mole2
Short Description
nmnm...
Description
CHEMISTRY
Mole Concept-2 Oxidation & Reduction Let us do a comparative study of oxidation and reduction : Oxidation 1. Addition of Oxygen e.g. 2Mg + O2 2MgO
Reduction 1. Removal of Oxygen e.g. CuO + C Cu + CO
2. Removal of Hydrogen e.g. H2S + Cl2 2HCl + S
2. Addition of Hydrogen e.g. S + H2 H2S
3. Increase in positive charge e.g. Fe2+ Fe3+ + e–
3. Decrease in positive charge e.g. Fe3+ + e– Fe2+
4. Increase in oxidation number (+2) (+4) e.g. SnCl2 SnCl4
4. Decrease in oxidation number (+7) (+2) – e.g. MnO4 Mn2+
5. Removal of electron e.g. Sn2+ Sn4+ + 2e–
5. Addition of electron e.g. Fe3+ + e– Fe2+
Oxidation Number
It is an imaginary or apparent charge developed over atom of an element when it goes from its elemental free state to combined state in molecules.
It is calculated on basis of an arbitrary set of rules.
It is a relative charge in a particular bonded state.
In order to keep track of electron-shifts in chemical reactions involving formation of compounds, a more practical method of using oxidation number has been developed.
In this method, it is always assumed that there is a complete transfer of electron from a less electronegative atom to a more electronegative atom.
Rules governing oxidation number The following rules are helpful in calculating oxidation number of the elements in their different compounds. It is to be remembered that the basis of these rule is the electronegativity of the element .
Fluorine atom : Fluorine is most electronegative atom (known). It always has oxidation number equal to –1 in all its compounds
Oxygen atom : In general and as well as in its oxides , oxygen atom has oxidation number equal to –2.
In case of
(i) peroxide (e.g. H2O2, , Na2O2 ) is –1, (ii) super oxide (e.g. KO2) is –1/2 (iii) ozonide (e.g. KO3) is –1/3 (iv) in OF2 is + 2 & in O2F2 is +1
Hydrogen atom : In general, H atom has oxidation number equal to +1. But in metallic hydrides ( e.g. NaH, KH), it is –1.
Halogen atom : In general, all halogen atoms (Cl, Br , I) have oxidation number equal to –1. But if halogen atom is attached with a more electronegative atom than halogen atom, then it will show positive oxidation numbers. 5
5
7
5
e.g. K ClO , HI O , HCIO , KBrO 3 3 4 3
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CHEMISTRY
Metals : (a) Alkali metal (Li , Na, K, Rb, .......) always have oxidation number +1 (b) Alkaline earth metal (Be , Mg , Ca .......) always have oxidation number +2. (c) Aluminium always has +3 oxidation number
Note : Metal may have negative or zero oxidation number Oxidation number of an element in free state or in allotropic forms is always zero 0
0
0
0
e.g. O , S , P , O 2 8 4 3
Sum of the oxidation numbers of atoms of all elements in a molecule is zero. Sum of the oxidation numbers of atoms of all elements in an ion is equal to the charge on the ion . If the group number of an element in modern periodic table is n, then its oxidation number may vary from (n – 10) to (n – 18) (but it is mainly applicable for p-block elements ) e.g. N- atom belongs to 15th group in the periodic table, therefore as per rule, its oxidation number may vary from 3
2
3
4
5
–3 to +5 ( N H3 ,NO , N 2 O3 , N O2 , N 2 O5 )
The maximum possible oxidation number of any element in a compound is never more than the number of electrons in valence shell.(but it is mainly applicable for p-block elements )
Calculation of average oxidation number : Example-1
Calculate oxidation number of underlined element : (a) Na2 S2O3 (b) Na2 S 4O6
Solution. (a)
Let oxidation number of S-atom is x. Now work accordingly with the rules given before . (+1) × 2 + (x) × 2 + (–2) ×3 =0 x=+2
(b)
Let oxidation number of S-atom is x (+1) × 2 + (x) × 4 + (–2) × 6 = 0 x = + 2.5 It is important to note here that Na2S2O3 have two S-atoms and there are four S-atom in Na2S4O6. However none of the sulphur atoms in both the compounds have + 2 or + 2.5 oxidation number, it is the average of oxidation number, which reside on each sulphur atom. Therefore, we should work to calculate the individual oxidation number of each sulphur atom in these compounds.
Calculation of individual oxidation number It is important to note that to calculate individual oxidation number of the element in its compound one should know the structure of the compound and use the following guidelines. Formula : Oxidation Number = Number of electrons in the valence shell – Number of electrons taken up after bonding Guidelines : It is based on electronegativity of elements. 1.
If there is a bond between similar type of atom and each atom has same type of hybridisation, then bonded pair electrons are equally shared by each element.
Example : Calculate oxidation number of each Cl-atom in Cl2 molecule
Structure :
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CHEMISTRY
Number of electrons in the valence shell = 7. Number of electrons taken up after bonding = 7. oxidation number = 7 – 7 = 0. 2.
:
:
similarly, oxidation number = 7 – 7 = 0
If there is a bond between different type of atoms : e.g. A – B (if B is more electronegative than A) Then after bonding, bonded pair of electrons are counted with B - atom . Example : Calculate oxidation number of each atom in HCl molecule
Structure :
Note : Electron of H-atom is now counted with Cl-atom, because Cl-atom is more electronegative than H-atom H : Number of electrons in the valence shell = 1 Number of electrons taken up after bonding = 0 Oxidation number of H = 1 – 0 = + 1 Cl : Number of electrons in the valence shell = 7 Number of electrons taken up after bonding = 8 Oxidation number of Cl = 7– 8 = – 1 Example-2
Calculate individual oxidation number of each S-atom in Na2S2O3 (sodium thiosulphate) with the help of its structure .
Solution.
Structure :
Note : (central S-atom) is sp 3 hybridised (25% s-character) and (terminal S-atom) is sp 2 hydbridised (33% s-character). Therefore, terminal sulphur atom is more electronegative than central sulphur atom. Now, the shared pair of electrons are counted with terminal Satom.
, S-atom : Number of electrons in the valence shell = 6 Number of electrons left after bonding = 0 Oxidation number of central S-atom = 6 – 0 = + 6 , S-atom : Number of electrons in the valence shell = 6 Number of electrons left after bonding = 8 Oxidation number of terminal S-atom = 6 – 8 = – 2
Now, you can also calculate Average Oxidation number of S =
6 ( 2 ) = + 2 (as we have calculated before) 2
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CHEMISTRY Miscellaneous Examples : In order to determine the exact or individual oxidation number we need to take help from the structures of the molecules. Some special cases are discussed as follows:
O
The structure of CrO5 is
O
Cr || O O O
From the structure, it is evident that in CrO5 there are two peroxide linkages and one double bond. The contribution of each peroxide linkage is –2. Let the oxidation number of Cr is x. x + (–2)2 + (–2) = 0 or x = 6 Oxidation number of Cr = + 6 Ans
O
The structure of H2SO5 is H
O
O
S
H
O
O
From the structure, it is evident that in H2SO5, there is one peroxide linkage, two sulphur-oxygen double bonds and one OH group. Let the oxidation number of S = x. (+ 1) + (– 2) + x + (–2) 2+ (–2) + 1 = 0 or x+2–8=0 or x–6=0 or x=6 Oxidation number of S in H2SO5 is + 6 Ans. Paradox of fractional oxidation number Fractional oxidation number is the average of oxidation state of all atoms of element under examination and the structural parameters reveal that the atoms of element for whom fractional oxidation state is realised a actually present in different oxidation states. Structure of the species C3O2, Br3O8 and S4O62– reveal the following bonding situations :
The element marked with asterisk (*) in each species is exhibiting different oxidation number from rest of the atoms of the same element in each of the species. This reveals that in C3O2, two carbon atoms are present in +2 oxidation state each whereas the third one is present in zero oxidation state and the average is + 4/3. However, the realistic picture is +2 for two terminal carbons and zero for the middle carbon. 2
0
2
O C C* C O Structure of C3O2 (Carbon suboxide) Likewise in Br3O8, each of the two terminal bromine atoms are present in +6 oxidation state and the middle bromine is present in +4 oxidation state. Once again the average, that is different from reality, is + 16/3.
In the same fashion, in the species S4O62–, average oxidation number of S is + 2.5, whereas the reality being +5,0,0 and +5 oxidation number respectively for respective sulphur atoms.
In general, the conclusion is that the idea of fractional oxidation state should be taken with care and the reality is revealed by the structures only.
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CHEMISTRY Oxidising and reducing agent Oxidising agent or Oxidant : Oxidising agents are those compounds which can oxidise others and reduce itself during the chemical reaction. Those reagents in which for an element, oxidation number decreases or which undergoes gain of electrons in a redox reaction are termed as oxidants. e.g. KMnO4 , K2Cr2O7 , HNO3, conc.H2SO4 etc are powerful oxidising agents . Reducing agent or Reductant : Reducing agents are those compounds which can reduce other and oxidise itself during the chemical reaction. Those reagents in which for an element, oxidation number increases or which undergoes loss of electrons in a redox reaction are termed as reductants. e.g. K , Na2S2O3 etc are the powerful reducing agents. Note : There are some compounds also which can work both as oxidising agent and reducing agent e.g. H2O2, NO2– HOW TO IDENTIFY WHETHER A PARTICULAR SUBSTANCE IS AN OXIDISING OR A REDUCING AGENT
Redox reaction A reaction in which oxidation and reduction simultaneously take place is called a redox reaction In all redox reactions, the total increase in oxidation number must be equal to the total decrease in oxidation number. 5
2
2
3
e.g. 10 Fe SO 4 + 2KMnO4 + 8H2SO4 5 Fe 2 SO 4 3 + 2 Mn SO 4 + K2SO4 + 8H2O
Disproportionation Reaction : A redox reaction in which same element present in a particular compound in a definite oxidation state is oxidized as well as reduced simultaneously is a disproportionation reaction. Disproportionation reactions are a special type of redox reactions. One of the reactants in a disproportionation reaction always contains an element that can exist in at least three oxidation states. The element in the form of reacting substance is in the intermediate oxidation state and both higher and lower oxidation states of that element are formed in the reaction. For example : 1
2
0
2H2O2 (aq) 2H2O () + O 2 (g) 0
2
0
1
2
S8 (s) + 12OH¯(aq) 4S 2 (aq) + 2S 2O 23 (aq) + 6H2O () 1
Cl2 (g) + 2OH¯(aq) ClO (aq) + Cl (aq) + H2O () Consider the following reactions : (a)
2KClO3 2KCl + 3O2 KClO3 plays a role of oxidant and reductant both. Here, Cl present in KClO3 is reduced and O present in KClO3 is oxidized. Since same element is not oxidized and reduced, so it is not a disproportionation reaction, although it looks like one.
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CHEMISTRY (b)
NH4NO2 N2 + 2H2O Nitrogen in this compound has -3 and +3 oxidation number, which is not a definite value. So it is not a disproportionation reaction. It is an example of comproportionation reaction, which is a class of redox reaction in which an element from two different oxidation state gets converted into a single oxidation state.
(c)
4KClO3 3KClO 4 + KCl
5
7
–1
It is a case of disproportionation reaction and Cl atom is disproportionating. List of some important disproportionation reactions 1.
H2O2 H2O + O2
2.
X2 + OH–(dil.) X¯ + XO¯
3.
X2 + OH–(conc.) X¯ + XO3¯
(X = Cl, Br, I)
F2 does not undergo disproportionation as it is the most electronegative element. F2 + NaOH(dil.) F– + OF2 F2 + NaOH(conc.) F– + O2 4.
(CN)2 + OH– CN– + OCN–
5.
P4 + OH– PH3 + H2PO2¯
6.
S8 + OH– S2– + S2O32–
7.
MnO42– MnO4¯ + MnO2
8.
NH2OH N2O + NH3 NH2OH N2 + NH3
9.
Oxyacids of Phosphorus ( +1, +3 oxidation number) H3PO2 PH3 + H3PO3 H3PO3 PH3 + H3PO4
10.
Oxyacids of Chlorine( Halogens)( +1, +3, +5 Oxidation number) ClO– Cl– + ClO2– ClO2– Cl– + ClO3– ClO3– Cl– + ClO4–
11.
HNO2 NO + HNO3 Reverse of disproportionation is called Comproportionation. In some of the disproportionation reactions, by changing the medium (from acidic to basic or reverse), the reaction goes in backward direction and can be taken as an example of Comproportionation reaction. ¯ + O3¯ + H+ 2 + H2O
Balancing of redox reactions All balanced equations must satisfy two criteria. 1. Atom balance (mass balance ) : There should be the same number of atoms of each kind on reactant and product side. 2.
Charge balance : The sum of actual charges on both sides of the equation must be equal. There are two methods for balancing the redox equations : 1. Oxidation - number change method 2. Ion electron method or half cell method Since First method is not very much fruitful for the balancing of redox reactions, students are advised to use second method (Ion electron method ) to balance the redox reactions
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CHEMISTRY Ion electron method : By this method redox equations are balanced in two different medium. (a) Acidic medium (b) Basic medium Balancing in acidic medium Students are adviced to follow the following steps to balance the redox reactions by Ion electron method in acidic medium Example-3
Balance the following redox reaction : FeSO4 + KMnO4 + H2SO4 Fe2(SO4)3 + MnSO4 + H2O + K2SO4
Solution.
Step– Assign the oxidation number to each element present in the reaction. 2 6 2
1 7 – 2
1 6 2
3
6 2
2 6 2
1 2
Fe S O 4 + K MnO 4 + H 2 S O 4 Fe 2 (S O 4 )3 + Mn S O 4 + H 2 O
Step : Now convert the reaction in Ionic form by eliminating the elements or species, which are not undergoing either oxidation or reduction. 7
Fe2+ + Mn O 4 Fe3+ + Mn2+ Step : Now identify the oxidation / reduction occuring in the reaction
Step V : Spilt the Ionic reaction in two half, one for oxidation and other for reduction. Fe2+ Fe3+ oxidation
Re duction MnO 4 Mn 2
Step V : Balance the atom other than oxygen and hydrogen atom in both half reactions Fe2+
Fe3+
MnO4 – Mn2+
Fe & Mn atoms are balanced on both side. Step V : Now balance O & H atom by H2O & H+ respectively by the following way : For one excess oxygen atom, add one H2O on the other side and two H+ on the same side. Fe2+ Fe3+ (no oxygen atom ) .................(i) + – 2+ 8H + MnO4 Mn + 4H2O ................(ii) Step V : Equation (i) & (ii) are balanced atomwise. Now balance both equations chargewise. To balance the charge, add electrons to the electrically positive side. oxidation Fe2+ ............(1) Fe3+ + e– 2+ Re duction 5e– + 8H+ + MnO4– Mn + 4H2O ............(2)
Step V : The number of electrons gained and lost in each half -reaction are equalised by multiplying both the half reactions with a suitable factor and finally the half reactions are added to give the overall balanced reaction. Here, we multiply equation (1) by 5 and (2) by 1 and add them : Fe2+ Fe3+ + e–
..........(1) × 5
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CHEMISTRY
5e 8H MnO 4 Mn 2 4H2O .........( 2) 1
5Fe 2 8H MnO 4 5Fe 3 Mn 2 4H2O
(Here, at his stage, you will get balanced redox reaction in Ionic form) Step X : Now convert the Ionic reaction into molecular form by adding the elements or species, which are removed in step (2). Now, by some manipulation, you will get : 5 FeSO4 + KMnO4 + 4H2SO4
5 1 Fe2 (SO4)3 + MnSO4 + 4H2O + K SO 2 2 2 4
or
10FeSO4 + 2KMnO4 + 8H2SO4 5Fe2(SO4)3 + 2MnSO4 + 8H2O + K2SO4.
Example-4 Solution.
Balancing in basic medium : In this case, except step VI, all the steps are same. We can understand it by the following example: Balance the following redox reaction in basic medium : ClO– + CrO2– + OH– Cl– + CrO42– + H2O By using upto step V, we will get : 1
Re duction Cl O Cl –
3
6
Oxidation Cr O 2 Cr O 24
Now, students are advised to follow step VI to balance ‘O’ and ‘H’ atom. 2H+ + ClO– Cl– + H2O | 2H2O+ CrO2– CrO42– + 4H+
Now, since we are balancing in basic medium, therefore add as many as OH– on both side of equation as there are H+ ions in the equation. 2OH– + 2H+ + ClO– Cl– + H2O +2OH– Finally you will get H2O + ClO– Cl– + 2OH–
...........(i)
4OH– + 2H2O + CrO2– CrO42– + 4H+ + 4OH– Finally you will get 4OH– + CrO2– CrO42– + 2H2O ........... (ii)
Now see equation (i) and (ii) in which O and H atoms are balanced by OH– and H2O Now from step VIII 2e– + H2O + ClO– Cl– + 2OH– 4OH– + CrO2–
CrO42– + 2H2O + 3e–
............. (i) ×3 ............. (ii) ×2
––––––––––––––––––––––––––––––––––––––––––––––––––– Adding : 3ClO– + 2CrO2– + 2OH– 3Cl– + 2CrO42– + H2O
Concept of equivalents Equivalent mass of element Number of parts by mass of an element which reacts or displaces from a compound 1.008 parts by mass of hydrogen, 8 parts by mass of oxygen and 35.5 parts by mass of chlorine, is known as the equivalent weight of that element. e.g.
2Mg + O 2 2MgO
48g 32g 12g 8g 32 g of O2 reacts with 48 g of Mg
8 g of O2 =
Equivalent weight of Mg = 12
48 8 = 12 g 32
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CHEMISTRY Similarly,
Zn + H2SO4 ZnSO4 + H2 65.5 g 32.75
Equivalent weight of Zn = Al
65.5 = 32.75 g 2
3 Cl AlCl3 2 2
+
3 × 71 g 2 111.5 g chlorine reacts with 27 g of Al. 27 35.5 35.5 chlorine reacts with = 9.0 g of Al 111 .5 27 Equivalent weight of aluminium = = 9.0 3 As we can see from the above examples that equivalent weight is the ratio of atomic weight and a factor (say n-factor or valency factor) which is in above three cases is their respective valencies. 27 g
Equivalent weight (E) : In general,
Eq. wt. (E) =
Atomic weight or Molecular weight valency factor ( v.f )
Mol. wt. M n factor x
mass of species Number of Equivalents = eq. wt. of that species For a solution, Number of equivalents = N1V1, where N is the normality and V is the volume in litres
Equivalent mass is a pure number which, when expressed in gram, is called gram equivalent mass.
The equivalent mass of substance may have different values under different conditions.
There in no hard and fast rule that equivalent weight will be always less than the molecular mass.
Valency factor calculation :
For Elements : Valency factor = valency of the element. For Acids : Valency factor = number of replaceable H+ ions per acid molecule
Example-5
HCl ,
Solution.
{see there are only two replaceable H+ions} 1 2 3 2 (assume 100% dissicoiation)
Valency factor
Eq. wt. (E)
M 1
H2SO4
M 2
H3PO4
M 3
H3PO3
M 2
Replaceable hydrogen atoms are those hydrogen atoms which are attached with the atoms of group VI and group VII i.e. O,S,Se,Te, & F, Cl ,Br ,I.
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CHEMISTRY For Bases : Valency factor = number of replacable OH– ions per base molecule. Example-6 Solution.
NaOH, 1
v .f.
KOH 1
M M 1 1 Bases may be defined as the substances in which OH group is/are directly attached with group I elements (Li,Na, K,Rb,Cs), group II elements (Be, Mg,Ca,Ba ) or group III elements (Al, Ga,In,Tl), transition metals, non-metallic cations like PH4+ , NH4+ etc. Eq. wt.
Acid - base reaction :
In case of acid base reaction, the valence factor is the actual number of H+ or OH– replaced in the reaction. The acid or base may contain more number of replaceble H+ or OH– than actually replaced in reaction. v. f. for base is the number of H+ ion from the acid replaced by each molecule of the base
2NaOH + H2 SO4 Na2 SO4 + 2H2O
Example-7 Solution.
Base Acid Valency factor of base = 1 Here, two molecule of NaOH replaced 2H+ ion from the H2 SO4. Therefore, each molecule of NaOH replaced only one H+ ion of acid, so v.f. = 1.
v. f. for acid is the number of OH– replaced from the base by each molecule of acid
Example-8 Solution.
NaOH + H2SO4 NaHSO4 + H2O Base Acid Valency factor of acid = 1 Here, one of molecule of H2SO4 replaced one OH– from NaOH. Therefore, valency factor for H2SO4 is one
Eq. wt. of H2SO4 =
Mol.wt 1
Salts : (a) In non-reacting condition
Valency factor = Total number of positive charge or negative charge present in the compound.
Example-9 Solution.
Na2 CO3 ,
Fe2(SO4)3
FeSO4.7H2O 2
V.f.
2
2×3 = 6
Eq.wt.
M 2
M 6
M 2
Note : In case of hydrated salt, positive/negative charge of water molecule is not counted. (b) In reacting condition Example-10
Na2 CO3 +
HCl NaHCO3 + NaCl
Base
Acid
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CHEMISTRY Solution.
It is an acid base reaction, therefore valency factor for Na2CO3 is one while in non-reacting condition, it will be two.
(c) Equivalent weight of oxidising / reducing agents in a redox reaction In case of redox change , v.f. = Total change in oxidation number per molecule .
Example-11
KMnO4 + H2O2 Mn2+ + O2
Solution.
Mn in KMnO4 is going from +7 to +2 , so change in oxidation number per molecule of KMnO4 is 5. So the valency factor of KMnO4 is 5 and equivalent weight is
M . 5
Normality : Normality of a solution is defined as the number of equivalents of solute present in one litre (1000 mL) solution. Let V mL of a solution is prepared by dissolving W g of solute of equivalent weight E in water.
Number of equivalents of solute = VmL of solution contain
W E
W equivalents of solute E
W 1000 E V equivalents of solute.
1000 mL solution will contain
Normality (N) =
Normality (N) = Molarity x Valency factor
N × V (in mL) = M × V (in mL) × n or milliequivalents = millimoles × n
Example-12 Solution.
W 1000 E V
Calculate the normality of a solution containing 15.8 g of KMnO 4 in 50 mL acidic solution. Normality (N) =
Here
W 1000 E V
W = 15.8 g ,
V = 50 mL
E=
molar mass of KMnO 4 = 158/5 = 31.6 Valency factor
So, normality = 10 N Example-13 Solution.
Calculate the normality of a solution containing 50 mL of 5 M solution of K2Cr2O7 in acidic medium. Normality (N) = Molarity × valency factor = 5 x 6 = 30 N
Law of Equivalence The law states that one equivalent of an element combine with one equivalent of the other. In a chemical reaction, equivalents and milli equivalents of reactants react in equal amount to give same number of equivalents or milli equivalents of products separately. Accordingly (i) aA + bB mM + nN meq of A = meq of B = meq of M = m.eq. of N (ii) In a compound MxNy meq of MxNy = meq of M = meq of N
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CHEMISTRY Example-14 Solution.
Example-15
Find the number of moles of KMnO4 needed to oxidise one mole Cu2S in acidic medium. The reaction is KMnO4 + Cu2S Mn2+ + Cu2+ + SO2 From law of equivalence, equivalents of Cu2S = equivalents of KMnO4 moles of Cu2S × v.f. = moles of kMnO4 × v.f. 1 × 8 = moles of KMnO4 × 5 moles of KMnO4 = 8/5 ( v.f. of Cu2S = 2 (2 – 1) + 1 (4 – (–2))) = 8 and v.f. of KMnO4 = 1 (7 –2) = 5) The number of moles of oxalate ions oxidized by one mole of MnO4– ion in acidic medium are : 2 3 5 (B) (C) 5 5 2 2– – Equivalents of C2O4 = equivalents of MnO4 x(mole) × 2 = 1 × 5 ( v.f. of C2O42– = 2 (4 – 3) = 2 and v.f. of MnO4– = 1 (7 – 2) = 5).
(A) Solution.
x=
(D)
5 3
5 mole of C2O42– ions. 2
Drawbacks of Equivalent concept
Since equivalent weight of a substance (for example oxidising or reducing agent) may be variable hence it is better to use mole concept. e.g.
Eq.wt of MnO4– = e.g.
5e– + 8H+ + MnO4– Mn2+ + 2H2O Mol. wt. of MnO 4 5
–
3e– + 2H2O + MnO4– MnO2 + 4OH–
Eq.wt of MnO4– =
Mol. wt. of MnO 4 3
Thus, the number of equivalents of MnO4– will be different in the above two cases but number of moles will be same.
Normality of any solution depends on reaction while molarity does not. For example : Consider 0.1mol KMnO4 dissolved in water to make 1L solution. Molarity of this solution is 0.1 M. However, its normality is NOT fixed. It will depend upon the reaction in which KMnO4 participates. e.g. if KMnO4 forms Mn2+, normality = 0.1 x 5 = 0.5 N. This same sample of KMnO4, if employed in a reaction giving MnO2 as product (Mn in +4 state), will have normality 0.1 × 3 = 0.3 N.
The concept of equivalents is handy, but it should be used with care. One must never equate equivalents in a sequence which involves same element in more than two oxidation states. Consider an example, KIO3 reacts with KI to liberate iodine and liberated Iodine is titrated with standard hypo solution. The reactions are : (i) O3– + ¯ 2 (ii) 2 + S2O32– S4O62– + ¯ meq of hypo = meq of I2 = meq of IO3 = meq of I
meq of hypo = meq of IO3. This is wrong. Note that I2 formed by equation (i) has v.f. = 5/3 & in equation (ii) has v.f. = 2. v.f. of I2 in both the equation are different, therefore we cannot equate milli equivalents in sequence. In this type of case, students are advised to use mole concept.
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CHEMISTRY Example-16 Solution.
How many millilitres of 0.02 M KMnO4 solution would be required to exactly titrate 25 mL of 0.2 M Fe(NO3)2 solution in acidic medium ? Method -1 : Mole concept method Starting with 25 mL of 0.2 M Fe2+, we can write : Millimoles of Fe2+ = 25 x 0.2 ........(1) and in volume V (in milliliters) of the KMnO4, Millimoles of MnO4¯ = V (0.02) ........(2) The balanced reaction is : MnO4¯ + 5Fe2+ + 8H+ Mn2+ + 5Fe3+ + 4H2O This requires that at the equivalent point,
m.moles of Fe 2 m. moles of MnO 4– = 5 1
( 25 )(0.2) V(0.02) = (from (1) & (2)) 5 1 V = 50 mL. Method -2 : Equivalent Method : At the equivalence point, milliequivalents of MnO4¯ = milliequivalents of Fe2+ M1 × vf1 × V1 = M2 × vf2 × V2
0.02 × 5 × V1 = 0.2 × 1 × 25
( MnO4– Mn2+ ; v.f. = 5, Fe2+ Fe3+ ; v.f. = 1)
V1 = 50 mL.
Titrations Titration is a procedure for determining the concentration of a solution by allowing a carefully measured volume to react with a standard solution of another substance, whose concentration is known. Standard solution - It is a solution whose concentration is known and is taken in burette. It is also called Titrant. There are two type of titrants : Primary titrants/standard - These reagents can be accurately weighed and their solutions are not to be standardised before use. Ex : Oxalic acid, K2Cr2O7, AgNO3, CuSO4, ferrous ammonium sulphate, hypo etc.
Secondary titrants/standard : These reagents cannot be accurately weighed and their solutions are to be standardised before use. Ex : NaOH, KOH, HCl, H2SO4, 2, KMnO4 etc. Titrate : Solution consisting of substance to be estimated, generally taken in a beaker . Equivalence point : It is the point when number of equivalents of titrant added becomes equal to number of equivalents of titrate. At equivalence point : n1V1M1 = n2V2M2 Indicator : An auxiliary substance added for physical detection of completion of titration at equivalence point. It generally show colour change on completion of titration. Type of Titrations :
Acid-base titrations (to be studided in Ionic equilibrium)
Redox Titrations
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CHEMISTRY Some Common Redox Titrations Table of Redox Titrations : (Excluding Iodometric / Iodimetric titrations) ––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––
Estimation of
By titrating with
Reactions
Relation*between OA and RA
––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––
1.
Fe2+
MnO4¯
Fe2+ Fe3+ + e– MnO4– + 8H+ + 5e– Mn2+ + 4H2O
5Fe2+ MnO4¯ Eq. wt. of Fe2+ = M/1
2.
Fe2+
Cr2O72–
Fe2+ Fe3+ + e– Cr2O72– + 14H+ + 6e– 2Cr3+ + 7H2O
6Fe2+ Cr2O72– Eq.wt. of Cr2O72– = M/6
3.
C2O42–
MnO4¯
C2O42– 2CO2 + 2e– MnO4– + 8H+ + 5e– Mn2+ + 4H2O
5C2O42– 2MnO4¯ Eq. wt. of C2O42– = M/2
4.
H2O2
MnO4¯
H2O2 2H+ + O2 + 2e– MnO4– + 8H+ + 5e– Mn2+ + 4H2O
5H2O2 2MnO4¯ Eq.wt. of H2O2 = M/2
5.
As2O3
MnO4–
As2O3 + 5H2O 2AsO43– + 10H+ + 4e– Eq. wt. of As2O3 = M/4 MnO4– + 8H+ + 5e– Mn2+ + 4H2O
6.
AsO33–
BrO3–
AsO33– + H2O AsO43– + 2H+ + 2e– BrO3– + 6H+ + 6e– Br– + 3H2O
Eq. wt. of AsO33– = M/2 Eq.wt. of BrO3– = M/6
Permanganate Titrations : KMnO4 is generally used as oxidising agent in acidic medium, generally provided by dilute H2SO4 . KMnO4 works as self indicator persistent pink color is indication of end point. Mainly used for estimation of Fe2+ , oxalic acid ,oxalates, H2O2 etc. Example-17
Write the balanced reaction of titration of KMnO4 Vs oxalic acid in presence of H2SO4.
Solution.
Reaction :
2KMnO4 + 3H2SO4 + 5H2C2O4 K2SO4 + 2MnSO4 + 8H2O + 10CO2
Redox Changes :
C23+ 2C4+ +2e
M EH2C 2O 4 2
5e + Mn7+ Mn2+
M EKMnO 4 5
Indicator :
KMnO4 acts as self indicator.
Example-18
Write the balanced reaction of titration of KMnO4 Vs ferrous ammonium sulphate in presence of H2SO4.
Solution.
Reaction :
2KMnO4 + 10[FeSO4(NH4)2SO4. 6H2O] + 8H2SO4 5Fe2(SO4)3 + 10(NH4)2SO4 + K2SO4 + 2MnSO4 + 68H2O
Redox Changes :
Fe2+ Fe3+ + e
M EFeSO 4 1
Mn7+ + 5e Mn2+
M EKMnO 4 5
Indicator : KMnO4 acts as self indicator
Iodometric/Iodimetric Titrations : Compound containing iodine are widely used in titrations. (i) Iodide ions can be oxidised to 2 by suitable oxidising agent 2¯ (aq) 2(s) + 2e¯ (ii) Iodine (V) ions, O3¯ , will oxidise ¯ to 2 O¯ (aq) + 5¯ (aq) + 6H+ (aq) 32(s) + 3H2O ()
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CHEMISTRY (iii) Thiosulphate ions, S2O32– , can reduce iodine to iodide ions. 2S2O (aq) (s) S4 O62– + 2– colourless black colourless Iodometric Titrations (Titration Solution is of Na2S2O3 . 5H2O) ––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––
S.No.
Estimation of
Reaction
Relation between O.A. and R.A.
––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––
1.
2
2 + 2Na2S2O3 2Na + Na2S4O6 or 2 + 2S2O32– 2¯ + S4O62–
2 2 2Na2S2O3 Eq.wt. of Na2S2O3 = M/1
2.
CuSO4
2CuSO4 + 4K 2Cu + 2K2SO4 + 2 or 2Cu2+ + 4¯ 2Cu + 2 white ppt
2CuSO4 2 2 = 2Na2S2O3 Eq.wt.of CuSO4 = M/1
3.
CaOCl2
CaOCl2 + H2O Ca(OH)2 + Cl2 Cl2 + 2K 2KCl + 2 Cl2 + 2¯ 2Cl¯ + 2
CaOCl2 Cl2 2 2 2Na2S2O3 Eq.wt. of CaOCl2 = M/2
4.
MnO2
MnO2 + 4HCl(conc.) MnCl2 + Cl2 + 2H2O MnO2 Cl2 2 2Na2S2O3 Cl2 + 2K 2KCl + 2 Eq.wt. of MnO2 = M/2 or MnO2 + 4H+ + 2Cl¯ Mn2+ + 2H2O + Cl2 Cl2 + 2¯ 2 + 2Cl¯
5.
O3¯
O3¯ + 5¯ + 6H+ 32 + 3H2O
O3¯ 32 6 6Na2S2O3 Eq.wt. of O3¯ = M/6
6.
H2O2
H2O2 + 2¯ + 2H+ 2 + 2H2O
H2O2 2 2 2Na2S2O3 Eq.wt. of H2O2 = M/2
7.
Cl2
Cl2 + 2¯ 2Cl¯ + 2
Cl2 2 2 2Na2S2O3 Eq.wt. of Cl2 = M/2
8.
O3
O3 + 6¯ + 6H+ 32 + 3H2O
O3 32 6 6Na2S2O3 Eq.wt. of O3 = M/6
9.
ClO¯
ClO¯ + 2¯ + 2H+ H2O + Cl¯ + 2
ClO¯ 2 2 2Na2S2O3 Eq.wt. of OCl– = M/2
10.
Cr2O72–
Cr2O72– + 14H+ + 6¯ 32 + 2Cr3+ + 7H2O
Cr2O72– 32 6 Eq.wt. of Cr2O72– = M/6
11.
MnO4–
2MnO4– + 10¯+ 16H+ 2MnO4– + 52 + 8H2O 2MnO4¯ 52 10 Eq.wt. of MnO4¯ = M/5
12.
BrO3–
BrO3– + 6¯ + 6H+ Br– + 32 + 3H2O
BrO3– 32 6 Eq.wt. of BrO3– = M/6
13.
As(V)
H2AsO4 + 2¯+ 3H+ H3AsO3 + H2O + 2
H3AsO4 2 2 Eq.wt. of H3AsO4 = M/2
14.
HNO2
2HNO2 + 2¯ 2 + 2NO + H2O
2HNO2 2 2 Eq.wt. of HNO2 = M/1
15.
HClO
HClO + 2¯ + H+ Cl¯ + 2 + H2O
HClOI22Na2S2O3 Eq.wt. of HClO = M/2
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CHEMISTRY Iodimetric Titrations ––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––
S.No.
Estimation of
Reaction
Relation between O.A. and R.A.
––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––
1.
H2S (in acidic medium) SO32– (in acidic medium)
2.
H2S + 2 S + 2¯ + 2H+ SO32– + 2 + H2O SO42– + 2¯ + 2H+
4.
Sn2+ Sn2+ + 2 Sn4+ + 2¯ (in acidic medium) As(III) (at pH 8) H2AsO3¯ + 2 + H2O HAsO42– + 2¯ + 3H+
5.
N2H4
3.
N2H4 + 22 N2 + 4H+ + 4¯
H2S 2 2 Eq.wt. of H2S = M/2 SO32– 2 2 Eq.wt. of SO32– = M/2 Sn2+ 2 2 Eq.wt. of Sn2+ = M/2 H2AsO3– 2 2 Eq.wt. of H2AsO3¯ = M/2 N2H4 = 22 4 Eq.wt. of N2H4 = M/4
Example-19
The sulphur content of a steel sample is determined by converting it to H2S gas, absorbing the H2S in 10 mL of 0.005 M I2 and then back titrating the excess I2 with 0.002 M Na2S2O3 . If 10 mL Na2 S2O3 is required for the titration, how many milligrams of sulphur are contained in the sample? Reactions : H2S + I2 S + 2I– + 2H+ I2 + 2S2O32– 2I– + S4O62–
Solution.
Used millimoles of I2 = (m.moles of I2 taken initially) – = 0.005 × 10 – 0.002 ×
m. moles of hypo used 2
10 2
= 0.04 = millimoles of H2S weight of sulphur = 0.04 × 10–3 × 32 × 103 mg = 1.28 mg.
Hydrogen peroxide (H2O2) H2O2 can behave both like oxidising and reducing agent in both the mediums (acidic and basic).
Oxidising agent : (H2O2 H2O) (a) Acidic medium : 2e– + 2H+ + H2O2 2H2O v.f. = 2 (b) Basic medium : 2e– + H2O2 2OH– v.f = 2 Reducing agent : (H2O2 O2) (a) Acidic medium : H2O2 O2 + 2H+ + 2e– v.f = 2 (b) Basic medium : 2OH– + H2O2 O2 + 2H2O + 2e– v.f = 2 Note : Valency factor of H2O2 is always equal to 2.
Volume strength of H2O2 : Strength of H2O2 is represented as 10V , 20 V , 30 V etc. 20V H2O2 means one litre of this sample of H2O2 on decomposition gives 20L of O2 gas at STP.
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CHEMISTRY Decomposition of H2O2 is given as : 1 H2O2 H2O + O 2 2 1 1 mole × 22.4 L O2 at STP 2 = 34g = 11.2 L O2 at STP To obtain 11.2 litre O2 at STP, at least 34 g H2O2 must be decomposed. 34 For 20 L O2 , we should decompose atleast ×20 g H2O2 11.2 34 1 L solution of H2O2 contains ×20 g H2O2 11.2
1 L solution of H2O2 contains
Normality of H2O2 =
MH2O2 =
M 34 = 17) 2 2
Volume strength of H 2 O 2 5.6
NH2O2 v.f.
=
NH2O2 2
Volume strength of H 2 O 2 11.2 Strength (in g/L) : Denoted by S Strength = Molarity × Mol. wt = Molarity × 34 Strength = Normality × Eq. weight = Normality × 17 Molarity of H2O2 (M) =
N KMnO4 for complete 12 oxidation. Final the strength of H2O2 solution. [Molar mass of H2O2 = 34]
Example-20
20 mL of H2O2 after acidification with dilute H2SO4 required 30 mL of
Solution.
meq. of KMnO4 = meq. of H2O2
( EH2O 2
20 34 20 × = 5 .6 11.2 17
Normality of H2O2 (N) =
34 20 × equivalents of H2O2 11.2 17
30 ×
1 = 20 × N 12
N =
30 1 = N 12 20 8
strength = N × equivalent mass =
1 × 17 = 2.12 g/L. 8
Hardness of water (Hard water does not give lather with soap) Temporary hardness - due to bicarbonates of Ca & Mg Permanent hardness - due to chlorides & sulphates of Ca & Mg. There are some method by which we can soften the water sample. (a)
By boiling
:
2HCO3– H2O + CO2 + CO32– or
By Slaked lime
:
Ca(HCO3)2 + Ca(OH)2 CaCO3 + 2H2O Ca2+ + CO32– CaCO3
:
CaCl2 + Na2CO3 CaCO3 + 2NaCl
(b)
By Washing Soda
(c) (d)
By ion exchange resins : Na2R + Ca2+ CaR + 2Na+ By adding chelating agents like (PO3–)3 etc.
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CHEMISTRY
Parts Per Million (ppm) When the solute is present in very less amount, then this concentration term is used. It is defined as the number of parts of the solute present in every 1 million parts of the solution. ppm can both be in terms of mass or in terms of moles. If nothing has been specified, we take ppm to be in terms of mass. Hence, a 100 ppm solution means that 100 g of solute is present in every 1000000 g of solution. ppmA =
mass of A 10 6 = mass fraction × 106 Total mass
Measurement of Hardness : Hardness is measured in terms of ppm (parts per million) of CaCO3 or equivalent to it. Hardness in ppm =
Example-21 Solution.
mass of CaCO 3 10 6 Total mass of solution
0.00012% MgSO4 and 0.000111% CaCl2 is present in water. What is the measured hardness of water and millimoles of washing soda required to purify water 1000 L water ? Basis of calculation = 100 g hard water MgSO4 = 0.00012g = CaCl2 = 0.000111g =
0.00012 mole 120
0.000111 mole 111
0.00012 0.000111 mole equivalent moles of CaCO3 = 111 120
0.00012 0.000111 × 100 = 2 × 10–4 g mass of CaCO3 = 111 120
Hardness (in terms of ppm of CaCO3) =
2 10 4 10 6 = 2 ppm 100
CaCl2 + Na2CO3 CaCO3 + 2NaCl NaSO4 + Na2CO3 MgCO3 + Na2SO4
0.00012 0.000111 mole Required Na2CO3 for 100g of water = 111 120
= 2 × 10–6 mole
Required Na2CO3 for 1000 litre water = =
2 10 6 2 10 6 mole 100 100
( d = 1g/mL)
20 mole = 20 m mole 1000
Strength of Oleum : Oleum is SO3 dissolved in 100% H2SO4. Sometimes, oleum is reported as more than 100% by weight, say y% (where y > 100). This means that (y 100) grams of water, when added to 100 g of given oleum sample, will combine with all the free SO3 in the oleum to give 100% sulphuric acid. Hence, weight % of free SO3 in oleum = 80(y 100)/18
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CHEMISTRY Example-22 Solution.
What volume of water is required (in mL) to prepare 1 L of 1 M solution of H2SO4 (density = 1.5g/mL) by using 109% oleum and water only (Take density of pure water = 1 g/mL). 1 mole H2SO4 in 1L solution = 98 g H2SO4 in 1500 g solution = 98 g H2SO4 in 1402 g water. Also, in 109% oleum, 9 g H2O is required to form 109 g pure H2SO4 & so, to prepare 98 g H2SO4, water needed is 9/109 × 98 = 8.09 g. Total water needed = 1402 + 8.09 = 1410.09 g = 1410.09 mL
Calculation of Available Chlorine from a sample of Bleaching Powder : The weight of available Cl2 released from the given sample of bleaching powder on reaction with dilute acids or CO2 is called available chlorine. CaOCl 2 H2 SO 4 CaSO 4 H2 O Cl 2 CaOCl 2 2HCl CaCl 2 H2 O Cl 2
CaOCl2 2CH3 COOH CaCH3COO2 H2O Cl2
CaOCl 2 CO 2 CaCO 3 Cl 2
Method of determination : CaOCl2
+ 2CH3COOH Ca(CH3COO)2 + H2O + Cl2
(Sample of bleaching powder)
Cl2 + 2KI I2 + v.f. = 2
2KCl + I2 Starch as indicator 2Na2S2O3 Na2S4O6 + 2Nal v.f. = 1
End point is indicated by disappearance of blue colour. Let M = Molarity of hypo (Na2S2O3) solution millimoles of Cl2 produced = m.moles of I2 used by hypo =
mass of Cl2 produced
or
M V 2
where
V = vol of hypo solution used in ml.
M V 10 3 × 71 2 = 35.5 × M × V × 10–3 =
35.5 M V 10 3 × 100 W where W = amount of belaching powder taken in g. 3.55 M V % of available Cl2 = W % of available chlorine
=
Example-23
3.55 g sample of bleaching powder suspended in H2O was treated with enough acetic acid and K solution. Iodine thus liberated required 80 mL of 0.2 M hypo for titration. Calculate the % of available chlorine.
Solution.
% of Cl2 =
3.55 0.2 80 = 16% 3.55
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CHEMISTRY MISCELLANEOUS SOLVED PROBLEMS (MSPS) 1.
Calculate individual oxidation number of each S-atom in Na2S4O6 (sodium tetrathionate) with the help of its structure .
Sol.
2. Sol.
Find the average and individual oxidation number of Fe & Pb in Fe3O4 & Pb3O4, which are mixed oxides. (i) Fe3O4 is mixture of FeO & Fe2O3 in 1 : 1 ratio so, individual oxidation number of Fe = +2 & +3 & average oxidation number =
1( 2) 2( 3) = 8/3 3
(ii) Pb3O4 is a mixture of PbO & PbO2 in 2 : 1 molar ratio so, individual oxidation number of Pb are +2 & +4 & average oxidation number of Pb = 3.
2( 2) 1( 4) 8/3 3
Balance the following equations : (a)
H2O2 + MnO4– Mn+2 + O2 (acidic medium)
(b)
Zn + HNO3 (dil) Zn(NO3)2 + H2O + NH4NO3
(c)
Cr3 + KOH + Cl2 K2CrO4 + KO4 + KCl + H2O.
(d)
P2H4 PH3 + P4
(e)
Ca3 (PO4)2 + SiO2 + C CaSiO3 + P4 + CO (a) 6H+ + 5H2O2 + 2MnO4– 2Mn+2 + 5O2 + 8H2O
Ans.
(b) 4Zn + 10HNO3 (dil) 4Zn(NO3)2 + 3H2O + NH4NO3 (c) 2Cr3 + 64KOH + 27Cl2 2K2CrO4 + 6KO4 + 54KCl + 32H2O. (d) 6P2H4 8PH3 + P4 (e) 2Ca3 (PO4)2 + 6SiO2 + 10C 6CaSiO3 + P4 + 10CO 4. Ans. 5. Ans. 6.
Find the valency factor for following acids (i) CH3COOH (ii) NaH2PO4 (iii) H3BO3 (i) 1 (ii) 2 (iii) 1 Find the valency factor for following bases : (i) Ca (OH)2 (ii) CsOH (iii) Al(OH)3 (i) 2 (ii) 1 (iii) 3
Ans.
Find the valence factor for following salts : (i) K2SO4 .Al2(SO4)3.24H2O (ii) CaCO3 (i) 8 (ii) 2
7.
Find the valency factor for following redox reactions : acidic Mn 2
(i)
neutral KMnO 4 MnO 2 alkaline K 2MnO 4
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CHEMISTRY
Ans. 8. Sol.
(ii)
K2Cr2O7 acidic
Cr3+
(iii)
C2O42
CO2
(iv) Fe2+ (i) 5, 3, 1;
(ii) 6 ;
Fe3+ (iii) 2 ; (iv) 1
Calculate the normality of a solution obtained by mixing 50 mL of 5 M solution of K2Cr2O7 and 50 mL of 2 M K2Cr2O7 in acidic medium. v.f. of K2Cr2O7 = 6 so
Nf = =
N1V1 N2 V2 V1 V2
5 6 50 2 6 50 = 21 N 50 50
9.
Calculate the normality of a solution containing 13.4 g of Sodium oxalate in 100 mL Sol.
Sol.
wt. in g / eq. wt Normality = vol of solution in litre Here, eq. wt. of Na2C2O4 = 134/2 = 67 so
10.
N=
13.4 / 67 = 2N 100 / 1000
The number of moles of ferrous oxalate oxidised by one mole of KMnO4 in acidic medium is : 2 5 (B) 5 2 Eq. of FeC2O4 = Eq. of KMnO4 moles of FeC2O4 × 3 = moles of KMnO4 × 5 so, moles of FeC2O4 = 5/3 Ans. (D)
(A) Sol.
11.
3 5
(D)
5 3
How many moles of KMnO4 are needed to oxidise a mixture of 1 mole of each FeSO4 & FeC2O4 in acidic medium ? 4 5
5 3 (C) 4 4 Eq. of KMnO4 = Eq. of FeSO4 + Eq. of FeC2O4 moles of KMnO4 × 5 = moles of FeSO4 × 1 + moles of FeC2O4 × 3 moles of KMnO4 = 4/5 Ans. (A) (A)
Sol.
(C)
(B)
(D)
5 3
12.
A sample of hydrazine sulphate [N2H6SO4] was dissolved in 100 mL water. 10 mL of this solution was treated with excess of FeCl3 Sol. Ferrous ions formed were estimated and it required 20 mL of M/50 KMnO4 solution in acidic medium. Fe3+ + N2H4 N2 + Fe2+ + H+ MnO4¯ + Fe2+ + H+ Mn2+ + Fe3+ + H2O (a) Write the balanced redox reactions. (b) Estimate the amount of hydrazine sulphate in one litre of Sol.
Sol.
(a) Given
4Fe3+ + N2H4 N2 + 4Fe2+ + 4H+
MnO4¯ +5Fe2+ + 8H+ Mn2+ + 5Fe3+ + 4H2O (b) In 10 mL solution, eq. of N2H6SO4 = Eq. of Fe2+ = Eq. of KMnO4 = 20 ×
1 × 5 × 10–3 = 2 × 10–3 50
v.f. of N2H6SO4 = 4 so,
weight of N2H6SO4 in 1 L solution =
2 10 3 1000 × 130 = 6.5 g. 4 10
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CHEMISTRY 13.
Write the balanced redox reaction and calculate the equivalent weight of oxidising agent and reducing agent for titration of K2Cr2O7 Vs Ferrous ammonium sulphate.
Ans.
The reaction :
6[FeSO4(NH4)2 SO4 . 6H2O] + K2Cr2O7 + 7H2SO4 3Fe2(SO4)3 + Cr2(SO4)3 + K2SO4 + 6(NH4)2SO4 + 43H2O
M M Redox changes : EFeSO 4 ; EK 2Cr2O7 1 6
14.
One litre of acidified KMnO 4 solution containing 15.8 g KMnO 4 is decolorized by passing sufficient SO2. If SO2 is produced by FeS2, what is the amount of FeS2 required to give desired SO 2 ?
Ans. Sol.
15 g. v.f. of KMnO4 = 5 & v.f. of SO2 = 2 Now, Eq. of KMnO4 = Eq. of SO2 15.8 = moles of SO2 × 2 158 / 5
15.
Sol.
so, Now,
moles of SO2 = 1/4 applying POAC on S, we get : 2 × mole of FeS2 = 1 × moles of SO2
so,
moles of FeS2 =
1 1 1 4 2 8
so,
weight of FeS2 =
1 × 120 = 15 g. 8
An aqueous solution containing 0.1 g KIO3 (formula weight = 214) and an excess of K was acidified with HCl. The liberated 2 consumed 45 mL of thiosulphate. The molarity of sodium thiosulphate solution is : The reaction involved is : O3– + – + H+ 2 + H2O (A) 0.0623 M (B) 0.0313 M O3– + 5– + H+ 32 + H2O 2Na2S2O3 + 2 2Na + Na2S4O6 Now, So,
(C) 0.126 M
(D) 0.252 M
0.1 214 0.1 Moles of 2 = 3 × 214
Moles of KO3 =
0.1 214
Now,
Moles of Na2S2O3 = 2 × 3 ×
M × VL = 2 × 3 ×
Now,
Molarity of hypo solution = 2 × 3 ×
0.1 214
M×
45 0.1 =2×3× 1000 214
0.1 1000 = 0.0623 M 214 45
Ans. (A)
16.
A fresh H2O2 solution is labelled 11.2 V. This solution has the same concentration as a solution which is : (A) 3.4% (w / w) (B) 3.4% (v / v) (C) 3.4% (w / v) (D) None of these
Sol.
Molarity of H2O2 = Now,
vol. strength 11.2 =1 11.2 11.2
wt. of solute in g %(w/v) = wt. of solution in mL × 100 = Molarity × Mol. wt. of solute × = 1 × 34 ×
1 = 3.4% 10
1 10
Ans. (C)
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CHEMISTRY 17.
Sol.
100 mL each of 1 N H2O2 and 11.2 V H2O2 solution are mixed, then the final solution is equivalent to : (A) 3 M H2O2 solution (B) 0.5 N H2O2 solution (C) 25.5 g/L H2O2 solution (D) 2.55 g/L H2O2 Sol.
Nfinal
= N1V1 N2 V2 V1 V2
11.2 1 100 100 5. 6 = 3/2 = 1.5 N 100 100
Normality 1.5 = 0.75 M v.f . 2 Strength of solution in g/L = Molarity × Mol. wt. = 0.75 × 34 = 25.5 g/L So,
18.
Molarity =
Ans. (C)
Ans.
Calculate the percentage of available chlorine in a sample of 3.55 g of bleaching powder which was dissolved in 100 mL of water. 25 mL of this solution, on treatment with KI and dilute acid, required 20 mL of 0.125 N sodium thiosulphate Sol. 10 %
Sol.
CaOCl2 + H2O Ca(OH)2 + Cl2 Cl2 + 2K 2KCl + 2 2 + 2Na2S2O3 Na2S2O6 + 2Na In 25 mL solution, moles of Na2S2O3 =
20 0.125 × = 25 × 10–4 1000 1
1 × moles of Na2S2O3 2
So,
moles of 2 =
So, So,
1 × 25 × 10–4 = 12.5 × 10–4 2 in 100 mL solution, moles of Cl2 = 4 × 12.5 × 10–4 = 50 × 10–4 –4 weight of Cl2 = 50 × 10 × 71 g =
% of available Cl2 =
50 10 4 71 × 100 = 10% 3.55
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