To prepare M/50 standard solution of oxalic acid. With its help, determine the molarity and strength of the given solution of potassium permanganate (KMnO4).

Aim- To prepare M/50 standard solution of oxalic acid. With its help, determine the molarity and strength of the given solution of potassium permanganate (KMnO₄).

Apparatus required- Chemical balance, weight box, conical flask, burette, Burette stand, 100 ml Beaker, 20 ml pipette, funnel, wire gauge, wash bottle and burner.

Chemical required- potassium permanganate, dilute sulphuric acid, distilled water and oxalic acid.

Theory-

     ·     Oxalic is a double salt and act as reducing agent against KMnO₄

    ·       Whereas potassium permanganate act as an oxidising agent and it oxidises iron present in Mohr’s salt. in the acidic medium KMnO₄ behave as a very strong oxidizing agent. So acidity is introduced by adding dil. H₂SO₄.

    ·       It oxidises oxalic acid into carbon-dioxide and water.

     ·       Hence the reaction between the titrant (KMnO₄) and the analyte (Mohr’s salt) is a redox reaction, Therefore the titration is called redox reaction.

     ·       The reaction between potassium permanganate and oxalic acid at room temperature is slow due to which oxalic after acidification in the conical is heated to about (50-60) C.

·       Potassium permanganate act as a self-indicator for this reaction.  

     ·       The colour at the end point changes from colourless to light pink. The endpoint of a titration is the point at which the reaction between the titrant and the analyte becomes complete.

     Chemical Reaction involved-

     (a)    Molecular equation

2KMnO₄   + 3H₂ SO₄    ------>   K₂SO₄   +   2MnSO₄   +   3H₂O +   5[O] -----(1)  

[C₂O₄H₂   +  [O] ------>   10CO_{2 +}  3H₂O ] x 5 ---------(2)                         

by combining of eq. (1) and (2), Then, Net reaction may be written as 

2KMnO₄ + 3H₂ SO₄  + 5C₂O₄H₂ .2H₂O----->  K₂SO₄ + 2MnSO₄ +8H₂O + 10CO_2 --- (3) 

(b. Ionic equation-

 MnO₄^-   +    8H⁺     + 5e^-    ------>  2Mn ²⁺    +   4H₂O] x2      ----------- (4) 

[C₂O₄^2-   ----->    CO₂   + 2e^-  ]   x5      ---------(5) 

By combing of equation (4) and (5), then equation may be written as-

2MnO₄^-   +  16H⁺   +  5 C₂O₄^2------>     2Mn ²⁺  +   8H₂O   +  5CO₂  -------(6)      

Procedure-

    ·       1.26g of oxalic acid are weighed and dissolved in 500ml of water in a measuring flask to prepare M/50 solution of it.

    ·       with the help of pipette, 20 ml oxalic acid is taken out in titration flask and one test tube (10ml) full of dilute H₂SO₄ is added to it. Since it makes acidic medium to the reaction .then it is heated about 50-60⁰C.

    ·       Burette is washed with distilled water and is filled with KMnO₄ solution given and its initial reading is noted.

    ·       Now we start adding KMnO₄ solution from the burette drop wise into titration conical flask already taken 20 ml oxalic acid and one full test tube of H₂SO₄ in it with continuous swirling until a permanent light pink just appear in the solution of titration conical flask.

    ·       Note down the reading of burette at which permanent light pink colour appeared in conical flask.

    ·       Thus, this process repeated 3-times to obtained three concordant reading.

Observation-

·  Molecular mass of oxalic acid=126g/mol

·  Molecular mass of KMnO₄ = 158 g/mol

·  Molarity of oxalic acid solution=M/50 = 1/50 mol/litre

·  Weight of oxalic acid dissolved in 500ml of distilled water = 1.26g.

Observation table-

s.no	Volume of oxalic acid solution taken by Pipette in ml (V1)	Volume of unknown KMnO4 solution taken by burette in (ml)			Concordant volume of KMnO4 used in ml (V2)
		Initial	Final	Volume of KMnO4 sol used
1.	20ml	0.0	12.7	12.7	12.62
2	20ml	0.0	12.6	12.6
3	20ml	0.0	12.6	12.6

                                            

Calculation-

Molarity of KmnO₄ solution from the balanced ionic Chemical equation given in theory. it clear that 1mol KMnO₄ reacts with 5 mole oxalic acid (according to molarity equation) 

Therefore,        

 Oxalic acid            Vs      KMnO₄            

n₂ M₁V₁     =    n₁M₂V₂

2 M₁V₁      =    5 M₂V₂                                                                              

Here

[ n₁ = no. of mole of oxalic acid= 5 mol                                                            

 M₁= no. of mole of oxalic acid = M/50

V_{1 =}volume of oxalic acid

n₂= no.  of mole of KMnO₄   = 2 mole                                                                                                                                                                            M₂= no. of mole of KMnO₄ =?

 V₂=volume of KMnO₄]

 So,   M₂  =  2M₁V₁/5V₂  

by putting the value of M_1, V₁ and V₂ in above formula

 M ₂   =   2 x 20 /50 x 5 x 12.62

  M ₂ =0.0126M                                                                         

Thus the molarity of KMnO₄ ( M ₂)     = 0.0126M 

then the strength of    KMnO₄   will be                                                                   

  = molarity of KMnO₄(M₂) x Molecular mass of KMnO₄

 = 0.0126 x 158 gm/litre

 = 2.003 gm/litre

Result – We found that molarity and strength of KMnO₄ as 0.0126 M   and 2.003gm/litre respectively.

Precautions –

    ·       We should Always add one test tube(10ml) dilute sulphuric acid before titration.

    ·        KMnO₄ should be added drop wise otherwise the solution will become brown due to MnO₂ form.

    ·   We should always take reading KMnO₄ solution in the burette from the lower meniscus.

    ·       Always heat the conical flask with its content to about 50-60⁰C.

·       Should always use distilled water for preparing solution and washing chemical apparatus.

    ·       Should always use distilled water for preparing solution and washing chemical apparatus.

To prepare M/20 solution of ferrous ammonium sulphate(Mohr’s salt).with its help, determine the molarity and strength of the given solution of potassium permanganate (KMnO4).

 

Aim- To prepare M/20 solution of ferrous ammonium sulphate (Mohr’s salt).with its help, determine the molarity and strength of the given solution of potassium permanganate (KMnO₄).

Apparatus required- Chemical balance, weight box, conical flask, burette, Burette stand, 100 ml Beaker, 20 ml pipette, funnel, wire gauge, wash bottle and burner.

Chemical required- potassium permanganate, dilute sulphuric acid, distilled water and ferrous ammonium sulphate.

Theory-

      ·   Ferrous ammonium sulphate is a double salt and act as reducing agent against KMnO₄

      ·   Whereas potassium permanganate act as an oxidising agent and it oxidises iron present in Mohr’s salt. in the acidic medium KMnO₄ behave as a very strong oxidizing agent. So acidity is introduced by adding dil. H₂SO₄.

     ·   Hence the reaction between the titrant (KMnO₄) and the analyte (Mohr’s salt) is a redox reaction, Therefore the titration is called redox reaction.

     ·   Potassium permanganate act as a self-indicator for this reaction.   

     ·  The colour at the end point changes from colourless to light pink. The endpoint of a titration is the point at which the reaction between the titrant and the analyte becomes complete.

 Chemical Reaction involved-

(a)Molecular equation-

Reduction half reaction –

2KMnO₄   + 3H₂ SO₄ ------>   K₂SO₄ + 2MnSO₄   +   3H₂O +   5[O] ---(1)

when dil. Sulphuric acid is added with KMnO₄, it produces [O]nascent oxygen.

·  Oxidation half reaction –[O] oxidises the Mohr’s salt

[   FeSO₄(NH₄)₂SO₄.6H₂O + 3H₂ SO₄   +  [O]---->  Fe₂ (SO₄)₃ + 2(NH₄)₂SO₄  +  13H₂O ] x5  ---(2)

by combining of eq. (1) and (2)

2KMnO₄   + 3H₂ SO₄  + 5x2FeSO₄(NH₄)₂SO₄.6H₂O +  5x3H₂ SO₄ + 5[O] ------------> 

K₂SO₄  + 2MnSO₄   +  3H₂O +  5[O] +  5Fe₂ (SO₄)₃+ 5x2(NH₄)₂SO₄  + 5x13H₂O

Then, Net reaction may be written as(Overall reaction) –

2KMnO₄   + 3H₂ SO₄  +  5x2FeSO₄(NH₄)₂SO₄.6H₂O +  5x3H₂ SO₄  -----> K₂SO₄   +   2MnSO₄   + 5Fe₂ (SO₄)₃ + 10(NH₄)₂SO₄    +  68H₂O  ……(3)

(b) Ionic equation-

MnO₄^-   + 8H⁺ +  5e^-    ------>   Mn²⁺ +  4H₂O  ……..(4)

[Fe²⁺ -------->   Fe²⁺  +  e^- ] x5  .....….(5)

by combining of equation (3) and (4), Net ionic equation ma be written as-

MnO₄^-   + 8H⁺    +   5Fe²⁺  +   5e^-    -------->  5Fe²⁺  +  Mn²⁺ +  4H₂O  +  5e^- ………..(6)

Procedure-

     ·       4.9g of Mohr’s salt crystal are weighed and dissolved in 250ml of water in a measuring flask to prepare M/20 solution of it.

     ·       with the help of pipette, 20 ml Mohr’s salt is taken out in titration flask and one test tube (10ml) full of dilute H₂SO₄ is added to it. Since it makes acidic medium to the reaction and prevents Mohr’ salt from hydrolysis.

     ·       Burette is washed with distilled water and is filled with KMnO₄ solution given and its initial reading is noted.

     ·       Now we start adding KMnO₄ solution from the burette drop wise into titration conical flask already taken 20 ml Mohr’s salt and one full test tube of H₂SO₄ in it with continuous swirling until a permanent light pink just appear in the solution of titration conical flask.

     ·     Note down the reading of burette at which permanent light pink colour appeared in conical flask.

·       Thus this process repeated 3-times to obtained three concordant reading.

Observation-

      ·       Molecular mass of Mohr’s salt =392g/mol

      ·       Molecular mass of KMnO₄ = 158g/mol

      ·       Molarity of the Mohr’s salt solution=M/20 = 1/20 mol/litre

      ·       Weight of Mohr’s salt dissolved in 250ml of distilled water = 4.9g.

Observation table-

s.no	Volume of Mohr’s salt solution taken by Pipette in ml (V1)	Volume of unknown KMnO4 solution taken by burette in (ml)			Concordant volume of KMnO4 used in ml (V2)
		Initial	Final	Volume of KMnO4 sol used
1.	20ml	0.0	9.5	9.5	9.43
2	20ml	0.0	9.4	9.4
3	20ml	0.0	9.4	9.4

                                            

Calculation-

Molarity of KmnO4 solution from the balanced ionic Chemical equation given in theory. it clear that 1mol  KMnO4 reacts with 5mol FeSO₄(NH₄)₂SO₄.6H₂O (according to molarity equation)

Therefore,        

 FeSO₄(NH₄)₂SO₄.6H₂O(Mohr’s salt)  Vs      KMnO₄            

  n₂ M₁V₁        =   n₁M₂V₂

  1 M₁V₁      =   5 M₂V₂               

 So,

 M₂  =  M₁V₁/5V₂                                 

Here-

[ n₁ = no. of mole of Mohr’s salt = 5mol     M₁= no. of mole of Mohr’s salt

V_{1 =}volume of Mohr’s salt

n₂ = no. of mole of KMnO₄   = 1mol                                                                                                    

M₂= no. of mole of KMnO₄ =?

V₂=volume of KMnO₄                                      

 So,      M₂  =  M₁V₁/5V₂  

by putting the value of M_1, V₁ and V₂ in above formula 

 M ₂     =  1 x 20 /20 x 5 x 9.43       

 M ₂ =  0.0212M                                                                          

Thus the molarity of KMnO₄ ( M ₂)   = 0.0212 M 

then, the strength of    KMnO₄   will be                                                                   

 = molarity of KMnO₄(M₂) x Molecular mass of KMnO₄

= 0.0212 x 158 gm/litre

= 3.349 gm/litre

Result – We found that molarity and strength of KMnO₄ as 0.0212M and 3.349gm/litre respectively.

Precautions –

     ·       We should Always add one test tube(10ml) dilute sulphuric acid before titration.

     ·        KMnO₄ should be added drop wise other wise the solution will become brown due to MnO₂ form.

     ·       We should always take reading KMnO₄ solution in the burette from the lower meniscus.

·       Should always use distilled water for preparing solution and washing chemical apparatus.                                               

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The chemistry of blood

The chemistry of blood 

Blood is a complex mixture. the chemicals in blood dictate its colour and some contribute its characteristics, slightly metallic odour. 

       here I am going to tell you about some of the difference of that determine a person’s blood type the colour of blood protein found in blood, built up smaller sub-unit containing ‘haems. these haems contain iron and their structure gives our blood its bright red colour when oxygenated, its non- oxygenated blood is a dark red colour not blue. blood lost due to bleeding gradually turn Brown as haemoglobin is oxidised to methaemoglobin. 

the smell of blood the compound that gives human blood its characteristic metallic odour is trans-4,5-epoxy-(E)-2-ethanol. 

Blood type  

It is determined by the presence of antigens are found on red blood cells surfaces that can bind to antibodies and stimulate immune response. Antibodies are protein in blood plasma that help fight infection. 

 The two most important blood group systems are as ABO (as A, B, AB and O) and RH (as Rh+ and Rh -)

There eight type of blood on the basis of presence of antigen and antibodies in it-

S.No.	A	B	AB	O
1	Antigen-A	Antigen-B	Antigen-A and B	none
2	Antibody-B	Antibody-A	none	Antibody-A and B