VOLUMETRIC ANALYSIS
Is a chemical procedure of determing the concentration of a solution.Volumetric analysis involves the techinque of titration.
Titration is the process of adding an acid into a base until an indicator show that the reaction is complete.
A solution with a known volume and known concentration called STANDARD SOOUTION.
Standard solution delivered from burette so that the volume addae is measured.
APPLICATION Of VOLUMETRIC ANALYSIS.
- It is used to determine unknown concentration of chemicals.- It is used to quantify the amount of the substance present in a solutions by anlytical
Procedure.
- it help in the standardization of acids and bases.
- used in preparation of standard solutions.
STANDARD SOLUTION.
Is the solution with known volume and known concentration. Standard solution categorized into two groups which are;1) primary standards
Standard solution are usually prepared from primary primary standards.
Standard solution should have the following characteristics
- high degree of purity
- stable which are not decompose with time
- have no water of hydration that is, they should not not be hygroscopic or efflorescent.
- not volatile so that looses due to evaporation do not occur.
- be highly soluable
- have high molecular mass
Examples of standard primary standards include;
- Sodium carbonate , for acid - base titration.
- Potassium dichromate for redox titrations.
2) Secondary standards
Some of acid and base such as hydrochloric acid, sulphuric acid and sodium hydroxide are secondary standards because they absorb water and vepourize during storage or prearation.
Their concentration in solution can be determined by standerdizing them with primary standards.
CONCENTRATION; is the amount of substance per 1 dm3 (L) of the solution.
- it is expressed in g/dm3 or mol/dm3.
Thus;. concentration = mass(gram)/ volume
CONCENTRATION = Molarity x Molar mass in g/dm3.
MOLARITY;
Is the amount of substance in moles per 1 dm3 of the solution
- The SI unit of Molarity is moles/ dm3
- The moles/dm3 is denoted by M.
INDICATOR
Indicators are substances which shows the end-point of the reactions.
OR
Indicators are substances that show definite colour change when they are in acid or base.
YTPES OF INDICATORS
There mainly four types of indicator as shown below.
1) ✓litmus paper; change red to blue in base.
✓litmus paper; . Change blue to red in acid.
2) ✓methyl orange; change to yellow or orange in base.
✓methyl orange; Change to pink in acid .
3)✓ phenolphthalein; change to pink in base.
✓phenolphthalen ; Change to colourless in acid
4) bromothymol blue. change to bule in base.
bromothymol blue; Change to yellow in acid
Look on graph below.
Choise for an indicator in acid-base base titration.
1) for strong acid and base you can use any indicator.
2) for weak acid and strong base indicator used is phenolphthalein indicator.
3) strong acid and weak base suitable indicator is methyl orange (m o) indicator.
4) weak acid and weak base, no suitable indicator.
ACID-BASE TITRATION
Titration is the process of adding an acid to a base until an indicator shows that the reaction is complete.
COndition necessary for a titration.
1) The reaction must be fast, so that the reaction can be performed at conviniet time.
2) The reaction should go to completion means that it should not be reversible.
3) The reaction should be free from side rractions, means that it can be represented by single chemical equaion.
4) The reaction should has a definite end point that can be determined by accuratelly.
MaVa/MbVb = na/nb
Where;
Ma is molar concentration of the acid
Va is volume of the acid used
na is the number of moles of the acid used
Mb is molar concentration of the base
Vb is volume of the base used
nb is the moles of the base used.
The table below shows sample results for the experiment during titration
Pipette used 25cm3.
NaOH + HCL > NaCL + H2O
The average volume of hydrochloric acid used is
( 23.85 + 23.80 + 23.80)/3 = 23.82cm3
Volume of acid used (Va) = 23.82cm3
Molarity of acid (Ma) = 0.1M
Namber of moles of acid (na) = 1
Volume of base used (Vb) = 25cm3
Molarity of base ( Mb). = ?
Number of moles of base (nb) = 1
Now from the formulas;
MaVa/MbVb = na/nb
Mb = MaVanb/Vbna
Mb = (0.1M x 23.82x1)/25cm3x1
Mb = 0.09528M.
Molarity of sodium hydroxide (NaOH) = 0.09528M.
There fore 25cm3, of 0.09528M sodium hydroxide is neutralized by 23.82cm3 of the hydrochloric acid.
EXERCISE.
The following data obtained in titration experiment, calculate concentration of sulphuric acid in.
a) g/dm3
b) mol/dm3
WATER OF CRYSTALIZATION.
Is the water that is bounded within the crystals of substances. This water can be removed by heating.-Water of crystals can be determined by titration methods.
Example
Determine the number of molecules of water of crystalization in the sample given.
Na2CO3.XH2O. If R.M.M is 286g/mol
Soln
Data given
R.M.M = 286g/mol
Na2CO3.XH2O
To find the value of X.
Then;
Na2CO3.XH2O = 286.
23X2 + 12 + 16X3 + X((2X1) + 16) = 286
106 + 18X = 286
18X = 286 - 106
18X/18 = 180/18
X = 10
There for;
Water of crystalization = 10.
Example: 2;
Calculate the number of molecules of water of crystalization in the hydrated sodium carbonate.
To determine the number of water of crystalization in the hydrated sodium carbonate sample with concentration of 14.3g/dm3 proceed as follow;
1) calculate the average volume of the hydrochloric acid used.
(23.85 + 23.80 + 23.80)/3 = 23.82 cm3
calculate the number of moles of HCl
Moles(n) = Volume x molarity
n =( 23.82/1000)dm3 x 0.1mol/dm3
n = 0.002382moles
3) write a balanced chemical equation to get mole ratio
Na2CO3 + 2HCl > 2NaCl + H2O + CO2
The ratio of Na2CO3 to HCl is 1:2
4) calculate the number of moles of sodium carbonate
0.002382/2 = 0.001191moles
5) Determine the mole concentration of sodium carbonate
Molarity = moles/volume
= 0.001191mole/0.02382
0.05M
6) calculate the R.M.M of the carbonate.
Molar mass = mass per litre/molarity
Nolar mass = 14.3g/dm3/0.05M
Molar mass=286g
7) Determine the number of molecules of water of crystalization in the sample.
Na2CO3.xH2O = 286g
23 x 2+12 + 16x3 +x((2x1) + 16) = 286g
58 + 48 + 18x = 286g
106g + 18x = 286g
18X = 286 - 106
18x = 180
18x/18 = 180/18
X = 10.
Percentage purity of a substance.
Is the percentage of the pure substance in a samplePercentage purity = ((Conc of pure substance)/(Conc of impure)) x 100%
PERCENTAGE IMPURITY OF A SUBSTANCE
Is the percentage of the impure substance in a sample.Percentage impurity =( 100% - %purity)
EXAMPLE.
25cm3 of 2M nitric acid reacted completely with 20cm3 of a solution containing 56.25g of impure sodium carbonate in 250cm3 of solution. Determine the percentage purity of the sodium carbonate
( N = 14, C = 12, Na = 23, O = 16, H = 1)
Soln
Data given
Volume of HNO3 (Va) = 25cm3
Molarity of HNO3 (Ma) = 2M
Volume of Na2CO3 (Vb) = 20cm3
Molarity of Na2CO3(Mb) = ?
Mass of impure Na2CO3 (m) = 56.25g in 250cm3
Reaction eqution
2HNO2 + Na2CO3 > 2NaNO3 + H2O + CO3
From the chemical equation
nb = 1
na = 2
Now;
From the molar ratio formula
((MaVa/na) = (MbVb/nb))
Mb = ((MaVanb/Vbna))
= ((2M x 25cm3 1) / (20cm3x20))
= 1.25M
Therefore;
Molarity of pure Na2CO3 is 1.25M
Now;
Conc of pure Na2CO3 = Molarity of pure Na2CO3 x Molar mass of Na2CO3.
= 1.25M x 106g/mol
= 132.5g/dm3
There fore;
Conc of pure Na2CO3 is 132.5g/mole
Conc of impure Na2CO3 = Mass/Volume
= 56.25g/0.25dm3.
= 225g/dm3
Now;
To calculate percentage purity of Na2CO3
% purity = (conc of pure Na2CO3/Conc of impure of Na2CO3) x 100%
= 132.5/225
= 58.9%
% purity of Na2CO3 is 58.9%
% of impurity = 100% - % purity
There fore;
% of impurity = 100% - 58.9%
= 41.1%
% of purity is 41.1%
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