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. It involves two solutions where one of them is a standard solution ( a solution with a know concentration) while you are required, to standardize the other one by titration.
The other aspect of quantitative analysis is gravimetric analysis,this aspect involves weighing and determining the masses of reactants and products of a chemical reaction The main technique used in Volumetric analysis is the acid – base titration and redox or precipitation reaction.
APPARATUS USED IN VOLUMETRIC ANALYSIS ARE:
The burette, pipette, beaker, flasks, funnel, wash bottle, chemical balance, dropping pipette and retort stand.
Basic volumetric analysis Terms
TITRATION: this is the method used for carrying out volumetric analysis.
CONCENTRATION: - this is the amount of solute in a given volume of the solution.
STANDARD SOLUTION: This is a solution with an accuratly known concentration.
that is, a known amount of solute in a known volume of solution. An example of a standard solution is containing 20grammes of a NaOH in 250cm³ of solution
Iv. MOLAR SOLUTION: It is a solution which contains one mole of solute or the molar mass in 1dm³ of solution.
An example is a solution containing 40grammes (molar mass) of NaOH in 1dm³ of solution.
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 changes according to the pH of the solution.
Indicators are widely used to monitor titrations involving colorless solutions of acids and bases.
INDICATORS AND THEIR COLOURS IN DIFFERENT MEDIA
Indicator Colour in acid Colour in alkaline Colour in neutral solutions
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 purpl
1 2 3 4 5 6 7 8 9 10 11 12 13 14
Acidity increases↼↼↼↼↼↼↼↼↼ ⇀⇀⇀⇀⇀⇀⇀⇀⇀⇀⇀⇀Alkalinity increases
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 strong 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 + Na2CO3 Phenolphthalein 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(3-1) O4(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
I. Standardize unknown solution
II. Calculate molar mass, water of crystallization and solubility.
III. Determine the purity of substances.
IV. Determine the masses of substances dissolved.
V. Faster and more convenient
.
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.
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
Amount of Na2CO3 IN 25.00 CM3 OF B used
Concentration of A in moldm-3
Concentration of A in gdm-3
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: concof 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 CA VA /CB VB = 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
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.]
STOICHIOMETRY OF REACTION: This is the study of the quantitative relationship implied by
chemical reactions. Stoichiometry is the term used to describe calculations involving mass –
volume relationship between atoms in a compound and between molecules and atoms
participating in a chemical reaction.
The stoichiometry of the reaction can also be defined as the mole ratio in which the reactants
combine and the products are formed.
Experiments are designed to determine the stoichiometry of reactions and these include.
Precipitation reaction
Displacement of hydrogen from acid
Displacement of metallic ions
Synthesis of metallic oxides
Reduction of metallic oxides
The titration experiment is used to study the stoichiometry of the reaction of acids with
bases/alkalis.
Mole ratio and mass relationship
Consider the equation;
Mg(s) + 2HCl(aq) MgCl(aq) + H2(g)
No. Of moles 1 2 1 1
Mole ratio 1 : 2 1 : 1
Molar mass 24g 36.5g 95g 2g
Reacting mass 24g 2 x (36.5) 95g 2g
From the relationship above;
Reacting mass = no of moles x molar mass
:- No. of moles = reacting mass ...................................... 1
Molar mass
Other formulae for moles are
No. of moles = no of elementary particles ......................... 2
Avogadro’s number
No. of moles = molar concentration x Volume (in cm3) ........ 3
1000
By combining formulae 1 and 3, an important formulae is obtained;
No. of mole = reacting mass = molar concentration x volume
Molar mass 1000
:- reacting mass = molar concentration x volume
Molar mass 1000
Hence; Molar concentration = reacting mass x 1000 ............................. 4
Molar mass x volume
Equation 4 is very important when calculating the concentration of a solution with volumes other
than 1000cm3 (1dm3)
Example
How many moles are contained in 50g of Magnesium trioxocarbonate (IV)? (Mg=24, C=12,
O=16)
Solution
Molar mass of MgCO3 = 24 + 12 + (3 x 16) = 84g/mol
Reacting mass = 50g
No. of moles = reacting mass/molar mass
= 50/84 = 0.6 moles
Calculate the number of moles of CaCl2 that can be obtained from 30g of CaCO3 in the presence
of excess HCl. [Ca=40, C=12, O=16, H=1, Cl=35.5]
Solution
CaCO3(s) + 2HCl(aq) CaCl2(aq) + H2O(l) + CO2(g)
Mole ratio 1 : 2 1 : 1 : 1
Molar mass of CaCO3 = 100g/mol
Molar mass of CaCl2 = 111g/mol
From the equation;
1 mole of CaCO3 produced 1 mole of CaCl2
100g of CaCO3 produced 111g of CaCl2
:- 30g of CaCO3 will produce 30 x 111 = 33.3g of CaCl2
100
Convert 33.3g of CaCl2 to moles;
No of moles = reacting mass/molar mass
= 33.3/111 = 0.3 moles
What mass of Na2CO3 is needed to prepare 1 dm3 of a 0.3M solution.
Solution
Molar mass of Na2CO3 = 106g/mol
Note: 0.3 M means that 0.3 mole of Na2CO3 is contained in 1dm3(1000cm3) solution.
:- No. of moles = reacting mass/molar mass
0.3 = mass/106
Mass = 0.3 x 106 = 31.8g
Calculations involving gas volume
In the industrial preparation of trioxonitrate (V) acid, ammonia gas is burned in oxygen in the
presence of a catalyst according to the following equation;
4NH3(g) + 5O2(g) 4NO(g) + 6H2O(g)
If 250cm3 of NH3 are burned completely, what volume of (a) oxygen is used up? (b) NO is produced
Solution
(a)
Equation of the reaction;
4NH3(g) + 5O2(g) 4NO(g) + 6H2O(g)
From the equation;
4 moles of NH3 ≡ 5 moles of O2
Thus 4 volume of NH3 ≡ 5 volumes of O2
Therefore 250 volumes of NH3 ≡ 250 x 5 = 312.50 cm3 of O2
4
(b)
From the equation;
4 volumes of NH3 ≡ 4 volumes of NO
:- 250 volumes of NH3 ≡ 250 volumes of NO
= 250cm3
What volume of dry oxygen gas (measured at s.t.p) will be produced from the decomposition of
3.50g potassium trioxochlorate (V)?
Solution
Equation of the reaction;
2KClO3(s) 2KCl(s) + 3O2(g)
From the equation;
2 moles of KClO3 produced 3 moles of O2
2 moles of KClO3 = 2 x 122.5g = 245g
Molar volumes of O2 at s.t.p. = 22.4dm3
:- 3 moles of O2 occupies 3 x 22.4 = 67.2dm3
Hence 245g of KClO3 produced 67.2dm3 of O2
:- 3.50g of KClO3 will produce 3.50 x 67.2 = 0.96dm3 of O2
245
Calculate the volume of carbon (IV) oxide at 350C and 720mmHg pressure which could be
obtained by heating 19g of copper (II) trioxocarbonate (IV) [ CuCO3 = 123.5, CO2 = 44, molar
volume of a gas at s.t.p. = 22.4dm3]
Solution
Equation of the reaction;
CuCO3(s) CuO(s) + CO2(g)
From the equation;
1 mole of CuCO3 ≡ 1 mole of CO2
Amount (in moles) of CuCO3 that reacted = 19/123.5 (i.e reacting mass/molar mass) = 0.15moles
:- 0.15 moles of CuCO3 yields 0.15 moles of CO2
But at s.t.p, 1 mole of CO2 occupies 22.4dm3
:- 0.15 moles of CO2 occupies 0.15 x 22.4 = 3.36dm3
1
Using the general gas equation;
P1V1 = P2V2
T1 T2
Where; P1 = 760mmHg, V1 = 3.36dm3 (i.e. 3360cm3), T1 = 273K
P2 = 720mmHg, V2 = ? T2 = 350C = 308K
V2 = P1V1T2 = 760 x 3360 x 308
T1P2 273 x 720
V2 = 4001cm3
Calculations involving liquid volumes
Calculate the volume of 0.1M ammonia which could be obtained by heating 2.7g of ammonium
chloride with excess sodium hydroxide and absorbing all the ammonia evolved.
Solution
Equation of the reaction;
NH4(s) + NaOH(aq) NaCl(aq) + NH3(g) + H2O(l)
From the equation
1 mole of NH4Cl produces 1 mole (22.4dm3) of NH3
Amount of NH4Cl that reacted = 2.7/53.5 (i.e. reacting mass/molar mass) = 0.05moles
:- 0.05 moles of NH4Cl produces 0.05 moles of NH3
Using the formula
Number of moles = molar conc. X volume (cm3)
1000
Where; number of moles = 0.05 moles
Molar conc. = 0.1M
Volume = ?
= 0.05 x volume
1000
Volume of NH3 = 0.1 x 1000 = 500cm3
0.5
Calculate the volume of 2M hydrochloric acid which could be obtained by dissolving 560cm3
hydrogen chloride gas (measured at s.t.p.) in water. [molar volume of gas at s.t.p = 22.4dm3]
Solution
22400cm3 hydrogen chloride gas = 1 mole of hydrogen chloride gas at s.t.p
:- 560 cm3 of hydrogen chloride gas = 560 x 1 = 0.025moles
22400
The number of moles of hydrogen chloride gas = 0.025 moles
The concentration of hydrochloric acid formed is = 2M
Volume of the acid formed = ?
Number of moles = molar conc. X volume (cm3)
1000
0.025 = 2 x volume
1000
Volume = 0.025 x 1000 = 12.5cm3
2
Calculation involving masses
Calculate the mass of carbon(IV) oxide produced by burning 104g of ethyne ( C=12, O=16, H=1)
Solution
Equation of the reaction
2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(g)
From the equation;
moles of ethyne produces 4 moles of CO2
i.e. 2 x 26g of ethyne produces 4 x 44g of CO2
52g of ethyne produces 176g of CO2
:- 104g of ethyne produces 104 x 176 = 352g of CO2
52
What mass of lead (II) trioxonitrate (V) would be required to yield 12g of lead (II) chloride on the
addition of excess sodium chloride solution? (Pb = 207, N=14, O=16, Na=23, Cl=35.5)
Solution
Equation of the reaction
Pb(NO3)2(aq) + 2NaCl(aq) PbCl(s) + 2NaNO3(aq)
331g 278g
From the equation
278g of PbCl is produced from 331g of Pb(NO3)2
:- 12g of PbCl will be produced 12 x 331 = 14.29g of Pb(NO3)2
278
Assignment
Answer question 6 (theory) on page 145 of Essential Chemistry
Tutorial questions
In a certain reaction, 15.0g of impure magnesium sample reacted with excess hydrochloric acid
liberating 8.6 dm3 of hydrogen gas at s.t.p.
Write a balanced equation for the reaction
Calculate the:
mass of pure magnesium in the sample
Percentage purity of the magnesium sample
Number of chloride ions produced in the reaction.
[ Mg = 24.0, volume at s.t.p. = 22.4dm3, Avogadro’s constant = 6.02 x 1023 mol-1]
What is the volume of 0.25moldm-3 solution of KOH that would yield 6.5g of solid KOH on
evaporation? [ K = 39.0, O = 16.0, H = 1.00]
Consider the reaction represented by the following equation:
Na2CO3(aq) + 2HCl(aq) 2NaCl(aq) + H2O(l) + CO2(g)
What volume of 0.02 moldm-3 Na2CO3(aq) would be required to completely neutralize 40cm3 of
0.10 moldm-3 HCl(aq)?
4a. Find the volume of oxygen produced by 1 mole of potassium trioxochlorate (V) at s.t.p in the
following reaction.
2KClO3(s) 2KCl(s) + 3O2(g)
b. Find the mass of sodium trioxocarbonate (IV) needed to give 22.4dm3 of carbon (IV) oxide at
s.t.p in the following reaction.
Na2CO3(s) + 2HCl(aq) 2NaCl(aq) + CO2(g) + H2O(l)
[ Na = 23, O = 16, C = 12, molar volume of gas at s.t.p = 22.4dm3]
What mass of anhydrous sodium trioxocarbonate (IV) is present in 500cm3 of 0.1moldm-3 of
the solution? [ Na = 23, C = 12, O = 16]
ACID-BASE REACTION
Determination of the composition of substances either qualitatively or quantitatively is very
important in chemistry. The two approaches to the quantitative analysis are volumetric and
gravimetric approaches. Volumetric analysis is commonly used because it is faster and more
convenient, although less accurate than gravimetric analysis.
TITRATION: it is the method employed in volumetric analysis. The method involves adding a
solution to a known volume of another solution until the chemical reaction between them is
completed. The completion is shown by a colour change in the resulting solution or in an added
indicator. Usually, a standard solution must be used to react with a solution of unknown
concentration. The reacting volumes of the solutions are the used to calculate the unknown
concentration of the solution.
Definition of terms in volumetric analysis.
End point: this is the point at which the neutralization reaction is just complete. The change in the
colour of the indicator is used to monitor the end point.
Indicators: They are weak organic acids or bases which produce different colours in solution
according to the hydrogen ion, H+ concentration in that solution.
Mass concentration: It is the amount in grams of solute present in 1 dm3 (1000cm3) of the
solution. It is expressed in g/dm3. It is calculated with the formula;
Mass conc. = molar concentration x molar mass
Molar concentration (Molarity): It is the amount in moles of solute per dm3 (1000cm3) of solution.
It is expressed in mol/dm3. It is calculated with the formula;
Molarity (Molar conc.) = mass conc
Molar mass
Standard solution: This is a solution of known concentration.
Molar solution: A molar solution of a compound is the one which contains one mole or the molar
mass of the compound in 1dm3 (1000cm3) of the solution.
Volumetric analysis usually involves the titration of;
Acid against base or trioxocarbonate (IV)
Oxidizing agent against reducing agent
One substance against another substance giving a precipitate.
Materials used in acid base titration include;
Weighing bottle
Chemical balance
Pipette
Burette
Retort stand
Filter paper
Funnel
White tile
Standard volumetric flask
Conical flask
CALCULATIONS IN VOLUMETERIC ANALYSIS
Volumetric analysis is used to:
Standardize unknown solution i.e. to get the concentration of a solution whose
concentration is not known
Determine the purity of substances
Calculate the molar mass of a compound
Calculate the molar concentration of a compound
Calculate the mass concentration of a compound
Number of ions/particles present in a solution
Water of crystallization in a solution
Solubility of a compound
The procedures for all these determinations are similar to that of acid-base titration. The following
formula is important in volumetric analysis.
CAVA = na
CBVB nb
Where:
CA = Molarity ( molar concentration) of acid
VA = Volume of acid
na = number of moles of acid
CB = Molarity (molar concentration) of base
VB = Volume of base
nb = number of moles of base
PRACTICAL CLASS
Experiment to determine the concentration of hydrochloric acid using standard sodium hydroxide
solution
Assignment.
What are the precautions observe during acid-base titration?
Calculations on molar solution
Calculate (a) the mass of anhydrous sodium trioxocarbonate (IV) present in 300cm3 of 0.1M
solution (b) the number of Na2CO3 particles present in the solution. (Na = 23, C = 12, O = 16)
Solution
(a)
Molar mass of Na2CO3 = 106g/mol
Molar conc. = 0.1 M
Mass conc. = ?
But mass conc = molar conc. X molar mass
:- mass conc = 0.1 x 106 =10.6g
This means that 10.6g of Na2CO3 is contained in 1000cm3 (1dm3) of 0.1 M Na2CO3 solution
:- in 300cm3 of the solution we have
300 x 10.6
1000
= 3.18g
Alternative method (formula method)
Molar mass of Na2CO3 = 106g/mol
Molar conc. = 0.1 M
Volume = 300cm3
Reacting mass = ?
Molar conc. = reacting mass x 1000
Molar mass x volume
= reacting mass x 1000
106 x 300
Mass = 106 x 300 x 0.1 =3.18g
1000
(b)
Number of moles (n) = number of elementary particles
Avogadro’s number
Number of particles = number of moles x Avogadro’s number
Since reacting mass = 3.18g
Number of moles = reacting mass = 3.18 = 0.03 moles
Molar mass 106
:- number of Na2CO3 particles = 0.03 x 6.02 x 1023 = 0.181 x 1023
Calculate the volume of 0.25M solution of H2SO4 that will contain a mass of 4.5g of the acid.
[H=1, S=32, O=16]
Solution
Molar conc. = 0.25M
Molar mass of H2SO4 = 98g/mol
Mass conc. = molar conc. X molar mass
= 0.25 x 98 = 24.5g/dm3
i.e. 24.5g of H2SO4 is contained in 1000cm3 of the solution
:- 4.5g of H2SO4 will be contained in
4.5 x 1000 = 184cm3
24.5
Alternative method (formula method)
Molar mass ofH2SO4 = 98g/mol
Molar conc. = 0.25 M
Volume = ?
Reacting mass = 4.5g
Molar conc. = mass x 1000
Molar mass x volume
0.25 = 4.5 x 1000
98 x vol
Volume = 4.5 x 1000 = 184 cm3
0.25 x 98
Calculate the volume of hydrogen chloride gas at s.t.p. that would yield 1.2dm3 of 0.15M
aqueous hydrogen chloride solution. (molar volume of all gases at s.t.p. = 22.4dm3)
Solution
Molar conc. Of HCl solution is 0.15M i.e. dm3 of the HCl solution contains 0.15 moles of HCl.
:- 1.2dm3 of the solution will contain
1.2 x 0.15 = 0.18moles of HCl
1
But 1 mole of HCl gas at s.t.p occupies 22.4dm3
:- 0.18 moles at s.t.p will occupy 0.18 x 22.4 = 4.03dm3
1
4.03 dm3 of HCl gas at s.t.p would be required to yield 1.2dm3 of 0.15M aqueous solution.
Alternative method (formula method)
Molar conc = 0.15M
Molar mass of HCl = 36.5g/mol
Volume = 1.2dm3 = 1200cm3
Reacting mass = ?
Molar conc. = mass x 1000
Molar mass x volume
0.15 = mass x 1000
36.5 x 1200
Mass = 0.15 x 36.5 x 1200 = 6.57g
1000
At s.t.p 36.5g of HCl occupies 22.4dm3
:- 6.57g of HCl will occupy 6.57 x 22.4 = 4.03dm3
36.5
Calculations on standardization
A is 0.05 mol/dm3 of H2SO4. B is sodium trioxocarbonate (IV) solution. If 37.5cm3 of the acid was
required to neutralize 25.00cm3 of the trioxocarbonate, calculate;
conc. Of B in mol/dm3
conc of B in g/dm3
volume of CO2 librated at s.t.p. during the titration
(Na = 23, C = 12, O = 16, S = 32)
Equation of the reaction;
H2SO4(aq) + Na2CO3(aq) Na2SO4(aq) + H2O(l) + CO2(g)
Solution
(a)
CA = 0.05mol/dm3 VA = 37.5cm3 na = 1
CB = ? VB = 25cm3 nb = 1
CAVA = na
CBVB nb
0.05 x 37.5 = 1
CB x 25 1
CB = 0.05 x 37.5 x 1 = 0.075mol/dm3
25 x 1
(b)
Mass conc. = molar conc. X molar mass
= 0.075 x 106
= 7.95g/dm3
Note: the molar mass of Na2CO3 is 106g/mol
(c)
From the equation;
mole of Na2CO3 liberated 1 mole of CO2
:- 0.075 moles of Na2CO3 would liberate 0.075 x 1 = 0.075 moles of CO2
1
At s.t.p, 1 mole of CO2 occupies 22.4dm3
:- 0.075 mole will occupy 0.075 x 22.4 = 1.68dm3
1
Hence, 1.68dm3 of CO2 was liberated during the reaction.
Ycm3 of hydrogen chloride gas at s.t.p. were passed into 60cm3 of 0.1moldm3 sodium
trioxocarbonate (IV) solution. The excess trioxocarbonate (IV) was neutralized by 20cm3 of 0.05
mol/dm3 H2SO4. Calculate (a) the mass of excess Na2CO3 in g/dm3 (b) the value of Y
[molar volume of gas at s.t.p = 22.4dm3, Na2CO3 = 106]
Solution
(a)
Equation of the reaction;
H2SO4(aq) + Na2CO3(aq) Na2SO4(aq) + H2O(l) + CO2(g)
From the equation;
Molar (i.e. 1000cm3 of 1mol/dm3) solution of H2SO4 ≡ 1 Molar solu on of Na2CO3
Thus; 1000cm3 1M H2SO4 ≡ 106g in 1000cm3 Na2CO3
1000cm3 0.05M H2SO4 ≡ 106 x 0.05g in 1000cm3 Na2CO3
:- 20cm3 0.05M H2SO4 ≡ 106 x 0.05 x 20g in 1000cm3 Na2CO3
1000
= 0.106g/dm3 Na2CO3
The mass concentration of the excess Na2CO3 which reacted with H2SO4 = 0.106g/dm3
1000cm3 1M Na2CO3 contains 106g Na2CO3
:- 1000cm3 0.1M Na2CO3 contains 10.6g Na2CO3
Hence 60cm3 0.1M Na2CO3 contains 10.6 x 60 g Na2CO3
1000
= 0.636g/dm3
The mass concentration of the original Na2CO3 solution = 0.636g/dm3
:- the mass concentration of Na2CO3 neutralized by Ycm3 HCl gas = 0.636 – 0.106 = 0.53g/dm3
(b)
Equation of reaction
Na2CO3(aq) + 2HCl(aq) 2NaCl(aq) + H2O(l) + CO2(g)
Form the equation
106g of Na2CO3 ≡ 2 x 22.4 dm3 of HCl gas at s.t.p
:- 0.53g of Na2CO3 ≡ 2 x 22400 x 0.53 cm3
106
= 224cm3
A is a dilute tetraoxosulphate (VI) acid. B contains 1.5g of sodium hydroxide per 250cm3 of
solution. 25cm3 of B requires 15.5cm3 of A for complete neutralization. Calculate
Concentration of B in mol/dm3
Concentration of A in mol/dm3
The number of hydrogen ion ions in 1.0dm3 of solution A
Solution
Equation of reaction
H2SO4(aq) + 2NaOH(aq) Na2SO4(aq) + 2H2O(l)
(a)
250cm3 of B contains 1.5g of NaOH
:- 1000cm3 of B will contain 1000 x 1.5g of NaOH
250
= 6.00g/dm3
Molar conc. Of B = mass conc/molar mass
The molar mass of NaOH = 40g/mol
:-Molar Conc. (Molarity) of B = 6/40
= 0.15mol/dm3
(b)
CAVA = na
CBVB nb
CA = ? VA = 15.5cm3 na = 1
CB = 0.15mol/dm3 VB = 25cm3 nb = 2
CA x 15.5 = 1
0.15 x 25 2
CA = 0.15 x 25 x 1 = 0.121mol/dm3
x 15.5
(c)
H2SO4 2H+ + SO4
2-
mole of H2SO4 yield 2 moles of H+
Since the Molarity of H2SO4 = 0.121mol/dm3
The concentration of H+ in the acid = 2 x 0.121 = 0.242mol/dm3
Number of H+ present = number of moles x Avogadro’s constant
= 0.242 x 6.02 x1023
= 1.46 x1023 hydrogen ions.
Assignment
A solution of trioxonitrate (V) acid contained 0.67g in 100cm3. 31.0cm3 of this solution
neutralized 25cm3 of a sodium trioxocarbonate (IV) solution. Calculate the concentration of the
trioxocarbonate (IV) solution. (HNO3 = 63, Na2CO3 = 106)
Tutorial questions
A solution of trioxonitrate (V) acid contained 0.67g in 100cm3. 31.0cm3 of this solution
neutralized 25cm3 of a sodium trioxocarbonate (IV) solution. Calculate the concentration of the
trioxocarbonate (IV) solution. (HNO3 = 63, Na2CO3 = 106)
2.Calculate the mass of pure sodium chloride that will yield enough hydrogen chloride gas to
neutralize 25cm3 of 0.5M potassium trioxocarbonate (IV) solution. ( NaCl = 58.5, HCl = 35.5,
K2CO3 = 138)
Calculations of percentage purity
A piece of limestone, CaCO3 was added to 1dm3 of 0.1mol/dm3 hydrochloric acid. After
effervescence had stopped, 31.25cm3 of the resulting solution required 25cm3 of 0.05mol/dm3
sodium hydroxide for complete neutralization. Calculate the mass of limestone added (CaCO3 =
100, HCl=36.5, NaOH = 40)
Solution
Equation of the reaction
CaCO3(s) + 2HCl(aq) CaCl2(aq) + H2O(l) + CO2(g)
100g 2 x 36.5g
NaOH(aq) + HCl(aq) NaCl(aq) + H2O(l)
Using CAVA = na
CBVB nb
Where CA = ? VA = 31.25cm3 na = 1
CB = 0.05moldm-3 VB = 25cm3 nb = 1
CA x 31.25 = 1
0.05 x 25 1
CA = 0.05 x 25 x 1 = 0.040mol/dm3
31.25 x 1
The mass concentration of the acid that react with NaOH therefore = 0.04 x 36.5 = 1.46g/dm3
But, the original concentration of the acid is 0.1mol/dm3 = 0.1 x 36.5 = 3.65g/dm3
:- the concentration of the acid that reacted with CaCO3 = 3.65 – 1.46 = 2.19g/dm3
From the equation;
CaCO3(s) + 2HCl(aq) CaCl2(aq) + H2O(l) + CO2(g)
100g of CaCO3 ≡ 73gof HCl
:- Yg of CaCO3 ≡ 2.19g of HCl
Y = 100 x 2.19 = 3g
73
The mass of CaCO3 added to 0.1mol/dm3 HCl is 3g
D. Calculations on molar mass, water of crystallization and solubility of substances
28.0cm3 hydrochloric acid of concentration 4.1g/dm3 neutralized 25.0cm3 of an unknown alkali
YOH, whose concentration was 7.0g/dm3. Calculate (a) the molar concentration of YOH (b) the
relative atomic mass of the element Y. Name the element Y if possible. (HCl = 36.5)
Solution
(a)
Equation of the reaction;
HCl(aq) + YOH(aq) YCl(aq) + H2O(l)
Molar conc. Of HCl = Mass conc/molar mass
= 4.1/36.5 = 0.112mol/dm3
CAVA = na
CBVB nb
CB = 0.112 x 28 x 1 = 0.125mol/dm3
25 x 1
:- molar conc. Of YOH = 0.125mol/dm3
(b)
Molar conc. = mass conc/molar mass
:- 0.125 = 7.0
Y + 16 + 1
Y + 17 = 7.0/0.125
Y = 56 – 17 = 39
The relative atomic mass of Y = 39
The element is Potassium and YOH is KOH
Some crystals of washing soda were exposed to the atmosphere for efflorescence to take place.
6.02g of this partly effloresced washing soda, Na2CO3.yH2O were then dissolved in 500cm3 of
water. 25cm3 of this trioxocarbonate (IV) solution required 32.10cm3 of 0.097mol/dm3
hydrochloric acid for complete neutralization. Calculate y. Hence, write the formula of the
effloresced washing soda. (Na = 23, H = 1, C = 12, Cl = 35.5, O = 16)
Solution
Equation of then reaction
Na2CO3.yH2O(s) + 2HCl(aq) 2NaCl(aq) + H2O(l) + CO2(g) + yH2O(l)
CAVA = na
CBVB nb
0.097 x 32.1 = 2
CB x 25 1
CB = 0.097 32.1 = 0.0623mol/dm3
25 x 2
6.02g of Na2CO3.yH2O is contained in 500cm3
:- in 1000cm3 we have 6.02 x 1000 = 12.04g
500
:- the mass concentration of Na2CO3.yH2O is 12.04g/dm3
Molar mass of Na2CO3.yH2O = (106+ 18y)gmol-1
Molar mass = mass conc./molar conc.
106 + 18y = 12.04/0.0623
106 + 18y = 193.26
18y = 193.26 – 106
18y = 87.3
Y = 4.9
Therefore the formula of the effloresced washing soda is Na2CO3.5H2O
A saturated solution of lead (II) trioxonitrate (V) Pb(NO3)2, was prepared at 220C. 27cm3 of this
solution required 46cm3 of sodium chloride, NaCl solution containing 96g/dm3 for complete
precipitation. Find the solubility of lead (II) trioxonitrate (V) at 220C in (a) mol/dm3 (b) g/dm3
(Na=23, Cl=35.5, Pb=207, N=14, O=16)
Solution
Equation of the reaction
2NaCl(aq) + Pb(NO3)2(aq) PbCl2(s) + 2NaNO3(aq)
(a)
Molarity of the NaCl solution = 96/58.5 = 1.641mol/dm3
CAVA = na
CBVB nb
Where A represent NaCl and B represents Pb(NO3)2
CA = 1.641mol/dm3 VA = 46cm3 na = 2
CB = ? VB = 27cm3 nb = 1
x 46 = 2
CB x 27 1
CB = 1.641 x 46 x 1 = 1.398mol/dm3
27 x 2
The solubility of Pb(NO3)2 at 220C = 1.398mol/dm3
(b)
Mass conc = molar conc x molar mass
= 1.398 x 331
= 462.7g/dm3
The solubility of Pb(NO3)2 at 220C = 462.7g/dm3
E. Calculations involving dilution of a solution
What volume of 2 mol/dm3 NaOH solution is required to prepare 100cm3 of a solution of NaOH
with a concentration of 0.2mol/dm3?
Solution
M1V1 = M2V2
Where; M1 = initial concentration = 2mol/dm3
V1 = initial volume = ?
M2 = final concentration = 100cm3
V2 = final Volume =0.2
V1 = M2V2 = 100 x 0.2
M1 2
V1 = 10cm3
Thus if 10cm3 of a 2mol/dm3 NaOH solution is diluted with water to 100cm3, it will produce a
0.2mol/dm3 solution of NaOH.
Assignment
Water is added to 100cm3 of a 0.25 mol/dm3 NaCl solution to make it up to 1.5dm3. Calculate
the concentration of the solution after dilution.
Tutorial questions
2g of a mixture of NaOH and NaCl (as impurity) were dissolved in 500cm3 of water. If 25cm3 of
this solution was neutralized by 21cm3 0.1mol/dm3 hydrochloric acid, calculate the percentage
of the NaCl impurity. (NaOH = 40, HCl = 36.5, NaCl = 58.5)
An excess of a divalent metal M was dissolved in a limited volume of hydrochloric acid. If
576cm3 of hydrogen were liberated at s.t.p, what was the mass of the metal that produced this
volume of hydrogen? (M = 24, H=1, molar volume of gas at s.t.p = 22.4dm3)
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