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# Volumetric Analysis

Volumetric analysis is a quantitative analytical method. Volumetric analysis is a method in which the amount of a substance is determined by measuring the volume that it occupies and the volume of a second substance that combines with the first in known proportions. This is also called titrimetric analysis or titration because volume is an important factor in titration. It’s a laboratory method of quantitative chemical analysis which is used to determine the unknown concentration of an analyte. Titration is a volumetric analysis because it’s based on the measurement of volumes.

One standard solution is used which is solution of a reagent, also called titrant or titrator. A known volume of titrant reacts with a solution of analyte or titrand to determine its concentration. Thus, the process involves measuring the volumes of two solutions which react to each other.The basic principles of volumetric analysis are given as below:
1. The one solution to be analyzed contains an unknown amount of chemicals.
2. The reagent of known concentration reacts with chemical of unknown amount in the presence of an indicator to show the end-point. This is the point which shows the completion of the reaction.
3. The volumes are measured by a titration which completes the reaction between reagent and solution.
4. The volume and concentration of reagent used in the titration give the amount of reagent in moles.
5. The amount of unknown chemical in the measured volume of solution is calculated by using the mole ratio of the equation.
6. The amount of unknown chemical in the original sample is calculated by the amount of unknown chemical in the measured volume.

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## Volumetric Analysis Acid-Base Titration

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The organic compounds containing the acidic group are analyzed by this titration. This is based on the neutralization reaction between acid and base. An acid-base indicator is used to detect the end point which indicates the completion of reaction between acid and base by changing the color of the mixed solution.

The selection of indicator is very important in acid-base titration because the equivalence point is determined by the stoichiometric of the reaction whereas the endpoint is measured by changing the color from indicator. The reaction is neutralized but it is not necessary to come to the end point at pH = 7. The completion of the reaction depends on the strength of acid and base.

For example: The acid-base titration between HCl (strong acid) and NaOH (strong base) in which phenolphthalein indicator is used.

Chemical reaction - HCl + NaOH NaCl + H2O

Similar titration between NH4OH (weak base) and HCl (strong acid) in which methyl orange is used as indicator.

NH4OH +HCl NH4Cl + H2O

### Steps for Acid base Titration

To determine the concentration of HCl, we take 20 ml of HCl and neutralize it with 0.150M of NaOH.

The steps for titration process are given below.
1. The titrant, NaOH solution of known concentration, is added from the burette.
2. 20ml HCl is taken in volumetric flask with the help of pipette and 2-3 drop of phenolphthalein indicator is added.
3. The solution is color less due to the acidic medium.
4. The solution of NaOH is added to the HCl drop by drop. As it comes in the contact with HCl acid, the color of solution becomes pink which quickly disappears. This is due to the formation of OH- ions from NaOH which reacts with indicator and changes the color. The solution becomes color less again due to the neutralization of OH- ions with H+ ions (from HCl).
5. After adding more NaOH, the equivalence point of titration is reached when the equal number of OH- ions and H+ ions react. At this point the pH of the solution in the flask is equal to 7 and the indicator is colorless.
6. After this, adding more NaOH drop by drop in the solution makes the solution pink from color less as the solution becomes basic now. So, indicator changes its color in basic medium. This permanent change shows the end point of titration. Note this reading which is the volume of base required to neutralize 20 ml of HCl acid. If 50ml is the volume of NaOH (end point reading from burette).
Now to calculate the concentration of HCl ;
HCl + NaOH NaCl + H2O
the molar ratio is 1:1 for this reaction

So mole of HCl = moles of NaOH
(molarity = moles/ volume)

[Molarity x Volume] of acid = [Molarity x Volume] of base
M x 20 = 0.150 x 50 ml
M = 0.150 x 50 /20 = 0.375M is the concentration of acid HCl.

## Indicators Used For Various Titration

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### Strong Acid Against a Strong Base

For example, titration of HCl and NaOH. At the initial stage of titration, the pH changes very slowly and increases up to pH= 4. But with addition of small amount of alkali like 0.01 mL approx., pH value reaches to about 7. This shows the neutralization of acid. Further addition of small amount of alkali increases the concentration of hydrogen ions and the pH value rises to about 9. There is a rapid increase of pH from about 4 to 9 when the end point approaches. Indicators like methyl orange, methyl red and phenolphthalein could be used for these kind of neutralization reactions which show the color change within the pH range of 4 to 10.

### Weak Acid against Strong Base

For example, titration of acetic acid against NaOH. The end point of this titration lies between pH 8 and 10 due to the formation of sodium acetate and water. So, phenolphthalein is a good indicator for this.

### Strong Acid against Weak Base

For example, the titration of ammonium hydroxide with HCl. The end point is in the range of 6 to 4 which shows the acidic pH of solution. Methyl orange is a good indicator for this titration.

 Indicator Color change on acidic side Range of color change Color change on basic side Methyl Violet Yellow 0.0 - 1.6 Violet Bromophenol Blue Yellow 3.0 - 4.6 Blue Methyl Orange Red 3.1 - 4.4 Yellow Methyl Red Red 4.4 - 6.3 Yellow Litmus Red 5.0 - 8.0 Blue Bromo thymol Blue Yellow 6.0 - 7.6 Blue Phenolphthalein Colorless 8.3 - 10.0 Pink Alizarin Yellow Yellow 10.1 - 12.0 Red

## Methods to Determine the Endpoint

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1. pH indicator - This is a substance which shows the chemical change by changing the color.
2. A potentiometer - This is used to measure the electrode potential of the solution. These are used for redox titration. They show the end point with changing potential of the working electrode.
3. pH meter - It is a ion-selective electrode. In pH meter the potential of electrode depends on the amount of H+ ion present in the solution. The pH of the solution can be measured in the whole titration. This gives more correct result than indicators.
4. Conductance - The conductivity of solutions is also changed in titration and it depends on the ions present in the solution, mobility of ions and ions concentration.
5. Color change - In the redox reactions, the color of solution changes without use of indicator. This is due to different oxidation states of the product.
The volumetric analysis can be categorized into four types which are based on the type of reaction involved in the process.

 Types Chemicals to be analyzed Reagents to be used Indicators to be used Acid / Base acid or base alkali or acid pH indicator Precipitation ion that form insoluble salt compound containingthe other ion needed toform the insoluble salt conductivity Redox oxidizing or reducing agent reducing oroxidizing agent natural color change orredox indicator Complexometric metal ion that form complexes complexing agent metal ion indicator

## Precautions of Volumetric analysis

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Some important points should be remembered in doing volumetric analysis to get the accurate results.
1. All the equipments like burette, beaker, pipette and volumetric flask should be washed properly with distilled water before taking them in use as the presence of any other chemical can be the reason for wrong measurement.
2. The process of filling the pipette should be accurate to avoid excess addition of solutions.
3. The flask should be shaking well after adding the indicator and also the titration flask with addition of each drop of solution from burette.
4. The addition of acid should be drop wise.
5. Contamination should be avoided.
6. The indicator should not be used in excess.
7. The flask should be removed as the indicator changes color.
8. Sometimes, an air bubble in the nozzle of the burette can be the reason for altering the readings, so, it must be removed before taking the initial reading.
9. The burette should not be leaked during titration.
10. Keep eyes in the level of the liquid surface during the time of taking the burette reading or measuring flask and pipette etc.
11. Lower meniscus and upper meniscus are always read in case of color less and colored solutions respectively.
12. Do not blow through the pipette to expel the last drop of solution from it; simply touch the inner surface of the titration flask with the nozzle of the pipette for this purpose.
13. Index finger should be used for pi-petting the solution.

## Volumetric Analysis Problems

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Below you could see problem

### Solved Examples

Question 1: Calculate the concentration of HCl acid if 50ml of HCl is required to neutralize 25ml of 1.00M NaOH in acid base titration.
Solution:

HCl + NaOH -> NaCl + H2O in this chemical reaction the molar ratio is 1:1 between
HCl and NaOH. So mole of HCl = moles of NaOH
[molarity = moles/volume]
MHCl x volume of HCl = MNaOH x volume of NaOH
MHCl = $\frac{M_{NAOH} \times volume\ of\ NaOH} {volume\ of\ HCl}$

MHCl = $\frac{25.00 ml \times 1.00 M }{50.00 ml}$

MHCl = 0.50 M HCl
So the concentration of HCl is 0.50 M.

Question 2: The titration curve of ammonia and HCl are given below. Find out all the unknown species A, B, C and D. Solution:
At point A, the pH is highly basic so its NH3. At point B, NH3 and NH4 +  ammonium ions are in the buffering region.
C = NH4+. This is equivalence point at which all the NH3 has been protonated and water molecules starts to take acidic protons.
D =H3O+ more acidic solution.

A = NH3
B = NH3 and NH4+ buffering solution
C = NH4+
D = H3O+

Question 3: Find out the molecular weight of the unknown mono protic acid whose 5 gm is neutralized with 26.23 ml of a 1.008 M NaOH solution in 150.2 ml of solution.
Solution:

Moles of NaOH = Molarity of NaOH x Volume of NaOH
So moles of NaOH = 1.008 x .02623 = 0.026439moles
As the acid is mono protic so the molar ratio of acid and base is 1:1.
So mole of mono protic acid = moles of NaOH
Moles of acid = 0.026439 moles.
Mole=$\frac{mass}{molecular\ mass}$

So molecular mass = $\frac{mass}{mole}$

Molecular mass of unknown mono protic acid = $\frac{5}{0.026439}$ = 189.1 gm/mole.

Question 4: If the 20 ml of an unknown acid is required 30ml of 0.125 N NaOH to get the equivalence point. Find out the normality of the unknown acid.
Solution:

equivalents of acid = equivalents of base

Vacid Nacid = Vbase Nbase

So Nacid = $\frac{(30.0 ml)(0.125 N)}{(20.0 ml)}$= 0.1875 N

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