3RD TERM

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VOLUMETRIC ANALYSIS
INTRODUCTION
Volumetric analysis is an aspect of quantitative analysis which involves the measurement of the volume of reacting solutions in order to find the masses of substances dissolved in them.in other words, two solutions are given; one is a solution of an acid and other a solution of a base .one of them is a standard solution and you are required, by titration to standardize the other.
The other aspect of quantitative analysis is gravimetric analysis, which involves weighing and determining the masses of reactants and products of a chemical reaction.
The most popular aspect of Volumetric analysis is the acid – base (neutralization) reactions. The technique of determination is by titration. Other aspects are redox and precipitation reaction.
APPARATUS USED IN VOLUMETRIC ANALYSIS ARE:
The burette, pipette, beaker, flasks, funnel, wash bottle, chemical balance, dropping pipette and retort stand.
Basic definitions in volumetric analysis
I.TITRATION: It is the technique for carrying out volumetric analysis.
II.CONCENTRATION: The concentration of a solution is the amount of solute in a given volume of the solution.
III.STANDARD SOLUTION:It is a solution containing a known amount of solute in a known volume of solution. An example of a standard solution is a molar solution.
IV.MOLAR SOLUTION: It is a solution which contains one mole of solute in 1dm3solution.
V.BUFFER SOLUTION: It is a solution which resists a change in pH or H+upon further addition of acid or alkali. Buffers are usually made of solutions of a weak acid and its conjugate base or a weak base and its conjugate acid. The weak acid usually has a pH value lower than 7 and is therefore used to buffer systems in the basic region. Examples of buffers solution and their pH ranges over which they are effective are given below:
BUFFER SYSTEM PH RANGE
Ethanoic acid/Sodium ethanoate(CH3COOH/CH3COONA) 4.25-5.25
Trioxocarbonate( iv)(carbonic) acid/Sodium hydrogentrioxocarbonate 5.87-6.87
Ammonia/Ammonium chloride(NH3/NH4Cl) 8.75-9.75
Sodiumtrioxocarbonate/sodiumhydrogentrioxocabonate 9.75-10.75

MOLAR MASS: This is the mass of one mole of a substance expressed in grams. Its unit is gram per mole.
END POINT: This is the point at which the chemical reaction is complete during titration. The end point is detected with the help of an indicator.
INDICATORS: These are weak organic acids or bases whose colors change with pH of the solution. Indicators are widely used in monitoring titration involving colorless solutions of acids and bases.
DIFFERENT INDICATORS AND THEIR COLOURS IN DIFFERENT MEDIA
INDICATOR COLOUR IN ACIDIC MEDIUM COLOUR IN ALKALINE MEDIUM COLOUR AT END POINT
Methyl Orange Pinkish red yellow Orange
phenolphthalein colorless red colorless
litmus Red blue Pink/purple
Methyl red Pink yellow orange
Screened methyl orange Purple(violet) green grey
Bromothymol blue yellow blue green

UNIVERSAL INDICATOR: This is a mixture of indicators used either as solution on a test paper to test the pH value of a solution. The pH scale is shown below:


Red pink yellow green blue indigo (deep blue) violet (bluish purple)
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14
Acidity increases Alkalinity decreases


EFFECT OF WRONG USE OF INDICATOR
The success of a titration exercises depend on the use of the correct indicator. Wrong use of indicator will definitely give wrong result. For instance, let’s consider a case of the titration of a solution of a wrong acid say HCl with that of a weak base say Na2CO3, methyl orange is the suitable indicator, but if phenolphthalein indicator is used instead, the end point will appear when only half of the weak base has been used up. This can then be represented with the following equation.
HCl + Na2CO3phenolphthalein NaHCO3 + NaCl
Indicator




This happened because the phenolphthalein is sensitive to a weak acid such as Na2CO3.
TITRATION EXAMPLE PH RANGE SUITABLE INDICATOR
1.Strongacid vs. strong base(3-11) H2SO4(aq) and KOH(aq) 3.5 – 9.5 Any indicator is suitable.
2.Weak acid vs. strong base(7-11) H2C2O4and NaOH 7.0-9.5 phenolphthalein
3.strong acid vs. Weak base(3-7) HCl(aq)andNH3(aq) or K2CO3(aq) or Na2CO3(aq) or Ca(OH)2(aq) 3.5-7.0 Methyl orange or screened Methyl orange
4 Weak acid vs. Weak base CH3COOH(aq) and NH3(aq) No sharp change No suitable indicator. Or phenol red.

IMPORTANCE OF VOLUMETRIC ANALYSIS
Standardize unknown solution
Calculate molar mass, water of crystallization and solubility.
Determine the purity of substances.
Determine the masses of substances dissolved.
Faster and more convenient
GENERAL PROCEDURES AND PRECUTIONS DURING ACID –BASE TITRATION

STEPS INVOLVES IN FILLING THE BURETTE WITH DILUTE ACIDES
STEP1. Clamp the burette in a vertical position to avoid error due to parallax while taking reading. Rinse it with the given acid solution, alloing part of the acid to pass through the tap and the jet. Then cloe the tap.
STEP2.With the funnel at the top of the burrette, fill the burettewith acid to a desired level ensure that the jet is filled with acid to avoid air bubbles inside it.tightly close the buette tap to avoid leakages. Remove the funnel toavoid droplets of acid from it.Hold a piece of white paper behind the burette so that the acid level is clearly shown and note the reading of the burette.
STEPS INVOLVES IN MEASURING 25.00Cm3 (20.00Cm3) of standard base solution.
STEP1.Collect four conical flasks and wash with distiled water only. Make sure that none of the flasks contain any acid or alkali.
STEP2.Rinse the pipette with the given base solution.
STEP3.suck the base solution into the pipette until the level is about 1cm3 or2cm3above the forefinger. Gently release the pressure until the bottom of the concave meniscus is at the graduation mark,with the eye level aligned.
STEP4.While filing the pipette ensure that the jet lies well inside the base solution to aviod air bubbles in the pipette.

STEP5.Run the base solution into one of the conical flasks.the sides of the flask should not be splased with the solution as the splasings may not be neutralised by acid later in the experiment. To removethe little alikali retained at the tip of the pipette after delivery,do not blow out thepipette or shake out the last drop of the solution but allow the pipette to drain for about 15 seconds and then allowthe tip of the pipette to touch bottom of the flask.
STEP6.Fill the remaing conical flasks as stated above.
(C )STEPS IN INVOLVES IN TITRATION OF THE ALIKALI SOLUTION WITH DILUTED ACID SOLUTION.
STEP1.Add 2 drops of the given indicator to each conical flask to obtain a shape end-point.


STEP2.Place one of the conical flasks on a piece of white paper or white tile beneath the burette to aideasy recognition of colour change at the end point.Run in the acid from the burette addingabout 1.00cm3 at a time.
Shake the conical flask gently during titration to ensure homogeneity of the two solutions but aviod spillage of the reacting solutiopns. Note the level of the acid when the colour of the indicator changes.this first titration is the trial or roug titration.

STEP3.Repeat the above procedures for the base in other three conical flasks.Take burette readings at the lowest concave meniscus for each titration.

These should be taken at the eyes levels to avoid error of parallax.the volmes of the acids used in the second,third and fourth accurate titrations should agree within ±0.20cm3 and should always be recorded to two places of decimals. These make a concordant orconcurrent results to be obtained.Record all your readings as they are taken and do not try to remember them or write on scripts of papers.

TITRATION PRECUTIONS
The burette must be clamped vertically or not tilted.
Wash the burette and pipette with water andrinse with distiled water.
Rinse the burette with acid,the pipette with base (alkali)before putting these solutions into them.
Ensure no air bubbles in the burette or pipette.
Remove the funnel after putting the acid into the burette(if a funnel is used).
The content of the pipette should be allowed to run into the conical flask without blowing air into it.
Use a drop or two (small amount)of indicators.
Read the lower meniscus.
Ensure the tap of the burette is not leaking.
NOTE:Most examination bodies have different ways of assessing students practical skills and report writing procedures. For instance, the following are some of the penalities.
i)use of pencil; -2marks
ii)Alteration(cancillation) on table of titrevalues; -2marks foreach,maximum of -4marks.
iii)No unit or wrong unit on table; -1mark.
iv)Non concordant values(i.e.difference greater than pluse or minus 0.2)for average titre value; -1mark.










STEPS IN RECORDING OBSERVATIONS MADE DURING ACID- BASE TITRATION EXPERIMENT
i.Construct the specimen table:Example
Burette Readings in cm3 Trial or Rough 1st 2nd
Final Burette reading 24.80 24.00 24.00
Initial Burette reading 0.00 0.00 0.00
Volume of Acid used 24.80 24.00 24.00



ii.Determine the average volume of acid used (i.e. average titer value).
Thus:Average Titre:- 24.00 + 24.00 = 24.00cm3
2
Depending on the differences in readings of the titre values. All your readings must have their units, i.e.cm3.
NOTE:This value and that of the table should be recorded to two places of decimal.
Iii. SIGNIFICANT FIGURES
Sudent should learn to round off their calculation numbers to three significant figures,since they are using foure figure table for their calculations.
Significant figures start from non-zero digit and all other digits that follow including zero are considered Significant. As usual the digits, 5,6,7,8,9 are rounded up while 1,2,3,4,are round off.


Examples:
Numbers Rounded off
299.30 299 to 3 S.f
0.0128 0.0128 to 3 S.f
6.0246 6.02 to 3 S.f
NOTE : for the sake of the examination,prepere your table on a sheet of paper and then neatly transfer it in INK to the answer sheet.

EXAMPLE 1
A is a solution of tetraoxosulphate {vi} acid.
B is a solution containing 0.0500 mole of anhydrous Na2CO3 per dm3.
(a)Put A in the burette and titrate 20.00 or 25.00 cm3 portions of B using methyl orange as the indicator. Record the size of your pipette. Tabulate the burette readings, and calculate the average volume of the acid used. (b) From your result and data provided, calculate the
I amount of Na2CO3 IN 25.00 CM3 OF B used
Ii concentration of A in moldm-3
Iii concentration of A in gdm-3
Iv number of hydrogen ions in 1.00dm3 of A. {Avogadro number = 6.02x1023 mol1}
The equation of reaction is: H2SO4{aq} +Na2CO3{aq] Na2SO4 {aq} + H2O +CO2
{H =1,O=16, S=32}






SOLUTION
A Volume of pipette: 25cm-3
Titration results {Hypothetical data}
Burette reading 1cm3 II cm3 IIIcm3

Final 24.75 49.15 25.70

Initial 0.00 24.75 1.35

Volume of acid used 24.75 24.40 24.35


Average volume of acid used from titrations II and III:
(24.40+24.35 )/2 = 48.75/2 = 24.38cm3

NOTES: Only the titre values from titrations I and II can be used in averaging, since they are within ± 0.20cm3 of each other. …Rough of first titre can also be used in averaging, if it is within ± 0.20cm3 of any other titre value, and is not crossed.-----Do not round up 24.38cm3 to 24.40cm3
(b )To calculate the amount of Na2CO3IN 25.00CM3 Given: conc of B = 0.050moldm3 :Volume = 25/1000dm3
Amount = Conc {moldm3 x Volume {dm3} = 0.050x25/1000 = 0.00125mol
To calculate the concentration of A in moldm3: The various titration variables are:
CA = xmoldm3; VA = 24.38CM3; Na = 1, CB = 0.050moldm3 VB = 25CM3: nB =1
Method 1: proportion method {from the first principle}
From the balanced equation of reaction:
1mol of Na2CO3 = 1 mol of H2SO4
. ∴ 0.00125mol Na2CO = 0.00125molH2so4
i.e. 24.38cm3 of A contained 0.00125 mol H2SO4
∴1000cm3 of A contained {0.00125 x 1000} /24.38 mol = 0.0513mol.Hence, concentration of A is 0.0513mol moldm3.
Method 2:
Mathematic formula method
From the data above, it is safe to use the mole ratio expression, in order to calculate the concentration CA of A, which the required variable.
Using CAVA/CBVB = nA/nB
Substituting; CA X 24.38/0.050 X25 =1/1
Making CA the subject of the formula
∴ CA =1 X0.050X25 /24.38 X1 = 0.0513 moldm-3
(iii) To calculate the concentration of A in gdm-3:
Using conc. {gdm-3} x Molar mass {gmol-1}
Concentration of H2SO4, moldm-3 =0.0513moldm-
Molar mass H2SO4, = 2 {1.0} + 32.0 + 4{16.0
= 2.0 +32.0 +64.0 = 98.0gmol-1
Substituting; Mass conc = 0.0513 x98 = 5.0274gdm-3.
=5.03gdm-3 {3 sig fig.}
(Iv).Number of hydrogen ions in 1.00dm3 of A
1 dm3 of A contained 0.0513mol of H2SO4.
H2SO4 ionizes in water completely thus:
H2SO4 (a q) 2H+SO42-(a q)
1mol 2mol
From the equation;
1 mole of H2SO4 produces2 x 0.0513 moles of H+ = 0.103 mol 0f H+
But 1 mole of H+ contains 6.02 x 1023 ions;
Therefore, 0.103 x 6.02 x 1023 ions =6.02 x10 23 ions.{3 sig fig.]


2. DETERMINE OF RELATIVE MOLAR MASS
In the determination of relative molar mass of an acid or bases by titration, the concentration of both acid and base, at least in gram per dm3, will be provided together with the balanced equation of reaction so as to establish the mole ratio
EXAMPLE 1.
A contains 1.6g of HNO3 in 250cm3 of solution
B contains 10gdm-3 of XHCO3
25cm3 portion of B requires an average of 24.90cm3 of A for complete neutralization. Calculatethe
Concentration of acid in moldm-3
Concentration of XHCO3 in B in moldm-3
Mass of XHCO3
Value of X
Equation of the reaction
HNO_3(aq) + XHCO_3(aq) →XNO_3(aq) +CO_2(g) +H_2 O_((l) )
(H=1;C=12;O=16

Solution
Ca =? Mole per dm3 Va = 24.90cm3; na =1
Cb =? Mole per dm3; Vb = 25cm3; nb = 1
To calculate the molar concentration of A.
The given mass concentration of A is
1.6g of HNO3 in 250cm3
i.e. 250cm3 of the solution contained 1.6g of acid
Therefore, 1000cm3 of A contained
(1.0×1000)/250=6.40gdm^(-3)
i.e. the mass concentration of A = 6.40gdm-3
But molar concentration=(mass concentration)/(molar mass)
Molar mass of HNO3 = (1x1) + (14x1) + (16x3) = 1+14+48 =63gmol-1
Molar conc C_a=6.4/63=0.102mol〖dm〗^(-3)
To calculate the molar concentration of B
Mass conc. Of XHCO3 in B = 10.0gdm-3 (given)
Since the value of X in the base XHCO3 is not known, the mole ratio expression must be used in order to find its molar concentration,

(C_a V_a)/n_a =(C_b V_b)/n_b
Substituting
(0.102 ×24.9)/1=(C_b×25)/1
Making Cb the subject of the formula
C_b=(0.102×24.9)/25=0.102moldm^(-3)
To find the mass of X in XHCO3
molar mass=(mass conc)/(molar conc)=(10.0gdm^(-3))/(0.102moldm^(-3) )=98.0gmol^(-1)
To find the value of X in XHCO3
Molar mass of XHCO3 =98.0gmol-1
i.e X +1 + 12 + (16 x 3 ) = 98
X + 1 + 12 + 48 = 98
X + 61 = 98
X = 98-61
Relative atomic mass of X is X = 37 (no unit)
2. DETERMINATION OF MOLE RATIO
In order to determine the mole ratio of acid to base (or base to acid) by titration, the solution of the acid and the base provided for the titration must be known concentrations (standard solution). However, the equation of reaction will not be provided.
SOLVED PROBLEM 1
A is a solution of 0.0500moldm-3 hydrochloric acid
B is 0.0250moldm-3 of a trioxocarbonate (iv) solution
25.00cm 3 portions of B required an average of 24.60cm3 of A for complete neutralization, using methyl orange as the indicator
calculate:
amount of acid in the average volume of A used
amount of trioxocarbonate (IV) in the volume of B used
mole ratio of the acid to trioxocarbonate (IV) solution in the reaction, express your answer as a whole number ratio of one
state whether the PH of the following would be equal to 7, greater than 7 or less than 7
solution A
solution B
titration mixture of A and B before end point
SOLUTION
(i) To calculate the amount of acid in A used
Amount of the acid = molar conc of A x volume in dm3.
= 0.050 x 24.60/1000 mol
= 0.00123mol
(ii) To calculate the amount of trioxocarbonate (iV) in B used
Amount of Base = molar conc. Of B x volume in dm3
= 0.025 x 25/1000 mol.
= 0.000625 mol
(iii) Mole ratio of acid to trioxocarbonate (iV)
Mole ratio of acid to base, A: B = 0.00123: 0.000625
A: B = 0.00123/0.000625 ∶ 0.000625/0.000625
A: B = 2 : 1
(i) PH of A is less than 7
(ii) PH of B is greater than 7
(iii) PH is titration mixture before end point is greater than 7
SOLVED PROBLEM 2
In an acid – base titration, 24.80cm3 of 0.05000moldm-3 of a mineral acid Y neutralized 25.00cm3 of a solution containing 5.83g of Na2 CO3per dm3.
From the information given above calculate:
Amount (in moles) of acid Y used
Amount in moles of Na2CO3 used
Mole ratio of the acid to Na2CO3 I the titration
i. What is the basicity of Y?
ii. Suggest what Y could be. Give reason for your answer.
From your answer, write a balance equation of the reaction
Solution
i. amount ny = conc (moldm-3) x volume (dm3)
= 0.0500 x 24.80/1000=0.00124mol
ii. Amount in mole of Na2CO3 used
first, calculate the concentration of Na2CO3 in moldm-3
Molar mass of Na2CO3 = 2 (23.0) + 12 + 3 (16)
= 46 + 12 + 48 = 106gmol-1
Conc of Na2CO3 =(mass concentration)/(molar mass)=5.83/106=0.0550moldm^(-3)
Amount of Na2CO3 , nz = 0.0550 ×25/1000=0.00138mol
mole ratio of the acid to Na2CO3 in the titration
Mole ratio of the acid to Na2CO3 = ny : nz = 0.00124: 0.00138
n_y:n_z=0.00124/0.00124:0.00138/0.00124
ny: nz = 1: 1

The basicity of Y is Y since one mole of Y requires one mole of Na2CO3 .
ii. Y is H2SO4. Reason: one mole of the acid Y produces two mole of H+.
H_2 SO_4+ Na_2 CO_3→〖Na〗_2 SO_4+H_2 O+CO_2
C=12.0.O=16.0,Na=23.0