TOPIC 5: VOLUMETRIC ANALYSIS
Standard Volumetric Apparatus
The Concept of Volumetric Analysis
Explain the concept of volumetric analysis
Volumetric
analysis is a quantitative analysis involving the measurement of
different solutions. These solutions are made to react completely and
the completion of the reaction is indicated by certain substances called
indicators. The quantitative composition of the solution is then determined.
analysis is a quantitative analysis involving the measurement of
different solutions. These solutions are made to react completely and
the completion of the reaction is indicated by certain substances called
indicators. The quantitative composition of the solution is then determined.
Important steps of volumetric analysis include:
- Weighing;
- Preparation of the solution;
- Titration; and
- Calculation
In
volumetric analysis, we deal with volumes of solutions. That is why
this quantitative determination of solutions of substances is called volumetric analysis.
volumetric analysis, we deal with volumes of solutions. That is why
this quantitative determination of solutions of substances is called volumetric analysis.
The
amount of a substance present in a solution is given in terms of its
volume and its concentration. The volume of a solution is usually given
in litres (dm3). The concentration of a solution is given in moles per litre (mol/dm3) or grams per litre (g/dm3).
amount of a substance present in a solution is given in terms of its
volume and its concentration. The volume of a solution is usually given
in litres (dm3). The concentration of a solution is given in moles per litre (mol/dm3) or grams per litre (g/dm3).
Volumetric
analysis is a means of finding the concentration of an unknown
solution. For example, the concentration of an unknown solution of an
acid can be found if it is reacted with a standard solution of an
alkali. A standard solution is one whose concentration is well known and does not change with time.
analysis is a means of finding the concentration of an unknown
solution. For example, the concentration of an unknown solution of an
acid can be found if it is reacted with a standard solution of an
alkali. A standard solution is one whose concentration is well known and does not change with time.
In
volumetric analysis, the reaction is carried out in a carefully
controlled way. The volumes are measured accurately using a pipette and
burette. The method is to add a solution of one reactant to the solution
of another reactant until the reaction is complete. When the reaction
is complete, we say the end-point has been
reached. If the reactants are acids and bases, completion (end-point) is
determined by the change in colour of an acid-base indicator. The
method is called titration. In other
reactions, completion is determined by a colour change of reactant(s).
The concentration of one of the reactant solutions must be known in
order to be able to find the concentration of unknown solution.
volumetric analysis, the reaction is carried out in a carefully
controlled way. The volumes are measured accurately using a pipette and
burette. The method is to add a solution of one reactant to the solution
of another reactant until the reaction is complete. When the reaction
is complete, we say the end-point has been
reached. If the reactants are acids and bases, completion (end-point) is
determined by the change in colour of an acid-base indicator. The
method is called titration. In other
reactions, completion is determined by a colour change of reactant(s).
The concentration of one of the reactant solutions must be known in
order to be able to find the concentration of unknown solution.
Significance of Volumetric Analysis
- Volumetric
analysis is used to quantify the amount of substances present in
solutions by analytical procedure, which involves precise measurements
of volumes of solutions and masses of solids. - Volumetric
analysis helps in the determination of the accurate volumes and
concentrations of the reacting substances, often solutions. - Volumetric analysis (titration) helps in the preparations of standard solutions.
- Volumetric analysis knowledge helps in the standardization of acids and bases.
Volumetric Apparatus
Use volumetric apparatus
We
have seen that volumetric analysis involves determinations of
quantities of substances, usually acids and alkalis, present in volumes
of solutions. This is usually done by using measuring apparatus.
have seen that volumetric analysis involves determinations of
quantities of substances, usually acids and alkalis, present in volumes
of solutions. This is usually done by using measuring apparatus.
Apparatus
used in volumetric analysis is based on volume measurements and since
the analysis demands high accuracy, the apparatus has to be calibrated
with the highest possible accuracy. It is for this reason that all
apparatus for volumetric analysis are specifically for this and not
other purposes.
used in volumetric analysis is based on volume measurements and since
the analysis demands high accuracy, the apparatus has to be calibrated
with the highest possible accuracy. It is for this reason that all
apparatus for volumetric analysis are specifically for this and not
other purposes.
Apparatus
used for volumetric analysis include, burette, pipette, burette stand,
white tile, conical flask, filter funnel, reagent bottle, watch glass,
beaker, measuring cylinder and measuring flask (or volumetric flask).
For approximate measurements, measuring cylinders may be used. For
accurate measurements of volumes, volumetric flasks are used.
used for volumetric analysis include, burette, pipette, burette stand,
white tile, conical flask, filter funnel, reagent bottle, watch glass,
beaker, measuring cylinder and measuring flask (or volumetric flask).
For approximate measurements, measuring cylinders may be used. For
accurate measurements of volumes, volumetric flasks are used.
Burette
This
is a long glass tube with a narrow lower part, which is fitted with a
tap that controls the amount of solution let out of the burette. This
instrument is calibrated from 0 to 50 cm3.Before measuring
the solution, rinse the burette with distilled water, then with the
solution it is going to hold. It has to be filled to the tip and all gas
bubbles removed. Thus, the burette is an apparatus used for
transferring the solution to the titration vessel (normally a flask).
is a long glass tube with a narrow lower part, which is fitted with a
tap that controls the amount of solution let out of the burette. This
instrument is calibrated from 0 to 50 cm3.Before measuring
the solution, rinse the burette with distilled water, then with the
solution it is going to hold. It has to be filled to the tip and all gas
bubbles removed. Thus, the burette is an apparatus used for
transferring the solution to the titration vessel (normally a flask).
Pipette
This
apparatus has a wider middle part with narrow parts at either ends. The
upper narrow part has a mark which marks the volume of all the space
below it. If, say, the pipette is one that is marked 25 cm3, we can say that a solution, when filled in the pipette up to this mark, will have a volume of 25 cm3.
apparatus has a wider middle part with narrow parts at either ends. The
upper narrow part has a mark which marks the volume of all the space
below it. If, say, the pipette is one that is marked 25 cm3, we can say that a solution, when filled in the pipette up to this mark, will have a volume of 25 cm3.
The
pipette is used in transferring a standard solution to the titration
flask. There are many types of pipettes depending on their volume
capacity. The common ones are the 25-cm3 and 20-cm3 capacity pipettes. Less common ones are the 10-cm3
capacity.Before measuring the solution, rinse the pipette several times
with distilled water and then with the solution to be measured; suck
the rinsing solution above the graduated mark, then discard the rinsing.
pipette is used in transferring a standard solution to the titration
flask. There are many types of pipettes depending on their volume
capacity. The common ones are the 25-cm3 and 20-cm3 capacity pipettes. Less common ones are the 10-cm3
capacity.Before measuring the solution, rinse the pipette several times
with distilled water and then with the solution to be measured; suck
the rinsing solution above the graduated mark, then discard the rinsing.
The
pipette is commonly filled by mouth suction but the use of pipette
fillers is highly recommended. When using a pipette, never blow out the
last drop.
pipette is commonly filled by mouth suction but the use of pipette
fillers is highly recommended. When using a pipette, never blow out the
last drop.
(a) A pipette (b) A pipette and pipette filler (used to fill and empty pipettes)
Measuring (Volumetric) flask
The
flask is made of glass and has a mark at the upper part of the narrow
tube. The space in the flask up to this mark represents a certain
volume. If a solution is filled up to this mark, the volume of the
solution is equal to the volume indicated by inscriptions on the flask
e.g. 50 cm3, 100 cm3, 150 cm3, 250 cm3, 500 cm3, etc.
flask is made of glass and has a mark at the upper part of the narrow
tube. The space in the flask up to this mark represents a certain
volume. If a solution is filled up to this mark, the volume of the
solution is equal to the volume indicated by inscriptions on the flask
e.g. 50 cm3, 100 cm3, 150 cm3, 250 cm3, 500 cm3, etc.
Filter funnel
A
filter funnel is required for effective transfer of the weighed solid,
liquid or solution into the volumetric flask or burette.
filter funnel is required for effective transfer of the weighed solid,
liquid or solution into the volumetric flask or burette.
A filter funnel
Wash bottle
Wash
bottle contains water and when squeezed, water squarts out. This is
used in washing down the remains of the weighed solid into the
volumetric flask.
bottle contains water and when squeezed, water squarts out. This is
used in washing down the remains of the weighed solid into the
volumetric flask.
A wash bottle
A weighing bottle
This is used in weighing the solute. It is a stoppered bottle. A watch glass can also be used to serve the same purpose.
Retort stand
A burette stand is used for holding the burette in place while carrying out volumetric analysis experiments.
A burette stand
Dropper
A dropper is used to add the indicator dropwise into the solution.
White tile or paper
A
white tile or piece of paper is placed under the flask to give a clear
background for accurate observation of the colour change at the end of
the reaction (end point).
white tile or piece of paper is placed under the flask to give a clear
background for accurate observation of the colour change at the end of
the reaction (end point).
Standard Solutions
The Steps for Preparation of Standard Solutions of Common Acids
Explain the steps for preparation of standard solutions of common acids
A standard solution is a solution of known concentration. For example, a solution containing 15g of sulphuric acid in 1 dm3 of solution is a standard solution.
It has now been approved that volumetric work should be based upon the molar (M) solution. A 1 molar (1M) solution of a compound is a solution which contains one mole of that compound in 1 dm3 of the solution. For example, 58.5g of sodium chloride (NaCl) dissolved in 1 dm3 of the solution makes a molar solution of sodium chloride (1M NaCl). Likewise, 106g of sodium carbonate (Na2CO3) in 1 dm3
of the solution gives a molar solution of sodium carbonate. Therefore, a
1 molar sodium carbonate solution contains 106g of the salt in 1 dm3 of the solution.
of the solution gives a molar solution of sodium carbonate. Therefore, a
1 molar sodium carbonate solution contains 106g of the salt in 1 dm3 of the solution.
1 molar solution of some compounds commonly used in titration contain the following masses of the compounds in 1 dm3 of solution:
| Compound | Relative molecular weight (1 mole) |
| Sodium hydroxide, NaOH | 40g |
| Potassium hydroxide, KOH | 56g |
| Sulphuric acid, H2SO4 | 98g` |
| Hydrochloric acid, HCl | 36.5g |
| Sodium carbonate, Na2CO3 | 106g |
| Sodium bicarbonate, NaHCO3 | 84g |
Derivative concentrations are also used e.g. 0.1M, 0.5M. 2M, etc.
Preparation of standard solutions
A
standard solution is required as a starting point for volumetric
analysis. We learned early that in order to find the unknown
concentration of a substance in volumetric analysis, the concentration
of one of the solutions must be known.
standard solution is required as a starting point for volumetric
analysis. We learned early that in order to find the unknown
concentration of a substance in volumetric analysis, the concentration
of one of the solutions must be known.
A
small range of substances are suitable for direct preparation of
accurately standard solutions. Substances that cannot be used for direct
preparation of standard solutions include sodium hydroxide, potassium
hydroxide and concentrated sulphuric acid. These substances absorb water
vapour from the air and hence cannot be weighed out precisely without
taking extra precautions. Apart from absorbing water vapour from the
air, sodium and potassium hydroxides react with carbon dioxide of the
air to form respective carbonates.
small range of substances are suitable for direct preparation of
accurately standard solutions. Substances that cannot be used for direct
preparation of standard solutions include sodium hydroxide, potassium
hydroxide and concentrated sulphuric acid. These substances absorb water
vapour from the air and hence cannot be weighed out precisely without
taking extra precautions. Apart from absorbing water vapour from the
air, sodium and potassium hydroxides react with carbon dioxide of the
air to form respective carbonates.
2NaOH(s) + CO2(g)→ Na2CO3(s) + H2O(l)
2KOH(s) + CO2(g)→ K2CO3(s) + H2O(l)
Some
solutions are volatile in nature and so are likely to change slowly in
concentration during ordinary use. These include concentrated
hydrochloric acid and ammonia.
solutions are volatile in nature and so are likely to change slowly in
concentration during ordinary use. These include concentrated
hydrochloric acid and ammonia.
A
compound commonly used for preparation of a precisely standard solution
is anhydrous sodium carbonate. It is best prepared from highly pure
sodium carbonate. This is achieved by heating sodium bicarbonate to
constant mass to make sure the compound is fully decomposed.
compound commonly used for preparation of a precisely standard solution
is anhydrous sodium carbonate. It is best prepared from highly pure
sodium carbonate. This is achieved by heating sodium bicarbonate to
constant mass to make sure the compound is fully decomposed.
2NaHCO 3(s) →Na2CO3(s) + H2O(g) + CO2(g)
The
sodium carbonate so formed is suitable for preparation of a standard
solution and can be weighed without undergoing any appreciable change in
composition.
sodium carbonate so formed is suitable for preparation of a standard
solution and can be weighed without undergoing any appreciable change in
composition.
Precautions to be observed while preparing a standard solution
- Transference
of the substance from the weighing bottle to the beaker or flasks
should be done with outmost care so that not a single particle of the
substance is lost. - Undissolved substance should not be
transferred to the measuring flask. Make sure all the solid dissolves
into solution before transferring the solution to flask. - During making up of the volume, the last drop of the water should be added carefully. Do not blow out the final drop.
Standard Solutions of Bases
Prepare standard solutions of bases
Preparation of 0.1M sodium carbonate solution
The molecular mass of sodium carbonate (Na2CO3) is 106g. Therefore, a molar (1M) solution of sodium carbonate contains 106g in 1 dm3 (1000 cm3) of solution.
In
order to prepare 0.1M solution of the carbonate, we have to weigh 10.6g
of the carbonate and put it into a volumetric flask, which has a
capacity of 1000 cm3.
order to prepare 0.1M solution of the carbonate, we have to weigh 10.6g
of the carbonate and put it into a volumetric flask, which has a
capacity of 1000 cm3.
However, normally 250 cm3 flasks are used. This means, in a 250 cm3 flask we have to add 10.6/4= 2.65g calcium of sodium carbonate.
Thus, 1000 cm3 ≡ 10.6g
250 cm3 ≡ X g
X = 250/1000×10.6 = 10.6/4 = 2.65g
The same procedure can be followed when preparing 0.25M, 0.5M, 2M, etc. of the solutions.
Procedure
- Weigh exactly 2.65g of sodium carbonate using a common balance and put it onto a watch glass.
- Transfer it slowly into a beaker of 500-cm3 capacity containing about 50 cm3 of hot distilled water.
- Wash
down the watch glass with a jet of hot distilled water from a wash
bottle and allow the washings to fall into the beaker (figure 5.7). Make
sure all the sodium carbonate is washed into the beaker. - Stir
with a glass rod until all the solid is completely dissolved, and then
cool the solution to room temperature. Leave the rod standing in the
solution. - Pour the solution carefully down the glass rod into a 250 cm3 measuring flask.
- Wash
the beaker out at least twice with jets of cold distilled water
directed round the slides and pour the washings down the glass rod into
the measuring flask (figure 5.8). - Shake the flask gently and fill it up with cold distilled water almost to the mark.
- Add more distilled water drop by drop from a pipette until the meniscus is on the graduation mark (figure 5.9).
- Stopper the measuring flask and shake well. The liquid should then be exactly 0.1M sodium carbonate solution.
Preparation
of standard solutions of other bases of different molarities e.g. 0.2M,
0.5M, 1.0M, 2.0M, etc. can be achieved by using the above procedures.
The only variable will be the weight of the solids and volume of water
as stated early.
of standard solutions of other bases of different molarities e.g. 0.2M,
0.5M, 1.0M, 2.0M, etc. can be achieved by using the above procedures.
The only variable will be the weight of the solids and volume of water
as stated early.
Washing the watch glass
Filling the washings into the flask
Correct reading of the liquid volume
Acid-base Titration Experiments
Carry out acid-base titration experiments
Preparation of 0.1M sulphuric acid solution
A
standard solution of sulphuric acid cannot be prepared directly because
concentrated sulphuric acid is hygroscopic in nature (it tends to
absorb water vapour from the air diluting itself) and is never reliably
pure. A solution a little above 0.1M is prepared and then standardized
and diluted with distilled water to exactly 0.1M.A molar solution of H2SO4 contains 98g of pure acid in 1 dm3. Therefore, the 0.1M acid contains 9.8g of the acid in 1 dm3 of the solution. The pure concentrated acid has a density (concentration) of about 1.8g/cm3. So, 9.8g of it occupy about 9.8/1.8 = 5.5cm3
standard solution of sulphuric acid cannot be prepared directly because
concentrated sulphuric acid is hygroscopic in nature (it tends to
absorb water vapour from the air diluting itself) and is never reliably
pure. A solution a little above 0.1M is prepared and then standardized
and diluted with distilled water to exactly 0.1M.A molar solution of H2SO4 contains 98g of pure acid in 1 dm3. Therefore, the 0.1M acid contains 9.8g of the acid in 1 dm3 of the solution. The pure concentrated acid has a density (concentration) of about 1.8g/cm3. So, 9.8g of it occupy about 9.8/1.8 = 5.5cm3
The preparation of a standard solution of sulphuric acid involves two stages:
- Diluting a concentrated solution of the acid to an approximate molarity.
- Finding
the exact concentration of the acid (standardizing it) by titrating it
against a standard solution of a base (previously prepared).
Dilution of concentrated sulphuric acid
Caution: Make sure you wear safety goggles and gloves before carrying out this experiment.
Procedure
- Cautiously, because the acid is very corrosive, take 5.5 – 6.0 cm3 of concentrated sulphuric acid in a small measuring cylinder.
- Pour the acid carefully, with stirring, into a 250-cm3 volumetric flask containing about 100 cm3 of cold distilled water.
- Pour this solution into, say, 700cm3 of cold distilled water in a measuring flask of capacity 1000cm3.
- Wash
out the acid solution remaining in the measuring cylinder with cold
distilled water twice and add the washings into the measuring flask. - Then add distilled water approximately to the mark on the measuring flask, stopper it, and shake well.
This
should give sulphuric acid of concentration a little above 0.1M. The
diluted acid is now standardized with the 0.1M sodium carbonate solution
prepared above.
should give sulphuric acid of concentration a little above 0.1M. The
diluted acid is now standardized with the 0.1M sodium carbonate solution
prepared above.
Determination
of the molarity (standardization) of sulphuric acid solution by
titrating it against 0.1M sodium carbonate solution
of the molarity (standardization) of sulphuric acid solution by
titrating it against 0.1M sodium carbonate solution
The estimation of the concentration of a solution of an acid by reacting the acid with a standard alkali solution is known as titration. The end-point of an acid-base reaction is commonly determined by using a substance known as an indicator.
Procedure
- Measure 25 cm3 of 0.1M sodium carbonate solution and transfer it into a conical flask, using a pipette. Add a few drops of methyl orange indicator. This will turn the sodium carbonate solution yellow.
- Set up the apparatus as shown in figure 5.10
- Pour the acid into a 50-cm3 burette. Read and note the level of the acid in the burette.
Titration setup
- By
means of a tap at the base of the burette, drip the acid slowly into
the conical flask, swirling the flask continuously until the colour of
the liquid in the flask turns orange. This is the end-point of titration. Record the new level of acid in the burette. - Repeat the titration three to four times, noting the initial and final reading of the burette each time.
- Find the volume of the acid as shown below:
Specimen readings
| Titration | Rough titration (Pilot) | Titre 1 | Titre 2 |
| Final burette reading | 24.20 | 23.65 | 23.55 |
| Initial burette reading | 0.00 | 0.00 | 0.00 |
| Volume of acid added | 24.20 | 23.65 | 23.55 |
Neglecting the first (rough) trial run, the average titration is 23.60 cm3
Calculation
The
first step in calculating the molarity of any solution from the results
of an acid-base titration is to write the equation for the reaction.
From the equation, find the number of reacting moles of the acid and
base.
first step in calculating the molarity of any solution from the results
of an acid-base titration is to write the equation for the reaction.
From the equation, find the number of reacting moles of the acid and
base.
Na2CO3(aq)1 mole+ H2SO4(aq)1 mole→Na2SO4(aq) + CO2(aq) + H2O(l)
Now we have the following data.
- Volume of acid, Va = 23.60 cm3
- Volume of base, Vb = 25.00 cm3 (this is the average amount of the base that was added to the flask in titration)
- Molarity of acid, Ma = ?
- Molarity of base, Mb = 0.1M
- Number of moles of acid, Na = 1
- Number of moles of base, Nb = 1
The molarity of the acid can be calculated from the following general formula:
The concentration of sulphuric acid is 0.106M.
To
make the molar concentration of the sulphuric acid solution exactly
equal to that of the sodium carbonate solution (0.1M), 23.6 cm3of the acid must be diluted to 25 cm3, that is, 1.4 cm3of distilled water must be added to 23.6 cm3of the acid.
make the molar concentration of the sulphuric acid solution exactly
equal to that of the sodium carbonate solution (0.1M), 23.6 cm3of the acid must be diluted to 25 cm3, that is, 1.4 cm3of distilled water must be added to 23.6 cm3of the acid.
Remember
it was stated early that in order to prepare 0.1M sulphuric acid
solution, you need to dissolve 9.8g of the acid in 1000 cm3 (1 dm3) of distilled water. Assume that some of the acid was wasted through spillage and mishandling and that only 920 cm3 of the acid was left. If, say, 920 cm3 of the acid was left, it can be made exactly 0.1M by the addition of 920×1.4/23.6 = 55cm3of distilled water. This gives exactly 0.1M of the acid. This is the same as saying that, if 23.6 cm3 of the acid were diluted with 1.4cm3 of distilled water, then 920cm3 of the acid would be diluted with
it was stated early that in order to prepare 0.1M sulphuric acid
solution, you need to dissolve 9.8g of the acid in 1000 cm3 (1 dm3) of distilled water. Assume that some of the acid was wasted through spillage and mishandling and that only 920 cm3 of the acid was left. If, say, 920 cm3 of the acid was left, it can be made exactly 0.1M by the addition of 920×1.4/23.6 = 55cm3of distilled water. This gives exactly 0.1M of the acid. This is the same as saying that, if 23.6 cm3 of the acid were diluted with 1.4cm3 of distilled water, then 920cm3 of the acid would be diluted with
If, for instance, the volume of acid left was, say, 850 cm3, the amount of distilled water to be added would be = 850×1.4/23.6 = 50.4cm3. This, also, would give exactly 0.1M of the acid.
In
principle, the amount of distilled water to be added is always
calculated based on the amount of the acid left as exemplified above.
principle, the amount of distilled water to be added is always
calculated based on the amount of the acid left as exemplified above.
These
two standard alkaline and acidic solutions can be used to standardize
other solutions, e.g. sodium hydroxide, potassium hydroxide,
hydrochloric acid, nitric acid, etc. You may dilute any base or
commercial acid to some required concentration e.g. 0.2M, 0.5M, 0.25M,
etc and then standardize it by similar procedures.
two standard alkaline and acidic solutions can be used to standardize
other solutions, e.g. sodium hydroxide, potassium hydroxide,
hydrochloric acid, nitric acid, etc. You may dilute any base or
commercial acid to some required concentration e.g. 0.2M, 0.5M, 0.25M,
etc and then standardize it by similar procedures.
Choice of indicators in acid-base titration
We
learned early that the estimation of the concentration of a solution of
an acid or base by reaction with a standard alkali or acid solution
respectively, is known as titration. The end-point of an acid-base
titration is commonly determined using substances known as indicators,
which usually portray certain characteristic colours when in alkaline or
acid solutions.
learned early that the estimation of the concentration of a solution of
an acid or base by reaction with a standard alkali or acid solution
respectively, is known as titration. The end-point of an acid-base
titration is commonly determined using substances known as indicators,
which usually portray certain characteristic colours when in alkaline or
acid solutions.
The
indicators in acid-base titrations must be chosen carefully because the
choice of an inappropriate indicator would lead to an incorrect result.
The choice of an indicator is based on the strength of an acid or base
involved in the reaction.
indicators in acid-base titrations must be chosen carefully because the
choice of an inappropriate indicator would lead to an incorrect result.
The choice of an indicator is based on the strength of an acid or base
involved in the reaction.
There
are three common indicators which are used in titration experiments
involving acids and bases namely, methyl orange, litmus and
phenolphthalein. The other indicators in less common use are as included
in the table below. The table shows the colours which each of these
indicators take up in acid or alkaline solution.
are three common indicators which are used in titration experiments
involving acids and bases namely, methyl orange, litmus and
phenolphthalein. The other indicators in less common use are as included
in the table below. The table shows the colours which each of these
indicators take up in acid or alkaline solution.
Colour of indicators in acid and alkaline solutions
| Indicator Colour of indicator |
| acid solution alkaline solution |
| Methyl orange pink yellow |
| Litmus red blue |
| Phenolphthalein colourless pink |
| Malachite green yellow blue/green |
| Thymol blue red yellow |
| Bromocresol green yellow blue |
| Bromothymol blue yellow blue |
Indicators suitable for particular types of acid-base reactions are as given in the table below:
Indicators suitable for different acid-base reactions
| Acid-base titration | Example | Choice of indicator |
| Strong acid/strong base | H2SO4 and NaOH | Any indicator |
| Weak acid/strong base | CH3COOH (ethanoic acid) and KOH | Phenolphthalein |
| Strong acid/weak base | HCl and NH3 | Methyl orange |
| Weak acid/weak base | CH3COOH and NH3 | No satisfactory indicator available |
Volumetric Calculations
Common Mineral Acids
Standardize common mineral acids
Data
for calculations of volumetric analysis problems are obtained from
volumetric analysis experiments. For any volumetric analysis problem, at
least one standard solution is required. A correctly balanced reaction
equation (from which moles ratios can be derived) is a prerequisite for
all these calculations. This is because the mole ratio is an integral
part of the general expressionused for all volumetric analysis
calculations. The general expression is given by:
for calculations of volumetric analysis problems are obtained from
volumetric analysis experiments. For any volumetric analysis problem, at
least one standard solution is required. A correctly balanced reaction
equation (from which moles ratios can be derived) is a prerequisite for
all these calculations. This is because the mole ratio is an integral
part of the general expressionused for all volumetric analysis
calculations. The general expression is given by:
where;
- VA = Volume of acid
- MA = Molarity of acid
- VB = Volume of base
- MB = Molarity of base
- NA = Number of moles of acid
- NB = Number of moles of base
The Relative Atomic Mass of Unknown Element in an Acid or Alkali
Find the relative atomic mass of unknown element in an acid or alkali
Example 1
12.5 cm3 of 0.5M sulphuric acid neutralized 50 cm3 of a given solution of sodium hydroxide. What is the molarity of the alkali?
Solution
Reaction equation is:
H2SO4(aq) + 2NaOH(aq) →Na2SO4(aq) + 2H2O(l)
From
the equation, 1 mole of sulphuric acid solution reacts with 2 moles of
sodium hydroxide solution. So, the number of moles of the acid, NA = 1 and the number of moles of base, NB = 2.The other data are as follows:
the equation, 1 mole of sulphuric acid solution reacts with 2 moles of
sodium hydroxide solution. So, the number of moles of the acid, NA = 1 and the number of moles of base, NB = 2.The other data are as follows:
VA = 12.5 cm3, MA = 0.5M and VB = 50 cm3
So, the molarity of the acid = 0.25M
Example 2
20 cm3 of a solution containing 7g/dm3 of a metal hydroxide, XOH, were exactly neutralized with 25 cm3 of 0.10M hydrochloric acid.
- Write a balanced chemical equation for the neutralization of the metal hydroxide, XOH, with hydrochloric acid.
- Calculate the concentration of the metal hydroxide in moles per dm3.
- (i) Calculate the molar mass of XOH (ii) Identify element X
Solution
<!–[endif]–>HCl(aq) + XOH(aq)→ XCl(aq) + H2O(l)
From the reaction equation, NA = 1 and NB = 1.
VA = 25 cm3, VB = 20 cm3, MA = 0.1M, MB = ?
X = 39g
Therefore, element X is potassium (K)
The Percentage Purity of an Acid or an Alkali
Calculate the percentage purity of an acid or an alkali
Example 3
5.1g of impure sodium carbonate solution was dissolved in water to make 500 cm3 of solution. 20 cm3 of this solution was titrated against 20.45 cm3 of 0.04M hydrochloric acid. Calculate the percentage purity of the sodium carbonate solid.
Solution
To calculate the concentration of impure Na2CO3 solution:5.1g per 500 cm3 = 10.2g dm-3Reaction equation is:
Na2CO3(aq) + 2HCl(aq) → 2NaCl(aq) + H2O(l) + CO2(g)
From the equation, NA =2 and NB = 1.
= 0.02045M
Therefore, molarity of the base = 0.02045M
To calculate the concentration of pure base (Na2CO3 solution) in g dm-3:
Concentration = Molarity × Molar mass
But molar mass of Na2CO3 = 106.
So, concentration (g/dm3) = 0.02045 × 106 = 2.1677g dm-3
The percentage purity of sodium carbonate = 21.25%
The Number of Molecules of Water of Crystallization of a Substance
Find the number of molecules of water of crystallization of a substance
Example 4
0.465g of a hydrated form of sodium carbonate exactly reacts with 75 cm3
of 0.10M hydrochloric acid. Calculate the number of molecules of water
of crystallization present in one mole of the hydrated salt.
of 0.10M hydrochloric acid. Calculate the number of molecules of water
of crystallization present in one mole of the hydrated salt.
Solution
The balanced reaction equation for the anhydrous salt is:
Na2CO3(s) + 2HCl(aq) → 2NaCl(aq) + CO2(g) + H2O(l)
Number of moles of HCl present in 75 cm3 = 75/1000×0.1 = 0.0075M
Mass of 0.00375 moles of Na2CO3 = 0.00375 × 106 = 0.3975g (= mass of anhydrous salt)
Molar mass of anhydrous salt = 106g
Mass of water contained in the hydrated salt = mass of hydrated salt – mass of anhydrous salt
= 0.465 – 0.3975
= 0.0675g
Therefore, the mass of water of crystallization = 0.0675g
But, remember that mass of the anhydrous salt = 0.3975g
Now, let the formula of the hydrated salt be Na2CO3.XH2O, where X is the number of moles of water crystallization
Therefore, the formula of the salt is Na2CO3.H2O
The Application of Volumetric Analysis in Real Life Situations
Explain the application of volumetric analysis in real life situations
Volumetric
analysis has a variety of laboratory and industrial applications in
everyday life. The following are just a few of the applications (uses)
of volumetric analysis in daily life:
analysis has a variety of laboratory and industrial applications in
everyday life. The following are just a few of the applications (uses)
of volumetric analysis in daily life:
- Use in preparation of standard solutions:Standard
solutions are prepared by applying the knowledge of volumetric
analysis. Volumetric analysis is used in school, college and university
chemistry laboratories to determine concentrations of unknown
substances. The titrant (the known solution) is added to a known
quantity of analyte (unknown solution) and a reaction takes place.
Knowing the volume of the titrant allows one to determine the
concentration of the unknown substance. - Use in environmental and water safety:Titration
is important in environmental chemistry, where scientists can use it to
analyze acid rain or contaminants in surface water samples.
Environmental studies usually involve an analysis of precipitation and
its response to pollution. To quantify the degree of contamination in
natural rainwater or snow, titration is used. The process is quick and
results are reliable. Since most titration processes do not require
expensive or specialized equipment, the test can be performed often and
in different areas with relatively little effort.The safety of water is
based on its chemical ingredients. By analyzing wastewater, the extent
of contamination and the requirements for filtering and cleaning can be
determined. Titration is a key mechanism in this analysis. Often, more
specialized titration equipment is used in this application, which
measures ammonia levels in combination with other reactants to quantify
other chemicals present. - Use in food and beverage industry:In
the food and beverage industry, manufacturers must ensure their
products meet certain quality criteria or contain standard
concentrations of specific additives, so titration is often used to
analyze the products before sale. Wine is often affected by its degree
of acidity. It is possible to improve wine production by measuring
acidity using titration. Simple, inexpensive titration kits are
available to winemakers for this purpose. The results of a titration
test on wine can suggest if additional ingredients are necessary to
maintain its quality.In general, all brewing industries and distilleries
apply the knowledge of volumetric analysis (titration) to determine the
acidity and alcohol contents of their beers and other alcoholic
beverages.The process also finds ample use in food industry. The
compounds which make up food products help determine their nutritional
implications. Titration is one technique that assists in these studies.
The acidity of orange juice, for example, is easily determined using a
standard titration process. In this process, an electrode is added to a
solution made up of orange juice and deionized water. The titrant
catalyst then measures the acidity of the juice. Manufacturers can use
the technique to vary this quality to satisfy customers or those with
special nutritional needs. - Use in agriculture:Volumetric
analysis technique is used to determine the soil pH. This is important
because, if the pH of a certain soil is found to be extremely low or
high, corrective measures are taken by adding the correct quantity of
agricultural limes or other chemicals to make the soil suitable for
plant growth.The method is also used by agronomists and farmers to
analyse the kind and amount of plant nutrient elements present in a
particular sample of soil, the knowledge of which helps determine soil
fertility.
Industrial and Laboratory Skills of Volumetric Analysis
Compare industrial and laboratory skills of volumetric analysis
The
knowledge of volumetric analysis (titration) is used in hospitals and
medical laboratories to carry out such duties as preparation of
solutions and suspensions, blood analysis, and diagnosis of certain
diseases and health problems. For example, when dissolving a solid drug
to make a solution for injection, utmost precision is required to
measure the correct volume of liquid to be used to dissolve a correct
amount of solid drug to prepare the solution of a given concentration to
inject to a patient.
knowledge of volumetric analysis (titration) is used in hospitals and
medical laboratories to carry out such duties as preparation of
solutions and suspensions, blood analysis, and diagnosis of certain
diseases and health problems. For example, when dissolving a solid drug
to make a solution for injection, utmost precision is required to
measure the correct volume of liquid to be used to dissolve a correct
amount of solid drug to prepare the solution of a given concentration to
inject to a patient.
Also
titration is very important in the pharmaceutical industry, where
precise measurements of quantities and concentrations are essential
throughout the manufacturing process. Titration is thus an important
part of the pharmaceutical industry to ensure quality control. Many
variations of the titration technique are used, and specialized
equipment for pharmaceutical titration is often developed to make the
process more efficient.
titration is very important in the pharmaceutical industry, where
precise measurements of quantities and concentrations are essential
throughout the manufacturing process. Titration is thus an important
part of the pharmaceutical industry to ensure quality control. Many
variations of the titration technique are used, and specialized
equipment for pharmaceutical titration is often developed to make the
process more efficient.
TOPIC 6: IONIC THEORY AND ELECTROLYSIS
Ionic Theory
The Mechanisms of Electrolysis
Electrolytic Cells of Different Electrolytes in the Molten and Aqueous States
Set up electrolytic cells of different electrolytes in the molten and aqueous states
The
conductivity of ionic compounds is explained by the fact that ions move
in a particular direction in an electric field. This can be shown in
experiments with coloured salts. For example, copper (II) chromate (VI)
(CuCrO4) dissolves in water to give a green
solution. This solution is placed in a U-tube. A colourless solution of
dilute hydrochloric acid (HCl) is then layered on top of the salt
solution in each arm. Graphite rods are fitted as shown in figure 13.3.
These rods (electrodes) carry the current into and out of the solution.
conductivity of ionic compounds is explained by the fact that ions move
in a particular direction in an electric field. This can be shown in
experiments with coloured salts. For example, copper (II) chromate (VI)
(CuCrO4) dissolves in water to give a green
solution. This solution is placed in a U-tube. A colourless solution of
dilute hydrochloric acid (HCl) is then layered on top of the salt
solution in each arm. Graphite rods are fitted as shown in figure 13.3.
These rods (electrodes) carry the current into and out of the solution.
After passing the current for a short time, the solution around the cathode becomes blue. Around the anode, the solution becomes yellow. These colours are produced by the movement (migration) of the ions in the salt. The positive copper ions (Cu2+) are blue in solution. They are attracted to the cathode (negative electrode). The negative chromate ions (CrO42-)
are yellow in solution. They are attracted to the anode (the positive
electrode). The use of coloured ions in solution has shown the direction
that positive and negative ions move in an electric field. Always
positive ions (cations) move to the cathode and negative ions (anions)
move to the anode.
are yellow in solution. They are attracted to the anode (the positive
electrode). The use of coloured ions in solution has shown the direction
that positive and negative ions move in an electric field. Always
positive ions (cations) move to the cathode and negative ions (anions)
move to the anode.
Ionic Migrations During Electrolysis and the Preferential Discharge of Ions at the Electrodes
Explain ionic migrations during electrolysis and the preferential discharge of ions at the electrodes
When
a salt such as sodium chloride is dissolved in water, its ions are set
free to move. So the solution can be electrolysed. Since the salts,
alkalis and acids are dissolved in water, most of the solutions are
aqueous. There is then a complication in electrolysis of such substances
in aqueous form. This is because the water used to dissolve them also
do ionize partially (it is a weak electrolyte)
a salt such as sodium chloride is dissolved in water, its ions are set
free to move. So the solution can be electrolysed. Since the salts,
alkalis and acids are dissolved in water, most of the solutions are
aqueous. There is then a complication in electrolysis of such substances
in aqueous form. This is because the water used to dissolve them also
do ionize partially (it is a weak electrolyte)
Then
at each electrode, we get more than one ion for discharge, but only one
is supposed to be discharged. Take an example of electrolysis of copper
(II) sulphate solution using platinum electrodes. By ionic theory, the
solution ionizes thus:CuSO4 (aq) → Cu2+ (aq) + SO42-(aq) (strong electrolyte)
at each electrode, we get more than one ion for discharge, but only one
is supposed to be discharged. Take an example of electrolysis of copper
(II) sulphate solution using platinum electrodes. By ionic theory, the
solution ionizes thus:CuSO4 (aq) → Cu2+ (aq) + SO42-(aq) (strong electrolyte)
During electrolysis, Cu2+ and H+ ions move to the cathode while SO42-and OH– ions move to the anode.
| Cathode | Anode |
| Cu2+ | SO42- |
| H+ | OH– |
In situations like this, the order of discharge of the ions at the electrode will depend on:
- the position of the metal ion or radical in the electrochemical series;
- the concentration or nature of the ions (or electrolyte) to be discharged; and
- the nature of the electrodes used.
The position of ion or radical in the electrochemical series
| Cations | Anions | ||
| K+ | ease of dischargeincreases | SO42- | ease of discharge increases |
| Ca2+ | NO3– | ||
| Na+ | Cl– | ||
| Mg2+ | Br– | ||
| Al3+ | I– | ||
| Zn2+ | OH– | ||
| Fe2+ | |||
| Pb2+ | |||
| H+ | |||
| Cu2+ | |||
| Ag+ |
The
arrangement of ions above is the same as that of the electrochemical
series. If all other factors are constant, any ion will be discharged
from solution in preference to those above it.
arrangement of ions above is the same as that of the electrochemical
series. If all other factors are constant, any ion will be discharged
from solution in preference to those above it.
| Cathode | Anode | |
| Cu2+ | SO42- | |
| H+ | OH– | |
| Cu2+ + 2e– → Cu(s)copper is discharged (loses its charge) | 4OH–→2H2O(l)+O2(g)+4e– hydroxyl ion is discharged |
In other words, we can say that the reaction that occurs at the cathode is reduction (electron gain) and that which occurs at the anode is oxidation (electron loss).
Result
- At cathode, copper is deposited
- At anode, oxygen is given out (liberated)
- The solution at the end of electrolysis is colourless and acidic because in the electrolyte there are left H+ and SO42- ions, which remain moving freely in the solution as ionic sulphuric acid.
The concentration or nature of electrolyte
Take an example of electrolysis of sodium chloride using platinum electrodes when in:
- aqueous solution;
- concentrated aqueous solution; or
- fused or molten state.
aqueous solution
NaCl → Na+ + Cl–
H2O⇔H+ + OH–
During electrolysis:
| Cathode | Anode |
| Na+and H+ | Cl–and OH– |
| H+ions are discharged in preference to Na+ ions: 2H+ + 2e– → H2(g) | OH–ions are discharged in preference to Cl–ions:4OH–→2H2O(l) +O2(g)+4e– |
Result
- At cathode, hydrogen is given out
- At anode, oxygen is out
- The concentration of the solution remains constant since all Na+ and Cl– ions remain undisturbed in the solution. This means that the Na+ and Cl– ions that are left in the solution are equivalent to sodium chloride solution.
<!–[endif]–>Concentrated aqueous solution
NaCl → Na+ + Cl–
H2O⇔H+ + OH–
| Cathode | Anode |
| Na+and H+H+ ions are discharged in preference to Na+ ions since Na+ and H+ are very far from each other in e. c.s2H+ + 2e– → H2(g) | Cl–and OH–Cl–ions are discharged in preference to OH– ions since Cl– and OH– ions are very close to each other in the e. c. s (and because there are more Cl– ions in the solution) 2Cl– → Cl2(g) + 2e– |
Results
- At cathode H2(g) is given off
- At anode Cl2(g) is given off
- The solution becomes progressively more alkaline as the electrolysis goes on because Na+ and OH–ions
remain in solution as caustic soda (sodium hydroxide) solution. This is
the main method used in the manufacture of sodium hydroxide in
industry. Na+(aq) + OH–(aq) → NaOH(aq)
Fused or molten sodium chloride
NaCl → Na+ + Cl–
During electrolysis:
| Cathode | Anode |
| Na+ + e– → Na(l) | 2Cl– → Cl2(g) + 2e– |
Result
- At cathode sodium is liberated (deposited)
- At anode chlorine is given off
The nature of electrodes (inert vs active electrodes)
<!–[endif]–>If dilute sulphuric acid iselectrolysed using platinum electrodes:
H2SO4 → 2H+ + SO42-
H2O⇔H+ + OH–
| Cathode | Anode |
| 2H+ + 2e– → H2(g) | SO42- and OH– 4OH–→2H2O(l)+O2(g)+4e– |
Result
- At cathode H2(g) is given out
- At anode O2(g) is given out
- The solution becomes acidic at the end of electrolysis because of the acidic ions (SO42-) left in the finala solution.
Charge flow during electrolysis
The
coulomb is the electrolytic unit of charge. A current of one ampere is
the rate of flow of charge equal to one coulomb per second.The charge is
calculated from the knowledge of the number of seconds for which a
steady current is passed.
coulomb is the electrolytic unit of charge. A current of one ampere is
the rate of flow of charge equal to one coulomb per second.The charge is
calculated from the knowledge of the number of seconds for which a
steady current is passed.
| Current in circuit | Time taken | Total charge |
| 1 ampere | 1 second | 1 coulomb |
| 1 ampere | 10 seconds | 10 coulombs |
| 20 amperes | 10 seconds | 200 coulombs |
| A amperes | t seconds | At coulombs |
Therefore charge = quantity of electricity (Q) = I×t
Flow of charge required to liberate 1 mole of element during electrolysis
Electrolysis always produces chemical reactions. Consider a reaction (at cathode) in which one mole of silver Ag+ ions is discharged and deposited.Ag+ + e– → Ag(s)
In this case, 1 mole of electrons (e–) is required to discharge 1 mole of Ag+ ions to produce 1 mole of silver atoms (Ag(s)).
1 mole of electrons is a large charge and experiments show that it is
equal to 96500 coulombs. Therefore 1 mole of electrons = 96500 coulombs.
This is called the faraday.
1 mole of electrons is a large charge and experiments show that it is
equal to 96500 coulombs. Therefore 1 mole of electrons = 96500 coulombs.
This is called the faraday.
The
number of faradays (moles of electrons) required to liberate 1 mole of
an element during electrolysis is deduced from the equation for the
electrode reaction.
number of faradays (moles of electrons) required to liberate 1 mole of
an element during electrolysis is deduced from the equation for the
electrode reaction.
Example 1
| Element | Electrode reaction | Faradays |
|
Na+ + e– → Na(s) | 1 |
|
Cu2+ + 2e– → Cu(s) | 2 |
|
Al3+ + 3e– → Al(s) | 3 |
|
2Cl– → Cl2(g) + 2e– | 2 |
That
is, 1 faraday is needed to deposit 1 mole of sodium atoms (23g), 3
faradays to deposit 1 mole of aluminium atoms (27g) and 2 faradays to
liberate 1 mole of chlorine gas (71g).
is, 1 faraday is needed to deposit 1 mole of sodium atoms (23g), 3
faradays to deposit 1 mole of aluminium atoms (27g) and 2 faradays to
liberate 1 mole of chlorine gas (71g).
- The mass of an element liberated by 1 coulomb of electricity during electrolysis is called electrochemical equivalent of that element.
- The mass of an element deposited or liberated by 1 faraday during electrolysis is called chemical equivalent of that element.
Experiments to Identify the Products of Electrolysis when Different Electrolytes are Used
Perform experiments to identify the products of electrolysis when different electrolytes are used
Activity 1
Perform experiments to identify the products of electrolysis when different electrolytes are used
Experiments to Identify the Products of Electrolysis when Different Electrodes are Used
Perform experiments to identify the products of electrolysis when different electrodes are used
Activity 2
Perform experiments to identify the products of electrolysis when different electrodes are used
Laws of Electrolysis
Experiments to Relate Masses Liberated and Quantity of Electricity Passed
Carry out experiments to relate masses liberated and quantity of electricity passed
The
laws expressing the quantitative results of electrolysis were first
stated by a British chemist called Michael Faraday. The laws assert that
the amount (expressed in moles) of an element liberated during
electrolysis depends on:
laws expressing the quantitative results of electrolysis were first
stated by a British chemist called Michael Faraday. The laws assert that
the amount (expressed in moles) of an element liberated during
electrolysis depends on:
- the time of passing the steady current;
- the magnitude of the steady current passed; and
- the charge on the ion of the element.
An Experiment to Verify Faraday’s First Law of Electrolysis
Carry out an experiment to verify Faraday’s First Law of Electrolysis
The
fact that the amount of a substance liberated during electrolysis
depends upon these factors can be proved by conducting experiments. The
product of time (measured in seconds) and the current passed (measured
in amperes) gives a measure of electricity known as the quantity of
electricity.
fact that the amount of a substance liberated during electrolysis
depends upon these factors can be proved by conducting experiments. The
product of time (measured in seconds) and the current passed (measured
in amperes) gives a measure of electricity known as the quantity of
electricity.
Quantity of electricity (Q) [coulombs]= current (I) [amperes]×time (t) [seconds]
Q = I × t
Because
of this relationship, factors (1) and (2) may be included in the same
experiment. The experiment to determine the effect of time on the amount
of element deposited or liberated is carried out by passing a steady
current through a solution of the compound of that element for different
lengths of time.
of this relationship, factors (1) and (2) may be included in the same
experiment. The experiment to determine the effect of time on the amount
of element deposited or liberated is carried out by passing a steady
current through a solution of the compound of that element for different
lengths of time.
The
following table summarises the specimen results obtained by passing a
steady current (0.21 amps) through a solution of copper (II) sulphate
for 15, 30, 45 and 60 minutes. The last column shows the mass of copper
deposited.
following table summarises the specimen results obtained by passing a
steady current (0.21 amps) through a solution of copper (II) sulphate
for 15, 30, 45 and 60 minutes. The last column shows the mass of copper
deposited.
Specimen results for electrolysis of copper (II) sulphate
| Current (amps) | Time(s) | Quantity of electricity (coulombs) | Mass of copper deposited (grams) |
| 0.21 | 15 × 60=900 | 900 ×0.21=189 | 0.063 |
| 0.21 | 30 × 60=1800 | 1800 ×0.21=378 | 0.129 |
| 0.21 | 45 × 60=2700 | 2700 ×0.21=576 | 0.187 |
| 0.21 | 60 × 60=3600 | 3600 × 0.21=756 | 0.250 |
The
relationship between the amount of copper deposited and the quantity of
electricity passed can be assessed by considering the values in the
last two columns in the table. This data may be represented in a graph
illustrated in figure 6.4. The shape of the graph shows a straight line
passing through the origin. From this fact, it is clear that the mass of
copper deposited is directly proportional to the quantity of
electricity passed. This is in accordance to what Faraday formulated in
his First Law.
relationship between the amount of copper deposited and the quantity of
electricity passed can be assessed by considering the values in the
last two columns in the table. This data may be represented in a graph
illustrated in figure 6.4. The shape of the graph shows a straight line
passing through the origin. From this fact, it is clear that the mass of
copper deposited is directly proportional to the quantity of
electricity passed. This is in accordance to what Faraday formulated in
his First Law.
Graph of mass deposited versus amount of electricity passed
Faraday’s First Law of Electrolysis states that the
mass of a substance liberated at (or dissolved from) an electrode
during electrolysis is directly proportional to the quantity of
electricity passing through the electrolyte.
mass of a substance liberated at (or dissolved from) an electrode
during electrolysis is directly proportional to the quantity of
electricity passing through the electrolyte.
The
quantity of electricity is measured in coulombs where a coulomb is the
passage of an electric current of one ampere for one second. Let
quantity of electricity is measured in coulombs where a coulomb is the
passage of an electric current of one ampere for one second. Let
- m be the mass of the substance liberated;
- I be the current passed in amperes; and
- t be the time in seconds.
We can therefore represent the first law mathematically as:m α I × t or m = Z × I × t where Z
is the proportionality constant referred to as electrochemical
equivalent of the substance liberated. Electrochemical equivalent is the
mass of a substance (element) liberated by 1 coulomb of electricity
during electrolysis.
is the proportionality constant referred to as electrochemical
equivalent of the substance liberated. Electrochemical equivalent is the
mass of a substance (element) liberated by 1 coulomb of electricity
during electrolysis.
An Experiment to Verify Faraday’s Second Law of Electrolysis
Carry out an experiment to verify Faraday’s Second Law of Electrolysis
The
third (3) factor mentioned previously as affecting the amount of
substance liberated during electrolysis may also be investigated
experimentally. Because our interest is the effect of the charge on the
ions present in solution, we need to keep the quantity of electricity
fixed whilst varying the types of the ions in solution. This may be
achieved by passing the same quantity of electricity through two cells,
with ions of different charges in each cell.
third (3) factor mentioned previously as affecting the amount of
substance liberated during electrolysis may also be investigated
experimentally. Because our interest is the effect of the charge on the
ions present in solution, we need to keep the quantity of electricity
fixed whilst varying the types of the ions in solution. This may be
achieved by passing the same quantity of electricity through two cells,
with ions of different charges in each cell.
Apparatus for confirmation of Faraday’s Second Law
The
experiment is conducted using two voltameters. The two voltameters are
connected in series as shown. The first one is a copper voltameter and
the second is a silver voltameter.The copper voltameter has copper electrodes in a solution of copper (II) sulphate. Hence, in this voltameter, copper ions are discharged and deposited at the cathode.
experiment is conducted using two voltameters. The two voltameters are
connected in series as shown. The first one is a copper voltameter and
the second is a silver voltameter.The copper voltameter has copper electrodes in a solution of copper (II) sulphate. Hence, in this voltameter, copper ions are discharged and deposited at the cathode.
The silver voltammeter has silver electrodes in a solution of silver nitrate. The discharged silver ions are deposited at the cathode.
Before
the experiment is started, each cathode electrode in the voltameters is
cleaned, dried and weighed. The electrodes are then connected to the
circuit, after which a suitable current is passed for a measured period
of time.
the experiment is started, each cathode electrode in the voltameters is
cleaned, dried and weighed. The electrodes are then connected to the
circuit, after which a suitable current is passed for a measured period
of time.
After
this, the cathode are removed from the voltameters, cleaned, dried and
reweighed. The increase in mass of the two cathode electrodes represents
the respective amounts of copper and silver deposited at the cathodes.
The quantity of electricity required to deposit one mole of each element
is calculated using Faraday’s Second Law of Electrolysis.
this, the cathode are removed from the voltameters, cleaned, dried and
reweighed. The increase in mass of the two cathode electrodes represents
the respective amounts of copper and silver deposited at the cathodes.
The quantity of electricity required to deposit one mole of each element
is calculated using Faraday’s Second Law of Electrolysis.
Specimen results
| Current flowing | = | 0.45A |
| Duration of current flow | = | 25 minutes |
| Mass of copper deposited | = | 0.221g |
| Mass of silver deposited | = | 0.755g |
The
results show that the masses of silver and copper deposited are
different. A comparison of the amounts of each of the elements deposited
can be made simple by calculating the number of moles of atoms of each
of the element deposited.
results show that the masses of silver and copper deposited are
different. A comparison of the amounts of each of the elements deposited
can be made simple by calculating the number of moles of atoms of each
of the element deposited.
Thus:
Amount of copper deposited = 0.221/63.5mole = 0.0035 mole
Amount of silver deposited = 0.755/107.8mole = 0.0070 mole
It
is seen that twice as many atoms of silver are deposited as atoms of
copper. The difference in amount of each element deposited arises from
the difference in charges on ions of the element concerned.
is seen that twice as many atoms of silver are deposited as atoms of
copper. The difference in amount of each element deposited arises from
the difference in charges on ions of the element concerned.
The
change on the copper ion is twice that on the silver ionand therefore
twice the quantity of electricity will be required to liberate one mole
of copper as for the liberation of one mole of silver.
change on the copper ion is twice that on the silver ionand therefore
twice the quantity of electricity will be required to liberate one mole
of copper as for the liberation of one mole of silver.
This
relationship is in accordance to Faraday’s Second Law of Electrolysis,
which describes the relationship between the amount of element deposited
and the charge on the ions of that element. Faraday’s Second Law of
Electrolysis states that when the same quantity of electricity is
passed through solutions of different electrolytes the relative numbers
of moles of the elements deposited are inversely proportional to the
charges on the ions of each of the elements respectively.
relationship is in accordance to Faraday’s Second Law of Electrolysis,
which describes the relationship between the amount of element deposited
and the charge on the ions of that element. Faraday’s Second Law of
Electrolysis states that when the same quantity of electricity is
passed through solutions of different electrolytes the relative numbers
of moles of the elements deposited are inversely proportional to the
charges on the ions of each of the elements respectively.
In
order to discharge one mole of monovalent ions such as hydrogen ion,
Sodium ion, Silver ion and Chlorine ion,96500 C of electricity are
required. This quantity of electricity has been experimentally
determined and is known as the Faraday constant.
order to discharge one mole of monovalent ions such as hydrogen ion,
Sodium ion, Silver ion and Chlorine ion,96500 C of electricity are
required. This quantity of electricity has been experimentally
determined and is known as the Faraday constant.
It
represents one mole of electrons, which is the same as the quantity of
electrons required to discharge one mole of Siliverions to give one mole
of silver atom. The validity of Faraday’s Second Law of Electrolysis is
evident from the following observations:
represents one mole of electrons, which is the same as the quantity of
electrons required to discharge one mole of Siliverions to give one mole
of silver atom. The validity of Faraday’s Second Law of Electrolysis is
evident from the following observations:
- One faraday (IF) discharges one mole of H+, Na+, Ag+, Cl– and OH– ions.
- Two Faradays (2F) discharges one mole of Cu2+, Pb2+, Mg2+, Ca2+, Fe2+, etc ions.
- Three Faradays (3F) discharge one mole of Al3+, Fe3+, etc. ions.
Relationship between the Chemical Equivalents of Elements and Quantity of Electricity Passed
Relate the chemical equivalents of elements and quantity of electricity passed
A
steady current of 4 amperes is passed through aqueous copper (II)
sulphate solution for 1800 seconds using platinum electrodes. Calculate:
steady current of 4 amperes is passed through aqueous copper (II)
sulphate solution for 1800 seconds using platinum electrodes. Calculate:
- mass of copper deposited
- mass of oxygen liberated
Given:
- Atomic weight of copper = 63.5
- Atomic weight of oxygen = 16
- 1 Faraday = 96500 C
Solution
<!–[endif]–>Cathode reaction: Cu2+ + 2e– → Cu(s)
In this case, 2 Faradays of electricity are required to deposit one mole of copper atom 63.5g.
This means 2 × 96500 C liberates 63.5g of copper.
Quantity of electricity passed = I×t = 4×1800C. So, if 2×96500 C liberates 63.5g, thenI×t = 4×1800Cwill liberate2.4g of copper.
Therefore, mass of copper deposited = 2.4g
Anode reaction: 4OH–→2H2O(l) + O2(g) + 4e–
The reaction shows that 4 moles of electrons (4 Faradays) are lost during the reaction process.
Therefore, 4 × 96500 C are needed to liberate one mole (32g) of oxygen.
<!–[endif]–>Quantity of electricity flowing = 4 × 1800 C
Therefore, mass of oxygen liberated = 0.6g
Electrolysis
has several uses in industry. Its main application has been in the
fields of manufacture of chemicals and in the purification of metals for
which other purification methods prove either too difficult or highly
expensive to apply. Some applications of electrolysis are as discussed
below:
has several uses in industry. Its main application has been in the
fields of manufacture of chemicals and in the purification of metals for
which other purification methods prove either too difficult or highly
expensive to apply. Some applications of electrolysis are as discussed
below:
The Industrial Purification of Copper by Electrolysis
Outline the industrial purification of copper by electrolysis
Some
metals can be purified by means of electrolysis. This process is used
in industry to purify copper, which must be very pure 99.9% for
electrical wiring. Copper made by roasting the sulphide ore is about
99.5% pure (so it has an impurity level of 0.5%). This level of impurity
cuts down electrical conductivity significantly.
metals can be purified by means of electrolysis. This process is used
in industry to purify copper, which must be very pure 99.9% for
electrical wiring. Copper made by roasting the sulphide ore is about
99.5% pure (so it has an impurity level of 0.5%). This level of impurity
cuts down electrical conductivity significantly.
This
is how the electrolytic purification (refining) process is carried
out:The anode is made of a large block of impure copper. The cathode is a
thin sheet of pure copper. The electrolyte is copper (II) sulphate
solution.During the refining process, the copper atoms of the impure
block become ions (the anode dissolves).Cu → Cu2+ + 2e–
is how the electrolytic purification (refining) process is carried
out:The anode is made of a large block of impure copper. The cathode is a
thin sheet of pure copper. The electrolyte is copper (II) sulphate
solution.During the refining process, the copper atoms of the impure
block become ions (the anode dissolves).Cu → Cu2+ + 2e–
The ions from the solution become atoms.
Cu2+ + 2e– → Cu(s)
They
stick onto the cathode. A layer of pure copper builds up on the
cathode. As electrolysis takes place, the cathode gains mass as copper
is deposited on it. As a result, the cathode gets smaller while the
cathode gets bigger as electrolysis proceeds. Eventually the whole
cathode dissolves.
stick onto the cathode. A layer of pure copper builds up on the
cathode. As electrolysis takes place, the cathode gains mass as copper
is deposited on it. As a result, the cathode gets smaller while the
cathode gets bigger as electrolysis proceeds. Eventually the whole
cathode dissolves.
Purification of copper by electrolysis
Only
pure copper sticks to the cathode. Most impurities fall to the bottom
of the electrolytic cell. They form a solid material (anode sludge or slime)
which contains small quantities of precious metals such as silver, gold
and platinum. The precious metals recovered from the slime are purified
and sold.
pure copper sticks to the cathode. Most impurities fall to the bottom
of the electrolytic cell. They form a solid material (anode sludge or slime)
which contains small quantities of precious metals such as silver, gold
and platinum. The precious metals recovered from the slime are purified
and sold.
An Experiment on Electroplating of Metallic Materials
Carry out an experiment on electroplating of metallic materials
Electroplating
is the coating of a metal with a layer of another metal by means of
electrolysis. Electrolysis can be used to coat a thin layer of a less
reactive metal onto a more reactive metal. The thin layer of less
reactive metal will provide protection from corrosion for the more
reactive metal underneath. It may also make the product more attractive.
is the coating of a metal with a layer of another metal by means of
electrolysis. Electrolysis can be used to coat a thin layer of a less
reactive metal onto a more reactive metal. The thin layer of less
reactive metal will provide protection from corrosion for the more
reactive metal underneath. It may also make the product more attractive.
The
object to be coated should be made the cathode and the coating material
should be the electrolyte. The most commonly used metals for
electroplating are copper, chromium, silver and tin.
object to be coated should be made the cathode and the coating material
should be the electrolyte. The most commonly used metals for
electroplating are copper, chromium, silver and tin.
Steel
can be electroplated with chromium or tin. This prevents the steel from
rusting and gives it a shiny, silver finish. This is also the idea
behind chromium-plating articles such as car bumpers, kettles, bath
taps, etc. Chromium does not corrode, it is a hard metal that resists
scratching and wear, and can also be polished to give an attractive
finish.
can be electroplated with chromium or tin. This prevents the steel from
rusting and gives it a shiny, silver finish. This is also the idea
behind chromium-plating articles such as car bumpers, kettles, bath
taps, etc. Chromium does not corrode, it is a hard metal that resists
scratching and wear, and can also be polished to give an attractive
finish.
Nickel
can be electroplated with silver. This will make nickel more
attractive.The diagram below shows how a steel jug is electroplated with
silver. The jug becomes the cathode of an electrolytic cell. The anode
is made of silver. The electrolyte is a solution of a silver compound,
for example silver nitrate.
can be electroplated with silver. This will make nickel more
attractive.The diagram below shows how a steel jug is electroplated with
silver. The jug becomes the cathode of an electrolytic cell. The anode
is made of silver. The electrolyte is a solution of a silver compound,
for example silver nitrate.
Silverplating a steel jug
At the anode: The silver dissolves, forming ions in solution:Ag → Ag+ + e–
At the cathode: The silver ions receive electrons, forming a coat of silver on the jug:Ag+ + e–→Ag (s)
When the layer of silver is thick enough, the jug is removed.In general, to electroplate any object with metal M, the set up is:
- Cathode – object to be electroplated
- Anode – metal M
- Electrolyte – solution of a soluble compound of M
CHEMISTRY FORM THREE ALL TOPICS
CHEMISTRY FORM THREE TOPIC 1 & 2.
CHEMISTRY FORM THREE TOPIC 3 & 4.
CHEMISTRY FORM THREE TOPIC 5 & 6.
CHEMISTRY FORM THREE TOPIC 7.
CHEMISTRY FORM THREE TOPIC 8.
O’LEVEL CHEMISTRY
CHEMISTRY STUDY NOTES, FORM FOUR.
CHEMISTRY STUDY NOTES, FORM THREE.
CHEMISTRY STUDY NOTES, FORM TWO.
CHEMISTRY STUDY NOTES FORM ONE.
