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ya ofcourse. Even i need a re-exam but it is not practically possible
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SEMI – MICRO QUALITATIVE ANALYSIS OF SIMPLE INORGANIC SALT
_______________________________________________________________
CATIONS: Pb 2+ ,Hg 2+ ,Cu 2+ ,Cd 2+ , Ag + , Fe 2+ , Fe 3+ , Al 3+ , Zn 2+ , Mn 2+ , Co 2+ , Sr 2+ , Ba 2+ , Mg 2+ , NH4 +
ANIONS: CO3 2- , S 2- , SO3 2- , SO4 2- , NO2 - , NO3 - , Cl - , Br 2- , PO4 3- , CH3COO -
_______________________________________________________________
PRELIMINARY TESTS
1. Note the state (amorphous or crystalline) and colour of salt.
2. Test the solubility of the salt in the following solutions:
i. Water (cold and hot)
ii. Dilute HCl (cold and hot)
iii. Dilute HNO3 (cold and hot)
_______________________________________________________________
IDENTIFICATION OF ACID RADICALS (ANIONS)
This part is the classified into three groups:
FIRST GROUP OF ACID RADICALS
The acid radicals involved in this group are carbonate(CO3 2- ), Sulphide(S 2- ), Sulphite(SO3 2- ),
Thiosulphate(S2O3 2- ) and nitrate(NO2 - ). The group reagent is dilute hydrochloric acid.
EXPERIMENT OBSERVATION INFERENCE
Effervescence or evolution of gases 1 st group of acid radicals is present.
(a) Colourless, odourless gas turns lime water
milky The acid radical may be CO3 2- or HCO3 -(
b) Colourless with rotten eggs smell and turns
lead acetate paper black. Sulphide(S 2- ) is confirmed.
(c) Colourless gas with suffocating smell. Heat
and pass the gas through acidified K2Cr2O7
solution. The solution turns green
SO3 2- is confirmed.
(d) Colourless gas with suffocating smell with a
yellow ppt. of sulphur in the test tube. Heat and
pass the gas through acidified K2Cr2O7 solution.
The solution turns dark blackish green.
S2O3 2- is confirmed.
(e) Colourless gas followed by brown gas and it
turns starch iodide paper blue. NO2 - is confirmed.
Salt + dil. HCl
(f) Colourless gas with the smell of vinegar. The acid radical may be CH3COO --
1 -?Tests to distinguish between CO3 2- and HCO3 -
EXPERIMENT OBSERVATION INFERENCE
(a) Lime water does not turn milky. CO3 2- is confirmed. Salt + water, boil and pass the
gas through lime water (b) Lime water turns milky. HCO3 - is confirmed.
Confirmation test for Acetate
EXPERIMENT OBSERVATION INFERENCE
Salt solution + Neutral FeCl3 sol. Appearance of blood red colour. Acetate is confirmed
SECOND GROUP OF ACID RADICALS
The acid radicals involved in this group are Cl - , Br - and NO3 - . The group reagent is concentrated
sulphuric acid (H2SO4).
EXPERIMENT OBSERVATION INFERENCE
Effervescence with colourless or coloured gases 2 nd group Acid radical is
present
(a) Colourless gas with a pungent smell and gives dense
white fumes when a glass rod dipped in ammonium
hydroxide (NH4OH) is exposed.
The Acid radical may be Cl -(
b) Brown gas and the solution is not blue. The acid radical may be Br -
Salt +
Concentrated
H2SO4 and heat
if necessary.
(c) Light brown gas and brown gas with pieces of copper
turnings and the solution turns blue in the test tube. The Acid radical may be NO3 -Confirmation
test for Chloride
EXPERIMENT OBSERVATION INFERENCE
(a) Chromyl – Chloride test: Salt + few
K2Cr2O7 crystals + conc. H2SO4 and heat
Pass the vapours through the test tube
which contains NaOH solution.
To this yellow solution, add dilute
CH3COOH and lead acetate solution.
Red vapours are obtained.
The solution turns yellow.
Yellow ppt. is formed
Chloride is confirmed.
(b) Silver Nitrate test: Salt solution +
AgNO3 solution + dilute HNO3
White ppt. is formed
which is soluble in NH4OH. Chloride is confirmed.
- 2 -?Confirmation test for Bromide
EXPERIMENT OBSERVATION INFERENCE
(a) Globule Test: Aqueous solution of salt + two
drops of CCl4 + 1ml chlorine water and shake
well.
An orange-red globule is
obtained Bromide is confirmed.
(b)Silver Nitrate test: Salt solution + AgNO3
solution + dilute HNO3
A pale yellow ppt. slightly soluble
in NH4OH is obtained. Bromide is confirmed.
Confirmation test for Nitrate
EXPERIMENT OBSERVATION INFERENCE
Brown ring test: Strong solution of the substance + 2 or
3 drops of conc. H2SO4, and cool. Add freshly prepared
FeSO4 solution on the sides of the test tube.
A brown ring is formed
at the junction of two
liquids.
Nitrate is confirmed.
THIRD GROUP OF ACID RADICALS
The basic radicals involved in this group are SO4 2- and PO4 3- .
EXPERIMENT OBSERVATION INFERENCE
Test for Sulphate: Aqueous solution of salt + dilute
HCl + BaCl2 solution.
A white ppt. insoluble in dilute
HCl is obtained
Sulphate (SO4 2- ) is
confirmed.
Ammonium – Molybdate Test: 2ml of soda extract 1 +
dilute HNO3 + Ammonium Molybdate solution.
A canary yellow precipitate is
obtained.
Phosphate (PO4 3- ) is
confirmed.
1 Soda extract to be prepared if the given salt is partially soluble in water.
- 3 -?ANALYSIS OF BASIC RADICALS OR CATIONS
This is classified into 6 groups. They are mentioned as below:
GROUP RADICALS GROUP REAGENTS
I Pb 2+ ,Ag + Dilute Hydrochloric acid(HCl)
II Pb 2+ ,Hg 2+ ,Cu 2+ ,Cd 2+ Dilute HCl + H2S gas.
III Al 3+ ,Fe 2+ , Fe 3+ NH4Cl(s) + NH4OH
IV Zn 2+ ,Mn 2+ ,Co 2+ ,Ni 2+ NH4Cl(s) + NH4OH + H2S gas
V Ca 2+ ,Sr 2+ ,Ba 2+ NH4Cl(s) + NH4OH + (NH4) 2CO3
VI Mg 2+ , NH4 + -Nil-
Preparation of Original solution: A suitable solvent is found for the salt as given in the preliminary test. A
few drops are used as solution.
Test for NH4 + Radicals: This test is carried out before starting the analysis for the basic radicals.
EXPERIMENT OBSERVATION INFERENCE
(a) Salt + NaOH solution and heat it.
Expose a glass rod dipped in conc.
HCl to the gas and moist red litmus
paper.
Colourless gas with a pungent smell is
obtained.
Dense white fumes and moist red
litmus paper turns blue
NH4 + may be present in the
salt
(b) Original solution + Nesseler’s
reagent. Brown solution or ppt. is obtained NH4 + is confirmed.
FIRST GROUP OF BASIC RADICALS
The basic radicals of this group are Pb 2+ & Ag + .The group reagent is dilute HCl.
EXPERIMENT OBSERVATION INFERENCE
White ppt. is formed. May be Pb 2+ & Ag +
(a) ppt. dissolves in hot water. May be Pb 2+
(1) Original solution + Dilute HCl
Filter and add water to ppt.
and heat (b) ppt. is insoluble in water. May be Ag +
- 4 -?(a) A yellow ppt. is formed. Pb 2+ is confirmed. (2) Original solution + Potassium
Chromate solution. (b) A brick red ppt. is formed. Ag + is confirmed.
(3) Original solution + KCl sol.
ppt. obtained + water and
heat it.
A yellow ppt. soluble in hot
water, on cooling reappears as
golden yellow spangles. Pb 2+ is confirmed. The yellow ppt. is insoluble
in hot water.
SECOND GROUP OF BASIC RADICALS
The basic radicals of this group are Pb 2+ , Hg 2+ , Cu 2+ and Cd 2+ .The group reagent is Dilute HCl + H2S gas.
EXPERIMENT OBSERVATION INFERENCE
May be Pb 2+ Hg 2+ & Cu 2+ (a)A Black ppt. is observed. Original solution + Dilute HCl +
H2S gas. (b)A yellow ppt. is observed. May be Cd 2+
The identification of Pb 2+ , Hg 2+ , Cu 2+ and Cd 2+ are as follows:
EXPERIMENT OBSERVATION INFERENCE
(a) Original solution + Potassium
Chromate. A yellow ppt. is observed. Pb 2+ is confirmed.
(b)Original solution + NH4OH
A bluish white ppt. soluble in excess
of NH4OH is observed which gives
rise to a deep blue solution.
May be Cu 2+
(c) Original solution + dil. acetic
acid + potassium Ferro cyanide sol. A chocolate red ppt. is observed. Cu 2+ is confirmed.
(d) Dissolve the black ppt. in aqua
regia 2 + 1ml of distilled water,
heat & add 2 drops of SnCl2
A white ppt. is observed which turns
grey. Hg 2+ is confirmed.
(e) Original solution + dilute acetic
acid + H2S gas. A yellow ppt. is formed. Cd 2+ is confirmed.
2 Aqua regia is mixture of 3 drops of HCl + 1 drop of HNO3
- 5 -?THIRD GROUP OF BASIC RADICALS
The basic radicals of this group are Al 3+ , Fe 2+ and Fe 3+ .The group reagent is NH4Cl + NH4OH
EXPERIMENT OBSERVATION INFERENCE
May be Al 3+ (a) A white gelatinous ppt. is obtained.
(b) A dirty green ppt. is obtained May be Fe 2+
(1) Original solution +
NH4Cl(s) + NH4OH in
excess (c) A reddish brown ppt. is obtained May be Fe 3+
(a) White gelatinous ppt. soluble in excess of NaOH
is obtained which gives a colourless sol. The basic radical is Al 3+
(b) Dirty green ppt. insoluble in excess of NaOH The basic radical is Fe 2+ (2) Original solution +
NaOH solution
(c) Reddish brown ppt. insoluble in excess of NaOH The basic radical is Fe 3+
(3) Clear solution of
2(a) + solid NH4Cl and
heat
The white gelatinous ppt. reappears. Al 3+ is confirmed.
(4) Original solution +
acidified KMnO4 solution
which is added drop wise.
(a) The pink colour of KMnO4 is not discharged. Fe 3+ is confirmed.
FOURTH GROUP OF BASIC RADICALS
The basic radicals of this group are Zn 2+ , Mn 2+ , Co 2+ & Ni 2+ .The group reagent is NH4Cl + NH4OH
And H2S(g).
EXPERIMENT OBSERVATION INFERENCE
May be Zn 2+ , Mn 2+ or Co 2+ (a) a white ppt. is obtained
(1) Original solution + NH4Cl(s)
+ NH4OH in excess + H2S(g) (b) Buff or pale pink or flash coloured ppt.
soluble in dilute HCl is obtained. May be Mn 2+
(a) The white ppt. is soluble in excess of
NaOH giving a colourless solution. Zn 2+ is confirmed
(2) Original solution + NaOH
solution+. (b) The white ppt. insoluble in excess of
NaOH but turns brown. Mn 2+ is confirmed
- 6 -?A Blue colour in the alcohol
layer is obtained.
FIFTH GROUP OF BASIC RADICALS
The basic radicals of this group are Ca 2+ , Sr 2+ & Ba 2+ .The group reagent is NH4Cl(s) + NH4OH and
(NH4)2 CO3.
EXPERIMENT OBSERVATION INFERENCE
(1) Original solution + NH4Cl(s) + NH4OH in
excess + (NH4)2 CO3 solution. (a) A white ppt. is obtained. May be Ca 2+ , Sr 2+ or
Ba 2+
A yellow ppt. is obtained. May be Ba 2+ .
A thin white ppt. is obtained. May be Sr 2+ .
(2) Dissolve the white ppt. obtained in small
amounts of dilute acetic acid. Divide the
solution in three parts.
Part (1) + Potassium chromate solution.
Part (2) + Calcium Sulphate Sol. and heat.
Part (3) + Ammonium exalate solution. A white ppt. is obtained. May be Ca 2+ .
(a) Apple green coloured flame is
obtained. Ba 2+ is confirmed.
(b) Crimson red coloured flame is
obtained. Sr 2+ is confirmed.
(3)Flame tests: Make a paste of the given
salt with few drops of Conc. HCl in a watch
glass. Heat a Pt. wire in non luminous flame
till no colour is imparted to the flame. Dip
the Pt. wire in the paste and hold it in the
flame. (c) Brick red coloured flame is
obtained, Ca 2+ is confirmed.
SIXTH GROUP OF BASIC RADICALS
EXPERIMENT OBSERVATION INFERENCE
(1) Original solution + excess of NH4OH +
Ammonium Hydrogen Phosphate. A white ppt. is obtained. Mg 2+ is confirmed.
______________________________________________________________
*****THE END*****
Co 2+ is confirmed
(3) Dissolve the 1(a) in aqua regia, evaporate to
Dryness, add 1ml of distilled water and divide
the resultant product into two parts.
Part (1) + 1 ml of amyl alcohol + 100 mg of solid
NH4SCN and stir it.
Part (2) + 5 drops of dimethyl glyaxime reagent +
NH4OH. A pink ppt. is obtained. Ni 2+ is confirmed.

Community shelf Community shelf -> salt analysis -> Go to message
This Post 5 points    (Olaaa!! Perrrfect answer.   in 1 votes )   [?]

SEMI – MICRO QUALITATIVE ANALYSIS OF SIMPLE INORGANIC SALT
_______________________________________________________________
CATIONS: Pb 2+ ,Hg 2+ ,Cu 2+ ,Cd 2+ , Ag + , Fe 2+ , Fe 3+ , Al 3+ , Zn 2+ , Mn 2+ , Co 2+ , Sr 2+ , Ba 2+ , Mg 2+ , NH4 +
ANIONS: CO3 2- , S 2- , SO3 2- , SO4 2- , NO2 - , NO3 - , Cl - , Br 2- , PO4 3- , CH3COO -
_______________________________________________________________
PRELIMINARY TESTS
1. Note the state (amorphous or crystalline) and colour of salt.
2. Test the solubility of the salt in the following solutions:
i. Water (cold and hot)
ii. Dilute HCl (cold and hot)
iii. Dilute HNO3 (cold and hot)
_______________________________________________________________
IDENTIFICATION OF ACID RADICALS (ANIONS)
This part is the classified into three groups:
FIRST GROUP OF ACID RADICALS
The acid radicals involved in this group are carbonate(CO3 2- ), Sulphide(S 2- ), Sulphite(SO3 2- ),
Thiosulphate(S2O3 2- ) and nitrate(NO2 - ). The group reagent is dilute hydrochloric acid.
EXPERIMENT OBSERVATION INFERENCE
Effervescence or evolution of gases 1 st group of acid radicals is present.
(a) Colourless, odourless gas turns lime water
milky The acid radical may be CO3 2- or HCO3 -(
b) Colourless with rotten eggs smell and turns
lead acetate paper black. Sulphide(S 2- ) is confirmed.
(c) Colourless gas with suffocating smell. Heat
and pass the gas through acidified K2Cr2O7
solution. The solution turns green
SO3 2- is confirmed.
(d) Colourless gas with suffocating smell with a
yellow ppt. of sulphur in the test tube. Heat and
pass the gas through acidified K2Cr2O7 solution.
The solution turns dark blackish green.
S2O3 2- is confirmed.
(e) Colourless gas followed by brown gas and it
turns starch iodide paper blue. NO2 - is confirmed.
Salt + dil. HCl
(f) Colourless gas with the smell of vinegar. The acid radical may be CH3COO --
1 -?Tests to distinguish between CO3 2- and HCO3 -
EXPERIMENT OBSERVATION INFERENCE
(a) Lime water does not turn milky. CO3 2- is confirmed. Salt + water, boil and pass the
gas through lime water (b) Lime water turns milky. HCO3 - is confirmed.
Confirmation test for Acetate
EXPERIMENT OBSERVATION INFERENCE
Salt solution + Neutral FeCl3 sol. Appearance of blood red colour. Acetate is confirmed
SECOND GROUP OF ACID RADICALS
The acid radicals involved in this group are Cl - , Br - and NO3 - . The group reagent is concentrated
sulphuric acid (H2SO4).
EXPERIMENT OBSERVATION INFERENCE
Effervescence with colourless or coloured gases 2 nd group Acid radical is
present
(a) Colourless gas with a pungent smell and gives dense
white fumes when a glass rod dipped in ammonium
hydroxide (NH4OH) is exposed.
The Acid radical may be Cl -(
b) Brown gas and the solution is not blue. The acid radical may be Br -
Salt +
Concentrated
H2SO4 and heat
if necessary.
(c) Light brown gas and brown gas with pieces of copper
turnings and the solution turns blue in the test tube. The Acid radical may be NO3 -Confirmation
test for Chloride
EXPERIMENT OBSERVATION INFERENCE
(a) Chromyl – Chloride test: Salt + few
K2Cr2O7 crystals + conc. H2SO4 and heat
Pass the vapours through the test tube
which contains NaOH solution.
To this yellow solution, add dilute
CH3COOH and lead acetate solution.
Red vapours are obtained.
The solution turns yellow.
Yellow ppt. is formed
Chloride is confirmed.
(b) Silver Nitrate test: Salt solution +
AgNO3 solution + dilute HNO3
White ppt. is formed
which is soluble in NH4OH. Chloride is confirmed.
- 2 -?Confirmation test for Bromide
EXPERIMENT OBSERVATION INFERENCE
(a) Globule Test: Aqueous solution of salt + two
drops of CCl4 + 1ml chlorine water and shake
well.
An orange-red globule is
obtained Bromide is confirmed.
(b)Silver Nitrate test: Salt solution + AgNO3
solution + dilute HNO3
A pale yellow ppt. slightly soluble
in NH4OH is obtained. Bromide is confirmed.
Confirmation test for Nitrate
EXPERIMENT OBSERVATION INFERENCE
Brown ring test: Strong solution of the substance + 2 or
3 drops of conc. H2SO4, and cool. Add freshly prepared
FeSO4 solution on the sides of the test tube.
A brown ring is formed
at the junction of two
liquids.
Nitrate is confirmed.
THIRD GROUP OF ACID RADICALS
The basic radicals involved in this group are SO4 2- and PO4 3- .
EXPERIMENT OBSERVATION INFERENCE
Test for Sulphate: Aqueous solution of salt + dilute
HCl + BaCl2 solution.
A white ppt. insoluble in dilute
HCl is obtained
Sulphate (SO4 2- ) is
confirmed.
Ammonium – Molybdate Test: 2ml of soda extract 1 +
dilute HNO3 + Ammonium Molybdate solution.
A canary yellow precipitate is
obtained.
Phosphate (PO4 3- ) is
confirmed.
1 Soda extract to be prepared if the given salt is partially soluble in water.
- 3 -?ANALYSIS OF BASIC RADICALS OR CATIONS
This is classified into 6 groups. They are mentioned as below:
GROUP RADICALS GROUP REAGENTS
I Pb 2+ ,Ag + Dilute Hydrochloric acid(HCl)
II Pb 2+ ,Hg 2+ ,Cu 2+ ,Cd 2+ Dilute HCl + H2S gas.
III Al 3+ ,Fe 2+ , Fe 3+ NH4Cl(s) + NH4OH
IV Zn 2+ ,Mn 2+ ,Co 2+ ,Ni 2+ NH4Cl(s) + NH4OH + H2S gas
V Ca 2+ ,Sr 2+ ,Ba 2+ NH4Cl(s) + NH4OH + (NH4) 2CO3
VI Mg 2+ , NH4 + -Nil-
Preparation of Original solution: A suitable solvent is found for the salt as given in the preliminary test. A
few drops are used as solution.
Test for NH4 + Radicals: This test is carried out before starting the analysis for the basic radicals.
EXPERIMENT OBSERVATION INFERENCE
(a) Salt + NaOH solution and heat it.
Expose a glass rod dipped in conc.
HCl to the gas and moist red litmus
paper.
Colourless gas with a pungent smell is
obtained.
Dense white fumes and moist red
litmus paper turns blue
NH4 + may be present in the
salt
(b) Original solution + Nesseler’s
reagent. Brown solution or ppt. is obtained NH4 + is confirmed.
FIRST GROUP OF BASIC RADICALS
The basic radicals of this group are Pb 2+ & Ag + .The group reagent is dilute HCl.
EXPERIMENT OBSERVATION INFERENCE
White ppt. is formed. May be Pb 2+ & Ag +
(a) ppt. dissolves in hot water. May be Pb 2+
(1) Original solution + Dilute HCl
Filter and add water to ppt.
and heat (b) ppt. is insoluble in water. May be Ag +
- 4 -?(a) A yellow ppt. is formed. Pb 2+ is confirmed. (2) Original solution + Potassium
Chromate solution. (b) A brick red ppt. is formed. Ag + is confirmed.
(3) Original solution + KCl sol.
ppt. obtained + water and
heat it.
A yellow ppt. soluble in hot
water, on cooling reappears as
golden yellow spangles. Pb 2+ is confirmed. The yellow ppt. is insoluble
in hot water.
SECOND GROUP OF BASIC RADICALS
The basic radicals of this group are Pb 2+ , Hg 2+ , Cu 2+ and Cd 2+ .The group reagent is Dilute HCl + H2S gas.
EXPERIMENT OBSERVATION INFERENCE
May be Pb 2+ Hg 2+ & Cu 2+ (a)A Black ppt. is observed. Original solution + Dilute HCl +
H2S gas. (b)A yellow ppt. is observed. May be Cd 2+
The identification of Pb 2+ , Hg 2+ , Cu 2+ and Cd 2+ are as follows:
EXPERIMENT OBSERVATION INFERENCE
(a) Original solution + Potassium
Chromate. A yellow ppt. is observed. Pb 2+ is confirmed.
(b)Original solution + NH4OH
A bluish white ppt. soluble in excess
of NH4OH is observed which gives
rise to a deep blue solution.
May be Cu 2+
(c) Original solution + dil. acetic
acid + potassium Ferro cyanide sol. A chocolate red ppt. is observed. Cu 2+ is confirmed.
(d) Dissolve the black ppt. in aqua
regia 2 + 1ml of distilled water,
heat & add 2 drops of SnCl2
A white ppt. is observed which turns
grey. Hg 2+ is confirmed.
(e) Original solution + dilute acetic
acid + H2S gas. A yellow ppt. is formed. Cd 2+ is confirmed.
2 Aqua regia is mixture of 3 drops of HCl + 1 drop of HNO3
- 5 -?THIRD GROUP OF BASIC RADICALS
The basic radicals of this group are Al 3+ , Fe 2+ and Fe 3+ .The group reagent is NH4Cl + NH4OH
EXPERIMENT OBSERVATION INFERENCE
May be Al 3+ (a) A white gelatinous ppt. is obtained.
(b) A dirty green ppt. is obtained May be Fe 2+
(1) Original solution +
NH4Cl(s) + NH4OH in
excess (c) A reddish brown ppt. is obtained May be Fe 3+
(a) White gelatinous ppt. soluble in excess of NaOH
is obtained which gives a colourless sol. The basic radical is Al 3+
(b) Dirty green ppt. insoluble in excess of NaOH The basic radical is Fe 2+ (2) Original solution +
NaOH solution
(c) Reddish brown ppt. insoluble in excess of NaOH The basic radical is Fe 3+
(3) Clear solution of
2(a) + solid NH4Cl and
heat
The white gelatinous ppt. reappears. Al 3+ is confirmed.
(4) Original solution +
acidified KMnO4 solution
which is added drop wise.
(a) The pink colour of KMnO4 is not discharged. Fe 3+ is confirmed.
FOURTH GROUP OF BASIC RADICALS
The basic radicals of this group are Zn 2+ , Mn 2+ , Co 2+ & Ni 2+ .The group reagent is NH4Cl + NH4OH
And H2S(g).
EXPERIMENT OBSERVATION INFERENCE
May be Zn 2+ , Mn 2+ or Co 2+ (a) a white ppt. is obtained
(1) Original solution + NH4Cl(s)
+ NH4OH in excess + H2S(g) (b) Buff or pale pink or flash coloured ppt.
soluble in dilute HCl is obtained. May be Mn 2+
(a) The white ppt. is soluble in excess of
NaOH giving a colourless solution. Zn 2+ is confirmed
(2) Original solution + NaOH
solution+. (b) The white ppt. insoluble in excess of
NaOH but turns brown. Mn 2+ is confirmed
- 6 -?A Blue colour in the alcohol
layer is obtained.
FIFTH GROUP OF BASIC RADICALS
The basic radicals of this group are Ca 2+ , Sr 2+ & Ba 2+ .The group reagent is NH4Cl(s) + NH4OH and
(NH4)2 CO3.
EXPERIMENT OBSERVATION INFERENCE
(1) Original solution + NH4Cl(s) + NH4OH in
excess + (NH4)2 CO3 solution. (a) A white ppt. is obtained. May be Ca 2+ , Sr 2+ or
Ba 2+
A yellow ppt. is obtained. May be Ba 2+ .
A thin white ppt. is obtained. May be Sr 2+ .
(2) Dissolve the white ppt. obtained in small
amounts of dilute acetic acid. Divide the
solution in three parts.
Part (1) + Potassium chromate solution.
Part (2) + Calcium Sulphate Sol. and heat.
Part (3) + Ammonium exalate solution. A white ppt. is obtained. May be Ca 2+ .
(a) Apple green coloured flame is
obtained. Ba 2+ is confirmed.
(b) Crimson red coloured flame is
obtained. Sr 2+ is confirmed.
(3)Flame tests: Make a paste of the given
salt with few drops of Conc. HCl in a watch
glass. Heat a Pt. wire in non luminous flame
till no colour is imparted to the flame. Dip
the Pt. wire in the paste and hold it in the
flame. (c) Brick red coloured flame is
obtained, Ca 2+ is confirmed.
SIXTH GROUP OF BASIC RADICALS
EXPERIMENT OBSERVATION INFERENCE
(1) Original solution + excess of NH4OH +
Ammonium Hydrogen Phosphate. A white ppt. is obtained. Mg 2+ is confirmed.
______________________________________________________________
*****THE END*****
Co 2+ is confirmed
(3) Dissolve the 1(a) in aqua regia, evaporate to
Dryness, add 1ml of distilled water and divide
the resultant product into two parts.
Part (1) + 1 ml of amyl alcohol + 100 mg of solid
NH4SCN and stir it.
Part (2) + 5 drops of dimethyl glyaxime reagent +
NH4OH. A pink ppt. is obtained. Ni 2+ is confirmed.

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This Post 5 points    (Olaaa!! Perrrfect answer.   in 1 votes )   [?]
 

Factors that Affect the Rate of a Reaction

There are 4 basic factors that can affect the rate of a chemical reaction:

1. Temperature

Temperature and rate at AS level is connected very closely with the Maxwell-Boltzmann Distribution.

An increase or decrease in temperature will change the shape of the curve which means fewer or more particles will have the required activation energy.

As the number of molecules is constant so the areas under the 2 curves are the same.

BUT, the average energy of the molecules in curve T2 is greater, so there are more particles with enough energy to react. So there are basically 2 factors to consider:

  • Increase in energy of particles due to increase in average energy of particles.

  • Increase in speed of particles do there are more successful collisions with particles having the required activation energy.

These two factors result in the rate of the reaction increasing.

2. Concentration and pressure

If the concentration or pressure of a chemical increases, there will be more particles within a given space. 

  • Particles with therefore collide more often

  • As concentration or pressure increases, the rate of reaction also increases.

3. Physical state

If particles are in the same phase (liquid/liquid) or (gas/gas), then it is very easy for them to mix with each other. This gives particles the maximum opportunity to collide.

BUT, if one of the reactants is a solid, then the reaction can only take place on the surface of the solid.

  • The smaller the size of the solid particles, the greater the area that the reaction can take place in.

So finely divided powder reacts more quickly than the same stuff in a great big lump!

A situation that can arise with a 2 liquids or 2 gases is if they are immiscible (can't mix), this will mean that the reaction can only occur at the interface between the two fluids.

4. Catalyst

What is a catalyst?

A catalyst is a substance that alters the rate of a chemical reaction without being used up or permanently changed chemically.

How does it work?

A catalyst works by changing the energy pathway for a chemical reaction. It provides an alternative route (mechanism) that lowers the Activation Energy meaning more particles now have the required energy needed to undergo a successful collision. 

The above graph demonstrates what a catalyst does to the reaction profile. You will be expected to produce one of these!

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Entropy and the Carnot Cycle

The efficiency of a heat engine cycle is given by

For the ideal case of the Carnot cycle, this efficiency can be written

Using these two expressions together

If we take Q to represent heat added to the system, then heat taken from the system will have a negative value. For the Carnot cycle

which can be generalized as an integral around a reversible cycle

Clausius Theorem

For any part of the heat engine cycle, this can be used to define a change in entropy S for the system

or in differential form at any point in the cycle

For any irreversible process, the efficiency is less than that of the Carnot cycle. This can be associated with less heat flow to the system and/or more heat flow out of the system. The inevitable result is

Clausius Inequality

Any real engine cycle will result in more entropy given to the environment than was taken from it, leading to an overall net increase in entropy.

 

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Song from Jab We Met by Sonu Nigam and Javed Ali. Notice how Shahid Kapoor's voice keeps changing back and forth between Sonu Nigam and Javed Ali. Hmmm
Catalogs Discussion Forums -> Organic Chemistry -> why is orhto nitro phenol steam volatile?? -> Go to message
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due to intermolecular hydrogen bonding
Catalogs Discussion Forums -> Organic Chemistry -> properties of chloroform -> Go to message
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ACTS AS:.Anesthetic.As a solvent.As a reagent in organic synthesis.

Properties
Molecular formula CHCl3
Molar mass 119.38 g/mol
Appearance Colorless liquid
Density 1.48 g/cm3
Melting point

-63.5 °C

Boiling point

61.2 °C

Solubility in water 0.8 g/100 ml at 20 °C

Chloroform undergoes further chlorination to give CCl4:

CHCl3 + Cl2 → CCl4 + HCl.

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ACTS AS:.Anesthetic.
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We begin with a common-sense geometrical fact:

 

somewhere between two zeros of a non-constant continuous function f, the function must change direction







For a differentiable function, the derivative is 0 at the point where f changes direction. Thus, we expect there to be a point c where the tangent is horizontal. These ideas are precisely stated by Rolle's Theorem:


Rolle's Theorem

Let f be differentiable on (a,b) and continuous on [a,b]. If f(a) = f(b) = 0, then there is at least one point c in (a,b) for which f¢(c) = 0.


Notice that both conditions on f are necessary. Without either one, the statement is false!

For a discontinuous function,
the conclusion of Rolle's Theorem may not hold:
Figure 2
   For a continuous, non-differentiable function,
again this might not be the case:
Figure 3

Though the theorem seems logical, we cannot be sure that it is always true without a proof.

 

Proof of Rolle's Theorem

The Mean Value Theorem is a generalization of Rolle's Theorem:

We now let f(a) and f(b) have values other than 0 and look at the secant line through (a, f(a)) and (b, f(b)). We expect that somewhere between a and b there is a point c where the tangent is parallel to this secant.

 

Figure 4
In Rolle's Theorem the secant
was horizontal so we looked
for a horizontal tangent.

That is, the slopes of these two lines are equal. This is formalized in the Mean Value Theorem.


Mean Value Theorem

Let f be differentiable on (a,b) and continuous on [a,b]. Then there is at least one point c in (a,b) for which

f¢(c) = f(b)-f(a)
b-a
.

Here, f¢(c) is the slope of the tangent at c, while (f(b)-f(a))/( b-a) is the slope of the secant through a and b. Intuitively, we see that if we translate the secant line in the figure upwards, it will eventually just touch the curve at the single point c and will be tangent at c. However, basing conclusions on a single example can be disastrous, so we need a proof.

 

Proof of the Mean Value Theorem

Consequences of the Mean Value Theorem

The Mean Value Theorem is behind many of the important results in calculus. The following statements, in which we assume f is differentiable on an open interval I, are consequences of the Mean Value Theorem:

  • f¢(x) = 0 everywhere on I if and only if f is constant on I.

  • If f¢(x) = g¢(x) for all x on I, then f and g differ at most by a constant on I.

  • If f¢(x) > 0 for all x on I, then f is increasing on I.
    If f¢(x) < 0 for all x on I, then f is decreasing on I.


Key Concept

Mean Value Theorem

Let f be differentiable on (a,b) and continuous on [a,b]. Then there is at least one point c in (a,b) for which

f¢(c) = f(b)-f(a)
b-a
.


Community shelf Community shelf -> the mean value theorem -> Go to message
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We begin with a common-sense geometrical fact:

 

somewhere between two zeros of a non-constant continuous function f, the function must change direction







For a differentiable function, the derivative is 0 at the point where f changes direction. Thus, we expect there to be a point c where the tangent is horizontal. These ideas are precisely stated by Rolle's Theorem:


Rolle's Theorem

Let f be differentiable on (a,b) and continuous on [a,b]. If f(a) = f(b) = 0, then there is at least one point c in (a,b) for which f¢(c) = 0.


Notice that both conditions on f are necessary. Without either one, the statement is false!

For a discontinuous function,
the conclusion of Rolle's Theorem may not hold:
Figure 2
   For a continuous, non-differentiable function,
again this might not be the case:
Figure 3

Though the theorem seems logical, we cannot be sure that it is always true without a proof.

 

Proof of Rolle's Theorem

The Mean Value Theorem is a generalization of Rolle's Theorem:

We now let f(a) and f(b) have values other than 0 and look at the secant line through (a, f(a)) and (b, f(b)). We expect that somewhere between a and b there is a point c where the tangent is parallel to this secant.

 

Figure 4
In Rolle's Theorem the secant
was horizontal so we looked
for a horizontal tangent.

That is, the slopes of these two lines are equal. This is formalized in the Mean Value Theorem.


Mean Value Theorem

Let f be differentiable on (a,b) and continuous on [a,b]. Then there is at least one point c in (a,b) for which

f¢(c) = f(b)-f(a)
b-a
.

Here, f¢(c) is the slope of the tangent at c, while (f(b)-f(a))/( b-a) is the slope of the secant through a and b. Intuitively, we see that if we translate the secant line in the figure upwards, it will eventually just touch the curve at the single point c and will be tangent at c. However, basing conclusions on a single example can be disastrous, so we need a proof.

 

Proof of the Mean Value Theorem

Consequences of the Mean Value Theorem

The Mean Value Theorem is behind many of the important results in calculus. The following statements, in which we assume f is differentiable on an open interval I, are consequences of the Mean Value Theorem:

  • f¢(x) = 0 everywhere on I if and only if f is constant on I.

  • If f¢(x) = g¢(x) for all x on I, then f and g differ at most by a constant on I.

  • If f¢(x) > 0 for all x on I, then f is increasing on I.
    If f¢(x) < 0 for all x on I, then f is decreasing on I.


Key Concept

Mean Value Theorem

Let f be differentiable on (a,b) and continuous on [a,b]. Then there is at least one point c in (a,b) for which

f¢(c) = f(b)-f(a)
b-a
.


Community shelf Community shelf -> The Mean Value Theorem -> Go to message
This Post 0 points    (Olaaa!! Perrrfect answer.   in 0 votes )   [?]

We begin with a common-sense geometrical fact:

 

somewhere between two zeros of a non-constant continuous function f, the function must change direction







For a differentiable function, the derivative is 0 at the point where f changes direction. Thus, we expect there to be a point c where the tangent is horizontal. These ideas are precisely stated by Rolle's Theorem:


Rolle's Theorem

Let f be differentiable on (a,b) and continuous on [a,b]. If f(a) = f(b) = 0, then there is at least one point c in (a,b) for which f¢(c) = 0.


Notice that both conditions on f are necessary. Without either one, the statement is false!

For a discontinuous function,
the conclusion of Rolle's Theorem may not hold:
Figure 2
   For a continuous, non-differentiable function,
again this might not be the case:
Figure 3

Though the theorem seems logical, we cannot be sure that it is always true without a proof.

 

Proof of Rolle's Theorem

The Mean Value Theorem is a generalization of Rolle's Theorem:

We now let f(a) and f(b) have values other than 0 and look at the secant line through (a, f(a)) and (b, f(b)). We expect that somewhere between a and b there is a point c where the tangent is parallel to this secant.

 

Figure 4
In Rolle's Theorem the secant
was horizontal so we looked
for a horizontal tangent.

That is, the slopes of these two lines are equal. This is formalized in the Mean Value Theorem.


Mean Value Theorem

Let f be differentiable on (a,b) and continuous on [a,b]. Then there is at least one point c in (a,b) for which

f¢(c) = f(b)-f(a)
b-a
.

Here, f¢(c) is the slope of the tangent at c, while (f(b)-f(a))/( b-a) is the slope of the secant through a and b. Intuitively, we see that if we translate the secant line in the figure upwards, it will eventually just touch the curve at the single point c and will be tangent at c. However, basing conclusions on a single example can be disastrous, so we need a proof.

 

Proof of the Mean Value Theorem

Consequences of the Mean Value Theorem

The Mean Value Theorem is behind many of the important results in calculus. The following statements, in which we assume f is differentiable on an open interval I, are consequences of the Mean Value Theorem:

  • f¢(x) = 0 everywhere on I if and only if f is constant on I.

  • If f¢(x) = g¢(x) for all x on I, then f and g differ at most by a constant on I.

  • If f¢(x) > 0 for all x on I, then f is increasing on I.
    If f¢(x) < 0 for all x on I, then f is decreasing on I.


Key Concept

Mean Value Theorem

Let f be differentiable on (a,b) and continuous on [a,b]. Then there is at least one point c in (a,b) for which

f¢(c) = f(b)-f(a)
b-a
.


Catalogs Discussion Forums -> Computer Science -> C++ program -> Go to message
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replace >> by <<.add ; after return 0.
Community shelf Community shelf -> L'Hospital's Rule -> Go to message
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Let lim stand for the limit lim_(x->c), lim_(x->c^-), lim_(x->c^+), lim_(x->infty), or lim_(x->-infty), and suppose that lim f(x) and lim g(x) are both zero or are both +/-infty. If

 lim(f^'(x))/(g^'(x))
(1)

has a finite value or if the limit is +/-infty, then

 lim(f(x))/(g(x))=lim(f^'(x))/(g^'(x)).
(2)

Historically, this result first appeared in l'Hospital's 1696 treatise, which was the first textbook on differential calculus. Within the book, l'Hospital thanks the Bernoulli brothers for their assistance and their discoveries. An earlier letter by John Bernoulli gives both the rule and its proof, so it seems likely that Bernoulli discovered the rule (Larson et al. 1999, p. 524).

Note that l'Hospital's name is commonly seen spelled both "l'Hospital" (e.g., Maurer 1981, p. 426; Arfken 1985, p. 310) and "l'Hôpital" (e.g., Maurer 1981, p. 426; Gray 1997, p. 529), the two being equivalent in French spelling.

LHospitalsRuleOscillatory

L'Hospital's rule occasionally fails to yield useful results, as in the case of the function lim_(u->infty)u(u^2+1)^(-1/2), illustrated above. Repeatedly applying the rule in this case gives expressions which oscillate and never converge,

lim_(u->infty)u/((u^2+1)^(1/2)) = lim_(u->infty)1/(u(u^2+1)^(-1/2))
(3)
= lim_(u->infty)((u^2+1)^(1/2))/u
(4)
= lim_(u->infty)(u(u^2+1)^(-1/2))/1
(5)
= lim_(u->infty)u/((u^2+1)^(1/2))
(6)
= ....
(7)

The actual limit is 1.

LHospitalsRule1

L'Hospital's rule must sometimes be applied with some care, since it holds only in the implicitly understood case that g^'(x) does not change sign infinitely often in a neighborhood of infty. For example, consider the limit f(x)/g(x) with

f(x) = x+cosxsinx
(8)
g(x) = e^(sinx)(x+cosxsinx)
(9)

as x->infty. While both f(x) and g(x) approach infty as x->infty, the limit of the ratio is bounded inside the interval [1/e,e], while the limit of f^'(x)/g^'(x) approaches 0 (Boas 1986).

LHospitalsRule2

Another similar example is the limit f(x)/g(x) with

f(x) = xsin(x^(-4))e^(-1/x^2)
(10)
g(x) = e^(-1/x^2)
(11)

as

x->0

. While both

f(x)

and

g(x)

approach 0 as

x->0

, the limit of the ratio is 0, while the limit

f^'(x)/g^'(x)

is unbounded on the real line.

Community shelf Articles -> impossible drawing -> Go to message
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challenge a friend to try this trick!




Show your friend this diagram. They must try and draw it without going over the same line twice and without taking the pencil off the paper .

They'll never do it until you show them the sneaky method!



The Secret:



Start doing the drawing like this:

( Obviously you make your line touch at the corners and in the centre. We've just put gaps here to show how to draw it. )

Once you've got this far then you're stuck... or are you?

To draw the last bit, fold the edge of the paper over !



Now you can finish the drawing and you haven't taken the pencil off the paper!

When you open out the paper again, you'll get the diagram as it should be.

This sneaky trick might seem obvious now you know what to do, but if you try it on someone they'll be worrying about it for hours!

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Corey-Winter Olefin Synthesis
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Community shelf Community shelf -> Magic Bottle Trick -> Go to message
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Effect: 

The magician asks for a volunteer from the audience who looks inside a bottle to make sure it's a normal, empty container. 

The volunteer returns the bottle and then examines a magic wand to ensure it is normal. 

The magician drops the wand into the bottle (noting how easily it falls in).

He/she turns bottle over and lets go of the wand.  

The wand magically remains suspended in the bottle.

Supplies:

~a magic wand (could use a pencil instead) that is taller than the bottle (when you drop the wand into the bottle, part of it should still be sticking up through the opening.
~a bottle that has an opening large enough to fit the wand in (but not too big).  The bottle cannot be see-through 
~an eraser

If you don't have a bottle that is dark, you can put some dark paint inside the bottle and shake it around so the inside is painted.

Cut a piece off the eraser, just large enough to wedge the wand into the opening of the bottle.  The eraser is what makes the trick work!

Secret:

Pass the bottle to the volunteer and ask them to make sure its empty.

Take back the bottle and give your friend the wand.  Meanwhile, slip the piece of eraser into the bottle without anyone seeing (you can have it in your pocket until this point).   You'll need to practice this a few times.

Take the wand back and drop it into the bottle.

Pick up the wand and bottle and turn them VERY SLOWLY upside down (mumble all sorts of enchantments while you're doing this).  Pull on the wand slightly when the bottle is turning over so the piece of eraser gets wedged into the opening (you'll need to practice this a few times too).

Let go of the wand... PRESTO!  It doesn't fall out.

Slowly turn upright again.  Let go of everything and PRESTO the wand remains suspended in the bottle (it doesn't fall back down).

To remove the wand, push it slightly to release the rubber and then take it out.

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This reaction occurs in two steps. The alkyl halide is treated with lithium metal, and solvated in ether, which converts the alkyl halide into an alkyl lithium compound, R-Li. The starting R-X can be primary, secondary or tertiary alkyl halide:

R-X + 2Li → R-Li + Li-X

The second step requires the alkyl lithium compound to be treated with cuprous iodide (CuI). This creates a lithium dialkyl cuprate compound. These compounds were first synthesized by Henry Gilman of Iowa State University, and are usually called Gilman reagents in honor of his contributions:

2RLi + CuI → R2CuLi + LiI

The lithium dialkyl cuprate is then treated with the second alkyl halide, which couples to the compound:

R2CuLi + R'-X → R-R' + RCu + LiX

If second alkyl halide is not the same as the first, then cross-products are formed.

It is important to note that for this reaction to work successfully, the alkyl halide must be a methyl halide, benzyl halide, primary alkyl halide or a secondary cyclo alkyl halide. The relative simplicity of this reaction makes it a useful technique for synthesizing organic compounds.

 
 
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