Titration

In order to utilize any matter just knowing a substance is not enough. One has to know about its properties and its characteristics. There are various methods to determine the properties of a
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.

What is Titration?

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.

Titration Equation

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.

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 = massmol mass

Normality(N) is defined as the ratio between the mass of the substance and its equivalent mass.

N = massEq mass

Equivalent mass is defined as the ratio between the molecular mass ans the valency of the substance

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.

Titration Formula

Since titrations are mostly meant to find the concentrations we can use a direct equation for titration formula. If C_k is the concentration of known solution the volume of which is found to be V_k. To find the concentration C_uk of V_uk volume of given unknown solution, the concentration.

C_uk = CkXVkVuk

This is titration formula.

End Point of Titration

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.

Titration Procedure

The titration procedure is very simple.

1. 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.
2. A solution of known concentration is made in a volumetric flask. This gives us the known volume and known normality or molarity.
3. This is filled in a burette and the readings are noted.
4. Care should be taken to avoid the presence of air gaps in the burette.
5. 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.
6. A few drops of suitable indicator is added to this solution in the flask.
7. 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.
8. At the end point the indicator changes its color. The same procedure is repeated several times until concurrent values are obtained.

Titration Examples

The common classroom examples of titrations are the determination of the strength of NaOH solution by titrating with Hydrochloric acid. It is not possible to make a standard solution of NaOH by molecular mass method since it is highly hygroscopic crystals.

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.

Titration Calculations

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) V_b mL
Normality of the base (To be foundout) N_b mL
Acid: Volume of the acid (Burette reading) V_a mL
Normality of the acid (Known) N_a mL

Normality of Base

N_b = VaXNaVb

Concentration of the base in given solution in percentage is

N_b X E.W. X 100/1000 = x%

Back Titration

1. 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.
2. The volatile component of the original compound whose concentration is to be established is allowed to react with the excess solution.
3. The remaining quantity of the known reagent is titrated which will give the value that is consumed by the volatile component.
4. 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).
5. Sodium hydroxide reacts with the salt and releases ammonia gas. The released gas is absorbed in known excess volume of sulfuric acid.
6. 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.

Redox Titration

Redox titration is also known as Oxidation reduction titration. In this type of titration the change in oxidation states give the amount of metal present in the given unknown quality substance.

Generally these titration use oxidizing agents like Potassium permanganate and Potassium di chromate in acid medium.

Titration Graph

• 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.

Titrations Problems

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:

• 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.

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:
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.

Analytical Chemistry

The development of chemistry is based on finding out the composition of various compounds. Not just the chemical aspect of it but the physical parameters and properties as well. The branch of chemistry which throws light on the various aspects involved in characterizing an element, compound or mixture is known as analytical chemistry. The importance of Analytical chemistry is the determination of the chemical composition of matter. This is done by the identification of a substance, the elucidation of its structure and the quantitative proportions of its composition. These are the aspects covered by analytical chemistry, which involve a wide variety of equipment and techniques.

Analytical chemistry is an interdisciplinary branch of science wherein a large number of inputs from different branches of sciences have contributed to its development. For instance, most of the chromatographic methods were invented by biochemists, or biological scientists, while contributions by physicists account for nuclear magnetic resonance and mass spectroscopy etc.

What is Analytical Chemistry?

Analytical chemistry can be defined as that branch of chemistry dedicated to the qualitative, quantitative, structural and other analysis of a substance by various experimental determinations. It defines the methods of isolation, identification and estimation of the quality and quantity.The importance of analytical chemistry is its many applications.

Methods of Quantitative Analysis

Analytical chemistry has a two step analysis. They are characterization and determination of the constituents of a compound. The identification step is called qualitative analysis, while the second step of quantitative analysis is more complicated. Quantitative analysis can be classified depending upon the method of analysis, or it can be categorized according to the scale of analysis.

Analytical Chemistry Impact Factor

Analytical chemistry has a lot of impact on the discovery, study and establishing of a substance. No other branch of science finds so many extensive applications as analytical chemistry purely for two reasons.
First analytical chemistry finds numerous applications in various disciplines of chemistry such as inorganic, organic, physical and biochemistry and secondly it finds wide applications in other fields of related sciences such as environmental science, agricultural science, biomedical and clinical chemistry, solid state research and electronics, oceanography, and space research.
The analysis of pesticides or insecticides from crops is done by gas chromatography or high performance liquid phase chromatography. Ascertaining the ratio of potassium to sodium in fertilizers is done by atomic absorption or flame emission methods. There are instances of the use of analytical chemistry in agricultural sciences like the mineral content in crops, estimation of amino acids and requirements of a crop under study.

The analysis of micro nutrients such as iron, copper, zinc, molybdenum, boron and manganese by spectrophotometer is another example.

Fundamentals of Analytical Chemistry

The fundamentals of analytical chemistry is sub divided into selection, experimentation, observation and establishment. Preliminary tests will give the nature of the substance under study and helps to select the process that needs to be undertaken. The next step in fundamentals of analytical chemistry is experimentation. The sample under study is subjected to a variety of tests. The results of all these required tests are recorded and carefully observed to consolidate an inference in the third step of analysis.
Finally after studying all the data proposing a method to identify that particular substance with some characteristic tests which will help in further research. The following are the topics listed under analytical chemistry.

• Titration
• Molecular spectroscopy
• Atomic spectroscopy
• Bio-analytical chemistry
• Material analysis
• Chemical analysis
• Gravimetric analysis
• Volumetric analysis
• Forensic toxicology
• Mass spectroscopy
• Spectrophotometry and calorimetry
• Chromatography and electrophoresis
• Crystallography
• Analytical electrochemistry

Trends in Analytical Chemistry

“Chemistry is at least as old as recorded history, but we recognized as experimental chemistry did not emerge until the end of the sixteenth century” are the words of R.Boyl in his book Sceptional Chemistry.
Gravimetry amounts to tracing the early history of chemistry by the study of quantitative estimations. There are three stages of the analysis.

1. The first phase is to encounter the new material while testing which is the early stages of analytical chemistry.
2. The second phase is to establish the structures.
3. The third phase, right now represents the broadening and deepening of the meaning of analysis.

As summarized from the literatures analytical chemistry as a science involves all techniques and methods for obtaining information regarding the composition, identity, purity and constitution of samples of matter in terms of the state, quantity, and grouping of atoms and molecules. It also involves the determination of those physical properties and behaviors that can be corrected with those objectives.

The future trends in analytical chemistry are to develop different techniques for high sensitivity studying at the atomic and molecular level. To meet the requirement for information, energy, environmental and life sciences and industry the future trends in analytical chemistry should focus on the present technological development and take the help of digitization, computerization, ultra microscopic study and spectroscopy etc.

Quantitative Analytical Chemistry

Basic analytical chemistry can be divided in to Qualitative and Quantitative analysis. In the first type, by the processes of analysis one can establish the presence of different units in the given substance. It will however not give the amounts of the units that compose the material.

In quantitative analytical chemistry, the stress is on establishing the amounts of those elements and compounds in the substance under test. There are two different varieties of quantitative analysis. Volumetric analysis or titrimetric analysis is one method to establish a compound. This technique involves making of a soluble and clear solution of the substance and then tit-rating it against the standard solutions, using an indicator.

The other method of quantitative analysis is the gravimetric or precipitate analysis. This method is considered much more accurate than the volumetric analysis. In this type of analysis the substance under study is subjected to a treatment where the anionic or cationic component forms an insoluble and stable compound.

For example if one has to estimate the BaCl₂ purity by gravimetric method, it can be done either by converting Ba²⁺ ions into insoluble BaSO₄ salt or converting chloride ion (Cl^-) into insoluble Silver chloride. This will give the purity of the compound.

Chemistry Problem Solver

Chemistry Problem Solver

One of the great joys of studying chemistry is learning to be a good problem solver. The ability to solve complicated problems is a skill which will help greatly throughout life.The basic steps in solving problems are.

1. Read: Read the problem carefully. Determine what is known and what is to be solved for and write them down. It is important to label all factors and measurements with the proper units.
2. Plan: Determine which principles are involved and which unit relationships are needed to solve the problem.
3. Setup: Set up the problem in a neat, organized and logical fashion, making sure all unwanted units cancel. Use the examples as guides for setting up the problem.
4. Calculate: Proceed with the necessary mathematical operations. make certain that the answer contains the proper number of significant figures.
5. Check: Check the answer to see if it is reasonable.

During problem solving should not allow any formal method of problem solving to limit the intuition. If a problem is clear and the solution seems to be simple by any other method by all means use it.

Chemistry Practice Problems

Calculate the specific heat of a solid in J/g^oC and cal/g^oC if 1638J raises the temperature of 125g of the solid from 25.0^oC to 52.6^oC.

Solution

1. Read:
Known 125g of the solid
Δt = 52.6 - 25.0 = 27.6^oC
Heat = 1638 J
Solving for: Specific heat of the solid

2. Plan:

Use the equation
(mass)(Specific heat)(t) = heat
solving for specific heat

3. Setup:

Specific heat = heatg×△t

4. Calculate:

Specific heat = 1639J125g×27.6oC = 0.475J/g^oC

Convert joules to calories using 1.00cal4.184J

specific heat = (0.475JgoC)(1.000cal4.184J) = 0.114cal/g^oC

5. Check:

Note that the units on the answer agree with the units for specific heat.

Mastering Chemistry Answers

Problem solving is an important part of chemistry. Problem solving is like learning to swim or play a musical instrument; it becomes easy only with practice.

Chemistry builds one concept on another and fact on fact. The secret of mastering chemistry answers is to know how to approach a new problem and how to put the pieces together.

The seven steps of scientific problem solving are as follows.

1. Identify the problem
2. Gather information relative to the problem.
3. Formulate tentative solutions (hypotheses)choose proffered solution.
4. Plan action to test suggested solution.
5. Experiment and observe the results.
6. Interpret the results (draw conclusion) understand what the results mean.
7. Evaluate the solution either concluding or revising the study to test the solution again if results are are unsatisfactory.

The use of complicated mix of inquiry, knowledge, intuition, logic, experience and common sense called critical thinking. The following are the steps should be taken care when a problem is solved.

• Read the problem carefully.
• Plan in detail just how the problem is to be solved.
• Specific definitely what each number represents and the units in which it is expressed when the mathematical operation is carried out.
• Having solved the problem examine the answer to see if it is reasonable and sensible.
• If the problem is not understood have it explained to you at the very earliest possible time.

Mastering Chemistry Help

Much of chemistry is concerned with the applications of concepts to practical problems.

Example

When 2.53 grams of metallic mercury is heated in air, which produces 2.73 grams of a red orange residue. Assume that the chemical change is the reaction of the metal with oxygen in air

Mercury + oxygen → red-orange residue

what is the mass of oxygen that reacts? When the red-orange residue is heated it decomposes to give back the mercury and release the oxygen when it is collected. What is the mass of collected oxygen?

Problem strategy

Apply the law of conservation of mass to the reaction.According to this law, the total mass remains constant during a chemical reaction; that is

mass of substance before reaction = mass of substance after reaction

Solution

From the law of conservation of mass

Mass of mercury + mass of oxygen = mass of red-orange residue

Substituting the values
2.53 grams + mass of oxygen = 2.73 grams
or
mass of oxygen = (2.73 - 2.53)grams
= 0.20grams

The mass of oxygen collected when the res-orange residue decomposes equals the mass of oxygen that originally reacted.