chemical substance.
One of the most common and most reliable methods is the method of titration. Titration can be broadly described as finding out the strength of an unknown solution with a standard solution whose strength is known.
Titration is a method of finding out the
strength of a solution either in the terms of molarity or normality or
molality or acidity or alkalinity or precipitatability.Titration
can also be used to find out the presence of elements either in their
elemental state or in their compound state. Some substances get adsorbed
on the surface of other substances and this can also be estimated by
the process of titration.
Formation of complexes is the characteristic of certain metals. Different varieties co complexes are formed by the group of same elements and these will have different properties. This is possible because of the change in the oxidation number of the metal which we call as variable valency.
By the method of titration the changes in the oxidation states can be established and in turn the strength of the solution. Estimation of amino acids are also done by the method of titration.
Titrations are done in liquid medium generally
between clear solutions. The solutions are measured in volumes and the
amount of the substance dissolved in the solvent gives the concentration
of the solution. The concentration of solution is in terms of molarity
or normality. Formation of complexes is the characteristic of certain metals. Different varieties co complexes are formed by the group of same elements and these will have different properties. This is possible because of the change in the oxidation number of the metal which we call as variable valency.
By the method of titration the changes in the oxidation states can be established and in turn the strength of the solution. Estimation of amino acids are also done by the method of titration.
Molarity (M) is defined as the number of moles present in one litre of solution. Mole is a number which is also called Avogadro number of molecules.
Number of moles of a substance is the ratio of the mass of the substance and the molecular mass.
M =
N =
Eq.mass(E) = Mol massValency
In titrations the concentration of one of the solutions is known which is called standard solution or titrant. The solution of which the concentration is to be found is called aliquot or titrate.
According to the Stoichiometric equation, a reaction takes place in equivalent proportions. Thus the product of Volume(V) and molarity(M) or the normality(N) of one of the solutions is equal to that of the other solution at the equivalence point or end point of the titration.
Indicator is used to get a sharp end point in a titration.
Thus the Titration equation is V1M1 = V2M2 Where V1is the volume and M1 is the molarity of standard solution. V2 is the volume of the unknown concentration solution of which the molarity is to be found.
Therefore
M2 = V1M1V2
Concentration of the unknown solution is M X mol.mass. This is the titration equation.Since titrations are mostly meant to find the concentrations we can use a direct equation for titration formula. If Ck is the concentration of known solution the volume of which is found to be Vk. To find the concentration Cuk of Vuk volume of given unknown solution, the concentration.
Cuk = CkXVkVuk
This is titration formula.
End point of titration is also known as equivalence point. At this point both the titrant and titrate are in stoichiometric proportions. If it is an acid base titration the endpoint of titration is thee point at which the acid and base are stoichiometrically balanced in reaction and the pH of the mixture is neutral.
The endpoint of titration is indicated sharply by a suitable indicator either added directly to the solution or used externally.
The titration procedure is very simple.
- The apparatus used are a conical flask of 25mL or 50 mL or 100 mL capacity depending on the type of titration, a pipette of suitable capacity marking, a volumetric flask suitable to make the standard solution, sensitive balance and a burner if heating is required.
- A solution of known concentration is made in a volumetric flask. This gives us the known volume and known normality or molarity.
- This is filled in a burette and the readings are noted.
- Care should be taken to avoid the presence of air gaps in the burette.
- With the help of a pipette a fixed volume of the unknown concentration solution is taken in a conical flask and this is called aliquot.
- A few drops of suitable indicator is added to this solution in the flask.
- Keeping the flask under the nozzle of the burette which is fixed vertically on a stand slowly the standard solution is run in to the flask with consistent stirring.
- At the end point the indicator changes its color. The same procedure is repeated several times until concurrent values are obtained.
Similarly hydrochloric acid is a liquid and difficult to get the exact strength. While to conduct the experiment first a standard sodium carbonate solution is made by weighing pure anhydrous sodium carbonate and dissolving it in known quantity of water. This standard solution is used to establish the strength of HCl, which in turn is used to determine the strength of NaOH.
For a typical acid base titration in which the strength of the base is to be found out, the calculations can be as follows.
Base: Volume of the base taken (Pipette reading) Vb mL
Normality of the base (To be foundout) Nb mL
Acid: Volume of the acid (Burette reading) Va mL
Normality of the acid (Known) Na mL
Normality of Base
Nb = VaXNaVb
Concentration of the base in given solution in percentage is
Nb X E.W. X 100/1000 = x%
- Back titration is a slightly complicated titration procedure than the simple or direct titration. In this one reagent with known concentration is added in excess.
- The volatile component of the original compound whose concentration is to be established is allowed to react with the excess solution.
- The remaining quantity of the known reagent is titrated which will give the value that is consumed by the volatile component.
- For example, to know the strength of Ammonium sulphate, the compound is mixed with excess sodium hydroxide in a specially designed glass flask (Kjehldahl flask).
- Sodium hydroxide reacts with the salt and releases ammonia gas. The released gas is absorbed in known excess volume of sulfuric acid.
- All the gas is driven out by heating the flask. The excess sulfuric acid is back titrated which will give the acid consumed by the ammonia gas which in turn gives the content of ammonia in the salt.
Generally these titration use oxidizing agents like Potassium permanganate and Potassium di chromate in acid medium.
- Titration is a technique of finding the concentration of a particular variety in a solution whose concentration is unknown.
- The process of titration involves two components titrant and titrate.
- Titrant is generally the liquid of unknown concentration and titrant is the solution the concentration of which is known and by adding which in small quantities the process of titration is performed. The point at which the change one is looking for arrives is called end point.
- There are varieties of titrations such as acid base titration, oxidation reduction (redox) titrations, precipitation titrations, adsorption titrations and complex titrations etc.,
- A titration graph or titration curve is obtained in a graph one the graph is plotted with the readings of a titration.
- Since it is a titration of volumes and one of the volumes is generally fixed, the other quality of the fixed volume is taken one axis.
- The other axis is for the volume that is added to achieve the end point. A line joining the volumes to the changing quality on the other axis is called the titration curve.
- The advantage of the titration curve is that it gives an accurate measure and value of the titration.
- There are different varieties of titration graphs. Acid-base graph, pH graph, pKa titration graph etc.
Solved Examples
Question 1: 25 mL of sulfuric
acid is found to be a deci-normal solution. When it is titrated against
NaOH of unknown strength, it took 28.8 mL to reach the neutralization
point. Calculate the normality of NaOH and How many grams of NaOH is
required to dissolve in 2 L to get this strength solution?
Solution:
Question 2: What will be
medium of the resultant solution when 200 mL of deci-normal solution of
sulfuric acid is mixed with 2.12 g of anhydrous sodium carbonate. Will
it be neutral, acidic or basic?Solution:
- Volume of acid = Va = 25 mL
- Normality of acid = Na = 1/10 = 0.1N
- Volume of alkali = Vb = 28.8 mL
- Normality of base = Nb = 25 X 0.1 / 28.8 = 0.0868 N
- Grams of NaOH per 1L = Normality X Mol.
- mass/ valency = 25X0.1X40/ 28.8X1 = 3.47222 g
- Grams of NaOH in 2 L = 3.472 X 2 = 6.974 g. Ans.
Solution:
Molarity M = N/2 . 0.1 N = 0.05 M. That is 0.05 moles in 1L. In 200 mL it will be 0.05 X200/1000 = 0.01 moles.
106 g of Na2CO3 is 1 mole. Then 2.12 g will be 2.12/106 = 0.02 moles.
The solution is basic in nature since the molar ratio is is 1:2; 2 being the base.
No comments:
Post a Comment