Volumetric analysis and titration


Volumetric analysis is the process of determination of concentration of a solution with the help of another solution of known concentration.

Volumetric analysis is experimental method of determination of volume of a solution of known concentration needed for a definite volume of another solution of unknown concentration.

Titration

Titration is the process of adding one solution to another solution until an indicator shows that the reaction is complete.

Titrant is the solution with a known concentration that will react with the analyte.

Analyte (titrand) is the solution with unknown concentration.

The end-point is the point in the titration at which the indicator changes colour.

Indicator is a chemical substance which changes colour at the end point.

The equivalence point (stoichiometric point) is the point at which chemically equivalent amounts/quantities of acid and base have reacted.

The functions of different apparatus during titration

Burette
It is used to deliver variable volumes of a solution accurately. It is usually calibrated from 0 to 50cm3.

It should be washed with water and rinsed with the solution it is going to hold.

When reading a burette one should read exactly at the bottom of meniscus.


Filter funnel
It is used to facilitate pouring solutions into the burette.
Pipette

It is used to measure the fixed or specific volume of a solution accurately.

Like burette, it has to be washed with water and rinsed with the solution it is going to hold.

Pipette filler – used to fill a pipette. In case it is not present, the pipette may be filled by mouth suction, where care must be taken to not drink/swallow the solution.


Conical flask used to hold and mix solutions thoroughly during titration.
Beaker used for holding solutions.

Volumetric flask used to prepare a standard solution accurately. Like pipette it has a single calibration at its neck.
It is filled until the bottom of the meniscus rests on the line of the neck of the flask.

Preparation of solutions

Standard solution is the solution whose concentration is accurately known.

How to prepare standard solution

(A). Solid bases/acids (Solutes)

  1. Determine the volume and concentration that you want to prepare.
  2. Calculate the mass of solute needed to give the required volume and concentration.
  3. Weigh the solute on the electronic balance.
  4. Dissolve the solute completely in distilled water and then transfer it to a volumetric flask partially filled with distilled water.
  5. Add the distilled water to the calibration mark of the volumetric flask.
  6. Invert the flask and shake it to make sure thoroughly mixing.

To determine the mass of the substance from the information given in the container

Mass = molarity x relative molecular mass x volume
                 % purity

The above volume should be in dm3 or litres.

(B)For liquid acids or bases (concentrated solution)

  1. Calculate the volume of a concentrated solution required to be diluted in order to give the desired concentration.
    Use the dilution law to get this volume
     McVc = MdVd
    	  where 
    	  Mc = Molarity of concentrated solution
    	  Vc = Volume of the concentrated solution
    	  Md = Molarity of dilute solution
    	  Vd = Volume of the dilute solution.	  
    Vc = the volume to be taken from the bottle in order to get the desired molarity.
  2. Put a small amount of distilled water in a beaker.
  3. Add the volume of the solution obtained (concentrated solution) in a beaker and stir it thoroughly to get a uniform solution.
  4. Pour the solution into distilled water which is in volumetric flask.
  5. Carefully add distilled water to the solution up to the calibration mark of the volumetric flask.
  6. Invert the volumetric flask several times to obtain a uniform solution.

To calculate the molarity from the information given in the acid/base bottle

Molarity = % purity x density(ρ) x 1000cm3
                    molecular weight

Density sometimes is known specific gravity

Remember: ACID SHOULD BE ADDED TO WATER AND NOT WATER TO ACID

When concentrated acid is mixed with water, it produces a lot of heat. The heat produced is so much that if the water is poured directly to the acid, it can produce acid sprays which are harmful.

Example

1. A commercial sulfuric acid is usually labeled as 98% of the acid by weight and having the density of 1.84 g/cm³.
(i) Calculate the molarity of this industrial acid.
(ii) What volume of this acid that will be required to prepare 2 dm³ of 0.5 M sulfuric acid?

Procedures for carrying out titration

  1. Clean the pipette and burette by washing them with water and rinsing them with the solutions they are going to hold.
  2. Support a cleaned and rinsed burette with a retort stand. Close burette with its tap or stopcock.
  3. With the help of a funnel, fill the burette with acid to just above the zero mark. Open the stopcock /tap to remove any air bubbles in the tip and adjust the acid to exactly zero mark.
  4. Pipette 25cm3 or 20cm3 of base and transfer the solution into a conical flask. Add two or three drops of an appropriate indicator into the flask.
  5. Note the initial reading of the burette and run out the solution from the burette into the flask drop by drop while shaking the flask until the colour of the indicator changes. Stop addition of the solution when the end-point is reached.
  6. Titration process
  7. Record the final burette reading. This is pilot titration (Rough titration).
  8. Calculate the differences between final burette reading and the initial burette reading. This is called TITRE volume; it is the volume of acid used for completion of the reaction.

  9. The results can be summarized in a tabular form as follows;

    Titration

    Burette readings Pilot 1 2 3
    Final reading (cm3)
    Initial reading (cm3)
    Volume used (cm3)
  10. Repeat the steps (iii) to (vii) three more times to get more accurate results.
  11. Calculate the concentration of the solutions by using acid-base mole ratio
    MaVa = na 
    MbVb   nb 

The choice of indicators

The choice of indicator depends on the strength of the solutions used during titration.
Acid-base Example of acid & base Indicator
1. strong acid versus strong base H2SO4 Vs. NaOH Any indicator
2.strong acid versus weak base HCl Vs. NH4OH Methyl orange
3.weak acid versus strong CHOOH Vs.KOH Phenolphthalein
4.weak acid versus weak base C2H5COOH Vs. NH4OH No suitable indicator

Changes of colors of indicators

Indicator Color of indicator in
Acid alkali Neutral
Methyl orange Pink Yellow Orange
Phenolphthalein Colourless Pink Colourless
Litmus Red BlueBlue or purple

Determination of Concentrations

  1. A student obtained the following results during titration of 25cm3 of 0.2M sodium hydroxide solution with sulphuric acid solution of unknown concentration. The following table of results summarize the data obtained:
    Experiment Pilot 1 2 3
    Final reading (cm3)25.4025.2024.9025.00
    Initial reading (cm3)0.00 0.00 0.000.00
    Volume used (cm3)
    1. Calculate the average volume of the acid that neutralized the base.
    2. Write the chemical equation for the acid-base reaction.
    3. Calculate the molarity of the acid.

    Soln

    (a) Average volume of acid used (cm3), Va:   
            Va   = 25.20cm3 + 24.90cm3 + 25.00cm3
                                3 
                 = 25.03cm3
                 
    Therefore, 25.03cm3 of the sulphuric acid neutralized 25cm3 0f 0.2M sodium hydroxide.
    
    (b) H2SO4 + 2NaOH(aq) → Na2SO4 + H2O
        na=1  nb = 2
        
    (c) Molarity of acid (Ma) = ?
        Volume of acid (Va) = 25.03cm3
        Number of moles of acid (na) = 1
        Molarity of base (Mb) = 0.2M
        Volume of base (Vb) = 25cm3
        Number of moles of base (nb) = 2
        
        Frome acid-base mole ratio:
              MaVa = na
              MbVb   nb
              
             Ma = MbVbna
                    Vanb
             Ma = 0.2mol/dm3 x 25cm3 1
                     25.03cm3 x 2
                     
                = 0.1mol/dm3
    Therefore, the molarity of the acid is 0.1mol/dm3
            
  2. 20cm3 of anhydrous sodium carbonate solution reacted completely with 25cm3 of 2M hydrochloric acid solution. Calculate the concentration of sodium carbonate solution in
    1. mol/dm3
    2. g/dm3

    Soln

    (a) Balanced chemical equation:
        2HCl(aq) + Na2CO3(aq) → 2NaCl(aq) + H2O(l) + CO2(g)
        na = 2 nb = 1
        
        Molarity of acid (Ma) = 2M
        Volume of acid (Va) = 25cm3
        Number of moles of acid (na) = 2
        Molarity of base (Mb) = ?
        Number of moles of base (nb) = 1
        Volume of base (Vb) = 20cm3
        
        From acid-base mole ratio:
             MaVa = na
             MbVb   nb
             
        Mb = MaVanb
              Vbna
           = 2mol/dm3 x 25cm3 x 1
                   20cm3 x 2
                   
           = 1.25mol/dm3
               
    (b) The molar mass of Na2CO3  = (23x2) + 12 + (16x3)
                                 = 106g/mol 
                                 
        Molarity = Concentration in g/dm3
                      Molar mass
               
        Therefore:
        Concentration in g/dm3 = Molarity x molar mass
                               = 1.25mol/dm3 x 106g/mol
                               = 132.5g/dm3
               
               
           

Determination of relative atomic mass of unknown element in acid/base

The relative atomic mass of an unknown element can be determined by titrating the acid or base containing the element with a standard solution.

Example:

25cm3 of XOH solution containing 32g/dm3 of the base was found to neutralize 10cm3 0f 1M sulphuric acid solution according to the equation:
H2SO4(aq) + 2XOH(aq) → X2SO4(aq) + 2H2O(l)
Calculate:
  1. Concentration of the base in mol/dm3
  2. Relative atomic mass of X
  3. Identify the element X

Soln

(a)Data given:
Molarity of acid (Ma) = 1M
Number of moles of acid (na) = 1
Volume of acid (Va) = 10cm3
Molarity of base (Mb) = ?
Volume of base (Vb) = 25cm3
Number of moles of base (nb)= 2

From acid-base mole ratio:

 From acid-base mole ratio:
         MaVa = na
         MbVb   nb
         
    Mb = MaVanb
          Vbna
       = 1mol/dm3 x 10cm3 x 2
            25cm3 x 1 
        = 0.8mol/dm3 
Therefore, the molarity of base is 0.8mol/dm3

(b) Molar mass = Concentration in g/dm3
                    Molarity

               = 32g/dm3
                0.8 mol/dm3
               = 40g/mol
               
    This means that 
    (X + 16 + 1) g/mol = 40g/mol
     X = 40 - 17
       = 23
(c) Since the relative atomic mass of X is 23, therefore the element X is sodium (Na).

Determination of water of crystallization

Water of crystallization is the definite amount of water which some substances chemically combine with when they form crystals from their saturated solution.

Water of crystallization plays a vital role in the formation and properties of many inorganic compounds.

When certain compounds crystallize (form crystals from their saturated solution), water molecules become incorporated within their crystal lattice, forming stable bonds with the compound's ions or molecules. This water is referred to as water of crystallization.

The presence of water of crystallization affects the physical properties of compounds. For example, it can influence the color of crystals, as seen in compounds like copper sulphate, which appears blue due to the water molecules in its crystal structure.

The water of crystallization also affects the melting point and solubility of compounds, as the water molecules need to be released or incorporated in order to change the state or dissolve the compound.

The removal of water of crystallization can be achieved through processes such as heating or desiccation. By applying heat, the water molecules are evaporated, leaving behind an anhydrous compound. This process is often reversible, and upon exposure to moisture, the compound can regain its water of crystallization.

There are various experimental techniques which are used to determine the water of crystallization in a given compound, including titration method, where a solution containing water of crystallization is titrated with a standard solution such as a base or an acid.

Hydrate is the compound that contains water of crystallization. They are also known as hydrated compound. Example of acid that is hydrated is oxalic acid (COOH)2.2H2O and base is Na2CO3.10H2O.

The compound with no water of crystallization is usually called anhydrous compound.

The amount of water in a hydrated acid or base can be determined through titration.

Concentration of hydrated compound(g/dm3)  =  Molar mass of hydrated compound 
Concentration of anhydrous compound (g/dm3)   Molar mass of anhydrous compound

Example 01
Sodium carbonate solution (Na2CO3.xH2O) was made by dissolving 179g of its crystals to make a litre of solution. 20cm3 of the solution was found to neutralize 25cm3 of a 1mol/dm3 solution of nitric acid. Calculate
  1. The concentration of base in
    1. mol/dm3
    2. g/dm3
  2. Value of x
  3. Write the formula of the compound
Solution
  1. Chemical equation:
    Na2CO3 + 2HNO3 → NaNO3 + CO2 + H2O na = 2 nb = 1 given: Ma = 1 mol/dm3 na = 2 nb = 1 Va = 25cm3 Vb = 20 cm3 Mb = ? From MaVa = na MbVb nb Mb = MaVanb Vbna = 1mol/dm3 x 25cm3 x 1 20cm3 x 2 = 0.625mol/dm3
The concentration in mol/dm3 is 0.625dm3

b. Concentration in g/dm3
Molarity = Concentration in g/dm3
            Molar mass

Concentration in g/dm3 = Molarity x Molar mass 
                      = 0.625mol/dm3 x 106g/mol
                      = 66.25g/dm3
The concentration in g/dm3 is 66.25g/dm3

The value of x can be calculated using the formula:

Concentration of hydrated compound(g/dm3)  =  Molar mass of hydrated compound 
Concentration of anhydrous compound (g/dm3)   Molar mass of anhydrous compound

179g/dm3 =  (106 + 18x)g/mol
66g/dm3         106g/mol

Cancelling the units and dealing with numbers:
        106 + 18x = 179 x 106
                     66
					 
	  18x = 287.5 - 106
	  
	  Dividing both sides by 18
	  18x = 181.5
          18      18

 x = 10
 The value of x in the given compound is 10.
c. The formula for the compound is Na2CO3.10H2O

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