiling.

 

Q68.    Can concentrated HCl be used instead of H₂SO₄ in the test of nitrogen?

Ans:     HCl can be used in nitrogen test as it also dissolves green ppt of ferrous hydroxide. However it is             better to use sulphuric acid because if HCl is used the colour of the solution turns yellow due to   the formation of ferric chloride,FeCl₃ instead of Prussian blue which may mislead our result.

 

Q69.    Why do we get Prussian blue colour if nitrogen is present?

Ans:     The Prussian blue colour is due to the formation of a complex called ferric ferrocyanide with             formula Fe₄[Fe(CN)₆)]₃     

 

Q70.    Why acetic acid is added before lead acetate in the test for sulphur?

Ans:     Acetic acid is added to neutralize NaOH present in alkaline sodium extract which (NaOH) would             otherwise react with lead acetate to form white ppt of lead   hydroxide.

             

	2NaOH  +   Pb(CH3COO)2	¾¾¾¾®	Pb(OH)2¯	+   2CH3COONa
			White ppt

                                                                            

Q71.    Why a black ppt appear in the test for sulphur?

Ans:     It is because of the formation of black coloured insoluble lead sulphide (PbS). It is noteworthy that all sulphides are black with the exception of CdS, SnS₂ and Sb₂S₃.

 

Q72.    Why does violet colour appear in the confirmatory test for sulphur?

Ans:     The violet colour appears in the C.T. for sulphur due to formation of violet coloured complexcommonly called Sodium sulphoprusside or Sodium thionitroprusside. Its IUPAC name is sodium pentacyanothionitroferrate(II) with formula, Na₄[Fe(CN)₅NOS]

 

Q73.    Why dilute nitric acid is added before adding silver nitrate in the halogen test?

Ans:     Dilute nitric acid is added to neutralize NaOH present in alkaline sodium extract which (NaOH)             would   otherwise react with silver nitrate to form brown ppt of silver hydroxide or silver oxide.             When Na-Extract containing Cl^¯ion is boiled with HNO₃ (dil), the remaining traces of CN^¯ and S^2-                  ions are removed. The presence of CN^¯ and S^2¯ ions in Na-Extract gives white ppt. of AgCN or   Black ppt. of Ag₂S with AgNO₃

              

	NaOH	+	AgNO3	¾¾¾¾®	AgOH¯	+	NaNO3
					Brown ppt
			2AgOH	¾¾¾¾®	Ag2O	+	H2O

 

Q74.    How do you differentiate the presence of chlorine, bromine and iodine in halogen test?

Ans:     In halogens test, the ppt of different colours and their reaction with ammonia to form complex   differentiate chlorine, bromine and iodine

 i. Chlorine is indicated by the formation of white ppt. of AgCl (with AgNO₃) which are soluble in  excess of NH₄OH due to formation of soluble Silver-Ammonia Complex called      diamminesilver(I) Chloride.

ii. Bromine is indicated by the formation of pale yellow ppt. of AgBr (with AgNO₃) which are  partially soluble in excess of NH₄OH due to formation of partially soluble Silver-Ammonia Complex called diamminesilver(I) bromide.

iii. Iodine is indicated by the formation of deep yellow ppt. of AgI (with AgNO₃) which are   insoluble in excess of NH₄OH

 

Q75.    What is layer test?   

Ans      Layer Test is a C.T. for Bromine and Iodine. Layer Test is based upon Displacement Reaction.

 

Q76.    What is the principal or basis of layer test?                    

Ans: The Layer Test is based upon the fact that a more electronegative element (i.e. Cl₂) displaces a weak electronegative element or ions (i.e. Br^- or I^-) from   its salts.  The displaced ions evolve in           the form of coloured vapours which dissolve in CCl₄/CHCl₃ to form coloured layer at the bottom.

 

Q77.What is the function of chlorine water in layer test?            

Ans: The Chlorine water acts as an oxidizing (displacing) agent and oxidizes (displaces) Br^¯ ion or iodide (I^¯) ion from its salt (NaBr or NaI) to Br₂ vapours or I₂ vapours.

 

Q78.    What is the function of chloroform or carbon tetrachloride in the layer test?

Ans:    The function of chloroform (CHCl₃) or carbon tetrachloride (CCl₄) is to dissolve Br₂ vapours or I₂ vapours displaced by Cl₂–water thereby giving brown or violet coloured layer bottom or top.

 

Q79.    Can any other liquid be used instead of carbon tetrachloride in layer test?

Ans:     Yes, any non-oxy organic solvent can be used like benzene, chloroform in layer test. However benzene and chloroform will form upper layer as they are lighter than water while carbon  tetrachloride forms lower layer as it is heavier than water.   

 

Boiling and melting point

 

Q80.    What is boiling?

Ans:     Boiling is the process of rapid conversion of liquid into vapours at specified temperature (i.e. at its       boiling point) when vapour pressure of the liquid becomes equal to atmospheric pressure that     takes place throughout the mass of liquid.

 

Q81.    What is evaporation?

Ans:     Evaporation is the slow conversion of liquid to gaseous state at room or all temperatures that      takes place only at the liquid’s surface.

 

Q82.    What is boiling point (b.p.)?

Ans:     The temperature at which vapour pressure of a liquid becomes equal to outside atmospheric       pressure (which is normally 760 torr) is called boiling point.

 

Q83.    What is normal boiling point (b.p.)?

Ans:     Normal boiling point is the temperature at which vapour pressure of a liquid becomes equal to       standard atmospheric pressure or 760 torr is called boiling point.

 

Q84.    What is the significance of boiling point?

Ans:     Boiling point is the criteria of purity of liquid because a pure liquid boils at a definite temperature.        Thus b.p. is a characteristic constant of a liquid which can be used for testing the purity of a       liquid.

 

Q85.    What is the effect of pressure on b.p?

Ans:     Boiling point increases with the rise of pressure and is lowered with the decrease in pressure i.e. b.p. is directly proportional to pressure.

 

Q86.    What is the effect of height on b.p?

Ans:     With increase in altitude (height), b.p. is reduced as the pressure decreases. That is why cooking food is difficult and takes longer time at high altitude mountain where pressure is below normal i.e. below 760 torr.

 

Q87.    What is the effect of impurities on b.p?

Ans:     Volatile impurities elevate the b.p. of a liquid.

 

Q88.    What is water bath?

Ans:     The beaker containing a non-volatile liquid (e.g. water) which is used for heating the volatile       liquids indirectly is known as water bath.

 

89.    Name some bath liquids

Ans:(i) Paraffin oil (ii)Concentrated sulphuric acid (iii) Glycerine      

 

Q90.    What is vapour pressure?

Ans:The pressure exerted by vapours in equilibrium with its pure liquid at a particular temperature is called vapour pressure or equilibrium pressure.

 

Q91.    What is the effect of temperature on vapour pressure?

Ans:     Vapour pressure increases with the rise of temperature and vice versa.

 

Q92.    What is latent heat of vaporization?

Ans:     The amount of heat energy required to vapourize one gram of a liquid at its b.p. without change in       temperature is called Latent Heat of Vaporization.

 

Q93.    Why at b.p., the temperature remains constant inspite of continuous supply of heat?

Ans:     It is because at b.p, molecules of liquid attain maximum average kinetic energy and further heat       provided at b.p. instead of raising the temperature of the liquid is used to overcome       intermolecular attractive forces and is utilized for converting liquid into vapours. This heat is       called Latent Heat of Vaporization. Thus average K.E. of liquid molecules or temperature of liquid       at b.p. remains constant at its maximum.

 

Q94.    What are the precautions taken while noting boiling point?

Ans:     Following precautions are taken while noting down boiling point:

(i)         The capillary tube should be long enough.

(ii)     The water bath should be heated slowly with constant stirring for uniform          distribution of temperature.

 (iii)       Reading of thermometer should be noted when steady stream of bubbles from the lower  end of capillary tube begin to arise continuously.

 

Q95.    What is melting or fusion?

Ans:     Melting or fusion is the characteristic of solid by virtue of which it is converted into liquid on       heating at its melting point.

 

Q96.    What is melting point (m.p.)?

Ans:     Melting point is the temperature at which both liquid and solid phases co-exist at   equilibrium. OR                         It is the temperature at which there is an equilibrium between solid and liquid phases of a solid.

 

Q97.    What is the significance of melting point?

Ans:     Melting point is the criteria of purity of solids as pure substances have sharp m.p.

 

Q98.    What is effect of impurity on m.p?

Ans:     Impurities (volatile) lower the m.p.

 

Q99.    What is the effect of pressure on m.p. of solid which expands on freezing?

Ans:     The substances which expand on freezing, have a fall in m.p with the increase in pressure i.e. m.p. decreases with pressure.

 

Q100.What is the effect of pressure on m.p. of solid which contracts on freezing?

Ans:     The substances which contract on freezing, have a rise in m.p. with the increase in pressure i.e. m.p. increases with pressure.