REDOX REACTION & EQUIVALENT CONCEPT

Transcription

1 REDX REACTIN & EQUIVALENT CNCEPT XIDATIN & REDUCTIN : Let us do a comparative study of oxidation and Reduction ; xi d a ti o n R e d u c t i o n (1) Addition of oxygen (i) Removal of oxygen e.g. Mg + Mg e.g. Cu + C Cu + C () Removal of Hydrogen (ii) Addition of Hydrogen e.g. H S + Cl HCl + S e.g. S + H H S () Increase in positive charge (iii) Decrease in positive charge e.g. Fe + Fe + + e e.g. Fe + + e Fe + () Increase in oxidation number (iv) Decrease in oxidation number (+) (+) (+7) (+) SnCl SnCl Mn Mn + (5) Removal of electron (v) Addition of electron e.g. Sn + Sn + + e e.g. Fe + + e Fe + xidation Number : It is an imaginary or apparent charge gained by an element when it goes from its elemental free state to combined state in molecules. It is calculated on basis of a arbitrary set of rules. It is a relative charge in a particular bonded state. Rule s governi ng oxidat ion number : The following rules are helpful in calculating oxidation number of the elements in their different compounds. It is remember that the basis of these rule is the electronegativity of the element. Fluorine atom : Fluorine is most electronegativity atom (known). It always has oxidation no. equal to 1 in all its compounds. xygen atom : In general and as well as in its oxides, oxygen atom has oxidation number equal to. In case of : (i) peroxide (e.g. H, Na ) is 1 (ii) Hydrogen atom : super oxide (e.g. K ) is ½ (iii) ozonide (K ) is 1 (iv) oxygen fluoride F is + & in F is +1 In general, H atom has oxidation number equal to +1. But in metallic hydrides (e.g. NaH, KH) it is 1. Halogen atom : e.g. In general, all halogen atom (Cl, Br, I) has oxidation number equal to 1. But if halogen atom is attached with an more electronegative atom than halogen atom then it will show positive oxidation numbers. 5 KCl Hl 5 7 HCl 5 KBr

2 Metals : Note : (a) Alkali metal (Li, Na, K, Rb,...) always have oxidation number +1. (b) Alkaline earth metal (Be, Mg, Ca...) always have oxidation number +. Metal may have negative or zero oxidation number. (c) Aluminium always have + oxidation number xidation number of an element in free state or in allotropic forms is always zero. e.g. e.g ,S,P, 8 Sum of the charges of elements in a molecule is zero. Sum of the charges of all elements in an ions is equal to the charge on the ion. If the group no. of an element in periodic table is n then its oxidation number may vary from n to n 8 (but it is mainly applicable in p-block elements) N-atom belongs to v group in the periodic table therefore as per rule its oxidation number may vary from to +5 5 (NH,N,N,N,N ) 5 Calculation of average oxidation number : Solved Examples : E x. Calculate oxidation number of underlined element Na S : S o l. Let oxidation number of S-atom is x. Now work accordingly with the rules given before. (+1) + (x) + () = 0 x = + E x. Na S 6 : S o l. Let oxidation number of S-atom is x (+1) + (x) + () 6 = 0 x = +.5 It's important to note here that Na S have two S-atom and there are four S-atom in Na S 6 but none sulphur atom in both the compound have + or +.5 oxidation number, it is the average charge (. No.) Which reside on each sulphur atom therefore we should work to calculate the individual oxidation number of each sulphur atom in these compound. E x. Calculate the.s. of all the atoms in the following species : (i) Cl, (ii) N, (iii) N (iv) CCl (v) K Cr and (vi) KMn S o l. (i) In Cl, the net charge on the species is 1 and therefore the sum of the oxidation states of Cl and must be equal to 1. xygen will have an.s. of and if the.s. of Cl is assumed to be 'x' then x should be equal to 1. x is + 1 (ii) N : ( ) + x = 1 (where 'x' is.s. of N) x = + (iii) N : x + ( ) = 1 (where 'x' is.s. of N) x = +5 (iv) In CCl, Cl has an.s. of 1 x + 1 = 0 x = + (where 'x' is.s. of C)

3 (v) K Cr : K has.s. of +1 and has.s. of and let Cr has.s. 'x' then, +1 + x + = 0 x = +6 (vi) KMn : +1 + x + ( ) = 0 x = +7 MISCELLANEUS EXAMPLES : (where x is.s. of Mn). 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 : The structure of Cr 5 is Cr From the structure it is evident that in Cr 5 there are two peroxide linkages and one double bond. The contribution of each peroxide linkage is. Let the.n. of Cr is x. x + () + () = 0 or x = 6.N. of Cr = +6 A n s. The strcuture of H S 5 is H S H From the structure, it is evident that in H S 5. there are one peroxide linkage, two sulphur - oxygen double bond and one H group. Let the.n. of S = x x + () + () + 1 = 0 or x + 8 = 0 or x 6 = 0 or x = 6.N. of S in H S 5 is +6 Ans Paradox of fractional oxidation number : Fractional oxidation state is the average oxidation state of the element under examination and the structural parameters reveal that the element for whom fractional oxidation state is realised is actually present in different oxidation states. Structure of the species C, Br 8 and S 6 reveal the following bonding situations. The element marked with asterisk in each species is exhibiting the different oxidation state (oxidation number) from rest of the atoms of the same element in each of the species. This reveals that in C, two carbon atoms are present in + oxidation state each whereas the third one is present in zero oxidation state and the average is /. However the realistic picture is + for two terminal carbons and zero for the middle carbon. 0 C C* C Structure of C (Carbon suboxide) Likewise in Br 8, each of the two terminal bromine atoms are present in +6 oxidation state and the middle bromine is present in + oxidation state. nce again the average, that is different from reality is 16/ = Br = Br * B = Structure of Br 8 (tribromooctaoxide)

4 In the same fashion, in the species S 6, is.5, whereas the reality being +5,0,0 and +5 oxidation number respectively for each sulphur S S* S* S Structure of S 6 (tetrathionate ion) XIDISING AND REDUCING AGENT : xidising agent or xidant : xidising agents are those compound which can oxidise others and reduced itself during the chemical reaction. Those reagents whose.n. decrease or which gain electrons in a redox reaction are termed as oxidants e.g. KMn, K Cr 7, HN, conc. H S etc, are powerful oxidising agents. Reducing agent or Reductant : Note : e.g. Redusing agents are those compound which can reduce others and oxidise itself during the chemical reaction. Those reagents whose.n. increase or which loses electrons in a redox reaction are termed as reductants. KI, Na S are powerful reducing agents. There are some compounds also which can work both oxidising agent and reducing agent. e.g. H, N HW T IDENTIFY WHETHER A PARTICULAR SUBSTANCE IS AN XIDISING R REDUCING AGENT : Find the.state of the central atom If.S. is=max..s. or (valence electron) If.S. =Minimum.S. If.S. is intermediate b/w max & minimum Its xidizing agent It's a reducing agent It can act both as reducing agent & oxidising agent It can disproportionate as well REDX REACTIN : e.g. A reaction in which oxidation and reduction simultaneously take place. In all redox reactions the total increase in oxidation number must equal the total decrease in oxidation number. 10 Fe S + 7 KMn Dispropor tionations reactions : + 8H S 5 Fe (S ) + Mn S + K S + 8H A redox reaction in which a same element present in a particular compound in a definite oxidation state is oxidized as well as reduced simultaneously is a disproportionation reactions. Disproportionation reactions are a special type of redox reactions. ne of the reactants in a disproportionation reaction always contains an element that can exist i n at least three oxidat ion state s. 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.

5 1 1 H 0 1 (aq) H () + 0 (g) S 8 (s) + 1H (aq) S (aq) + S (aq) + 6H () 0 Cl (g) + H (aq) 1 Cl (aq) + Consider following reactions : 1 Cl (aq) + H () (a) KCl = KCl + KCl plays a role of oxidant and reductant both. Because same element is not oxidised and reduced. Here, Cl present in KCl is reduced and present in KCl is oxidized. So its not a disproportion reaction although it looks like one. (b) NH N N + H Nitrogen in this compound has and + oxidation number so it is not a definite value, so its not a disproportion reaction. Its a example of comproportionation reaction which is a class of redox reaction in which a element from two different oxidation state gets converted into a single oxidation state. (c) KCl KCl + KCl Its a case of disproportionation reaction in which Cl is the atom disproportionating. List of some impor tant dispropor tionation reaction : 1. H H +. X + H (dil.) X + X. X + H (conc.) X + X F does not (can not) undergo dispropor tionation as it is the most electronegative element. F + NaH (dil) F + F F + NaH concentration (dil) F + 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. I + I + H + I + H (acidic) Some examples of redox reactions are : Sn + Hg Hg + Sn R Mn + 5Fe + 8H 5Fe + Mn R H Cr 7 + 6Fe + 1H 6Fe + Cr R Cu + N + 8H N + Cu R H + H If one of the half reactions does not take place, other half will also not take place. We can say oxidation and reduction go side by side.

6 Cl + 6H 5Cl + Cl.N.= R + H In the above reaction, we can see that Cl has been oxidized as well as reduced. Such type of redox reaction is called dispropor tionation reaction. Some More examples : Cu + Disproportionates Cu + Cu + R HCH + H CHH + HC R Cl Cl + Cl R Mn + H Mn + Mn + H R BALANCING F REDX REACTIN : All balanced equations must satisfy two criteria 1. Atom balance (mass balance) : That is there should be the same number of atoms of each kind in reactant and products side.. Charge bala nce : (a) (b) That is the sum of actual charges on both side of the equation must be equal There are two methods for balancing the redox equations xidation - number change method Ion electron method or half cell method ( a ) xidation number change method : (i) (ii) (iii) (iv) (v) This method was given by Jonson. In a balanced redox reaction, total increase in oxidation number must be equal to total decreases in oxidation number. This equivalence provides the basis for balancing redox reactions. The general procedure involves the following steps : Select the atom in oxidising agent whose oxidation number decreases and indicate the gain of electrons. Select the atom in reducing agent whose oxidation number increases and write the loss of electrons. Now cross multiply i.e. multiply oxidising agent by the number of loss of electrons and reducing agent by number of gain of electrons. Balance the number of atoms on both sides whose oxidation numbers change in the reaction. In order to balance oxygen atoms, add H molecules to the side deficient in oxygen. Then balance the number of H atoms by adding H + ions in the hydrogen.

7 E x. Balance the following reaction by the oxidaton number method : S o l. Cu + HN Cu(N ) + N + H Write the oxidation number of all the atoms Cu + HN Cu(N ) + N + H There is change in oxidation number of Cu and N. 0 + Cu Cu(N )...(1) (xidation no. is increased by ) +5 + HN N...() (xidation no. is decreased by 1) To make increase and decrease equal, eq. () is multiplied by. Cu + HN Cu(N ) + N + H Balancing nitrates ions, hydrogen and oxygen, the following equation is obtained. Cu + HN Cu(N ) + N + H This is the balanced equation. E x. Write the skeleton equation for each of the following processes and balance them by ion electron method : (i) (ii) (iii) Permagnet ion oxidizes oxalate ions in acidic medium to carbon dioxide and gets reduced itself to Mn + ions. Bromine and hydrogen peroxide react to give bromate ions and water. Chlorine reacts with base to form chlorate ion, chloride ion and water in acidic medium. S o l. (i) The skeleton equation for the process : Mn + C + H + Mn + + C + H Step (1) : Indicating oxidation number : 7 C Mn Mn C H Step () : Writing oxidation and reduction half reaction : C 7 C Mn Mn + (xidation half) (Reduction half) Step () : Adding electrons to make the difference in.n. C C + e 7 Mn + 5e Mn + Step () : Balancing '' atom by adding H molecules C C + e Mn + 5e Mn + + H Step (5) : Balancing H atom by adding H + ions C C + e Mn + 5e + 8H + Mn + + H Step (6) : Multiply the oxidation half reaction by and reduction half reaction by 5 to equalize the electrons lost and gained and add the two half reactions.

8 [C C + e ] 5 [Mn + 5e + 8H + Mn + + H ] (ii) Mn + 5C + 16H+ 10C + Mn + + 8H The skeleton equation for the given process is Br + H Br + H (in acidic medium) Step (1) : Indicate the oxidation number of each atom 0 1 Br H Br H Thus, Br and changes their oxidation numbers. Step () : Write the oxidation and reduction half reaction. 0 Br B 5 r (xidation half) 1 H H (Reduction half) Step () : Addition of electrons to make up for the difference in.n. 0 Br B 5 r +10e 1 H + e H Step () : Balance '' atoms by adding H molecules Br + 6H Br + 10e H + e H Step (5) : Equalize the electrons by multiplying the reduction half with 5 and add the two half reactions Br + 6H Br + 10e + 1H + [H + e + H + H ] 5 Br + 5H Br + H + H + (iii) The skeleton equation for the given process : Cl + H Cl + Cl + H Step (1) : Indicate the oxidation number of each atom 0 Cl H Cl + Cl H Thus, chlori ne is the only element which undergoe s the change i n oxidat ion number. It decreases its oxidation number from 0 to 1 and also increases its oxidation number from 0 to 5. Step () : Write the oxidation and reduction half reactions 0 Cl Cl 5 (xidation half) 0 Cl Cl (Reduction half) Step () : Add electrons to make up for the difference in.n. 0 Cl Cl e 0 Cl + e Cl

9 Step () : Balance atoms by adding H molecules Cl + 6H (Cl ) + 10e Cl + e Cl Step (5) : Since medium is basic, balance H atoms by adding H molecules to the side falling short of H atoms and equal number of H ions to the other side. Cl + 6H + 1H Cl + 10e + 1H Cl + e Cl Step (6) : Multiply the reduction half reaction by 5 and add two half reactions. Cl + 5H + H Cl + 10e + 1H [Cl + e Cl ] 5 Cl + 5Cl + 1H Cl + 10Cl + 6H or, 6Cl + 1H Cl + 10Cl + 6H or, Cl + 6H Cl + 5Cl + H Ex : Balance the following reaction by the oxidation number method : Mn + Fe + Mn + + Fe + Sol : Write the oxidation number of all the atoms. +7 Mn + Fe + Mn + + Fe + change in oxidation number has occurred in Mn and Fe Mn Mn...(1) (Decrement in oxidation no. by 5) Fe + Fe +...() (Increment in oxidation no. by 1) To make increase and decrease equal, eq. () is multiplied by 5. Mn + 5Fe + Mn + + 5Fe + To balance oxygen, H are added to R.H.S. and to balance hydrogen, 8H + are added to L.H.S. Mn + 5Fe + + 8H + Mn + + 5Fe + + H This is the balanced equation. E x. Balance the following chemical reaction by oxidation number method and write their skeleton equation : (i) (ii) Chloride ions reduce maganese dioxide to manganese (II) ions in acidic medium and get itself oxidized to chlorine gas. The nitrate ions in acidic medium oxidize magnesium to Mg + ions but itself gets reduced to nitrous oxide. S o l. (i) The skeleton equation for the given process is Mn + Cl Mn + + Cl + H Step (1) : Mn + Cl Mn Cl H Step () :.N. decreases by per Mn + Mn + Cl Mn + Cl + H +.N. increases by 1 per Cl

10 Step () : Equalize the increase/decrease in.n. by multiplying Mn by 1 and Cl by Mn + Cl Mn + + Cl + H Step () : Balance other atoms except H and. Here they are all balanced. Step (5) : Balance atoms by adding H molecules to the side falling short of atoms. Mn + Cl Mn + + Cl + H + H Step (6) : Balance H atoms by adding H + ions to the side falling short of H atoms. Mn + Cl + H + Mn + + Cl + H (ii) The skeleton equation for the given process is Mg + N Step (1) : Mg + + N + H 0 5 Mg ( ) N Mg N H Multiply N by to equalize N atoms Step () :.N. increases by per Mg atom 0 + Mg + N Mg + N + H = 8.N. decreases by per N atom +1 Step () : Equalize increase/decrease in.n. by multiplying Mg by and N by 1. Mg + N Mg + + N + H Step () : Balance atoms other than and H Mg + N Mg + + N + H Step (5) : Balance atoms Mg + N Mg + + N + H + H Step (6) : Balance H atoms as is done in acidic medium. Mg + N + 10H + Mg + + N + 5H ( b ) Ion electron met ho d or half cel l met ho d : By this method redox equation are balanced in two different medium (a) Acidic medium (b) Basic medium Solved Examples : Ex : Balancing in acidic medium Students are advised to follow the following steps to balance the redox reactions by ion electron method in acidic medium. Balance the following redox reaction. FeS + KMn + H S Fe (S ) + MnS + H Sol : Step-I assign the oxidation No. to each elements present in the reaction. 6 Fe S KMn H S 6 Fe ( S ) + 6 MnS + 1 H

11 Step-II Now convert the reaction in ionic form by eliminating the elements or species which are not going either oxidation or reduction Fe Mn Fe + + Mn + Step-III Now identify the oxidation / reduction occurring into the reaction. undergoes reduction Fe + Mn Fe + Mn undergoes oxidation Step-IV Split the ionic reaction in two half one for oxidation and other for reduction Fe + oxidation Fe + Mn Re duction Mn + Step-V Balance the atom other than oxygen and hydrogen atom in both half reactions Fe + Fe + Mn Mn + Fe & Mn atom are balanced in both side. Step-VI Now balance & H atom by H & H + respectively by the following way for one excess oxygen atom add one H on the other side and two H + on the same side. Fe + Fe + (no oxygen atom)...(i) 8H + + Mn Mn+ + H...(ii) Step VII Now see equation (i) & (ii) is balanced atomwise. Now balance both equations chargewise to balance the charge add electrons to the electrically positive side. Fe + oxidation Fe + + e...(1) 5e + 8H + + Mn Reduction Mn + + H...() Step VIII The number of electrons gained and lost in each half-reaction are equalised by multiply suitable factor in both the half reaction and finally the half reactions are added to give the over all balanced reaction. Here we multiply equation (i) by 5 and (ii) by one Fe + Fe + + e...(1) 5 5e + 8H + + Mn M + + H...() 1 5Fe + + 8H + + Mn 5Fe+ + Mn + + H (Here at this stage you will get balanced redox reaction in ionic form) Step IX Now convert the ionic reaction in to molecular form by adding the elements or species which are removed instep (II). Now by some manipulation you will get 5 FeS + KMn + H S 5 Fe (S ) + MnS + H or 10 FeS + KMn + 8H S 5Fe (S ) + MnS + 8H + K S Balancing in basic medium : In this case except step VI all the steps are same. We can understand it by following example balance the redox reaction in basic medium Ex : Cl + Cr + H Cl + Cr + H Sol : By using up to step V we will get

12 1 Cl Reduction Cl 6 xidation Cr Cr Now student are advised to follow step VI to balance '' and 'H' atom H + + Cl Cl + H H + Cr Cr + H + Now since we are doing balancing in basic medium therefore add as many as H on both side of equation as there are H + ions in the equation. H + H + + Cl Cl + H + H H + H + Cr Cr + H + + H Finally you will get Finally you will get H + Cl Cl + H...(i) H + Cr Cr + H...(ii) Now see equation (i) and (ii) in which and H atoms are balanced by H Now from step VIII e + H + Cl Cl + H...(i) and H H + Cr Cr + H + e...(ii) Adding : Cl + Cr + H Cl + Cr + H Equivalent weight (E) : Eq. wt (E) = Molecular weight valency factor (v.f) = Mol. wt. n factor no of Equivalents = mass of a sample eq. wt.of that species Equivalent mass is a pure number when expressed in gram, it is called gram equivalent mass. The equivalent mass of substance may have different values under different conditions. Valency factor calculation : For Acids : valence factor = number of replaceable H + Solved Examples : H P Ex : HCl, H S H P H P ions Sol : valence 1 factor (assume 100% Eq.wt. M 1 dissociation) M Self pract ice problems : 1. Find the valence factor for following acids (i) CH CH (ii) NaH P (iii) H B Answers : 1. (i) 1 (ii) (iii) 1 M M {see there are only two replaceable H + ions} H H

13 For Base : v.f. = number of replicable H ions Solved Examples : Ex : NaH, KH Sol : v.f. 1 1 M M E. 1 1 Self pract ice problems : 1. Find the valence factor for following bases (i) Ca(H) (ii) CsH (iii) Al(H) Answers : 1. (i) (ii) 1 (iii) Acid - base reaction : In case of acid base reaction, the valence factor is the actual number of H + or H replaced in the reaction. The acid or base may contain more number of replaceable H + or H than actually replaced in reaction. Solved Examples : v.f. for base is the number of H + ion form the acid replaced by per molecule of the base. Ex : NaH + H S Na S + H Base acid Sol : valency factor of base = 1 Ex : valency factor of acid = Here two molecule of NaH replaced H + ion from the H S therefore per molecule of NaH replaced only one H + ion of acid so v.f. = 1 v.f. for acid is number of H replaced for the base by per molecule of acid NaH + H S NaHS + H Base acid Sol : valence factor of acid = 1 here one of molecule of H S replaced one H from NaH therefore v.f. for H S is = 1 E = For Salts : mol. wt.of HS 1 ( a ) In non reacting condition Solved Examples : v.f. = Total number of positive charge or negative charge present into the compound. Ex : Na C Fe (S ) (Fe + + S ) FeS.7H Sol : V.f. = 6 E. M M 6 M

14 ( b ) Salt in reacting condition : Solved Examples : Ex : Na C + HCl NaHC + NaCl Base Acid Sol : It is a acid base reaction therefore v.f. for Na C is one while in non reaction condition it will be two. Ex : Note : ( c ) Eq. wt. of oxidisi ng / reduci ng agent s i n redox react ion : (a) In acidic medium The equivalent weight of an oxidising agent is that weight which accepts one mole electron in a chemical reaction. Equivalent wt. of an oxidant (get reduced) = Mol.wt. No. of electrons gained by one mole 6e + Cr 7 + 1H + Cr + + 7H Eq. wt. of K Cr 7 = (b) Mol. wt of K Cr 7 Mol. wt. 6 6 [6 in denominator indicates that 6 electrons were gained by Cr 7 as it is clear from the given balanced equation] Similarly equivalent wt. of a reductant (gets oxidised) = Mol. wt. No. of electrons lost by one mole E x : In acidic medium, C C + e Here, Total electrons lost = (c) So, eq. wt. = Mol. wt. In different condition a compound may have different equivalent weights. Because, it depends upon the number of electrons gained or lost by that compound in that reaction. Ex : (i) Mn Mn + (acidic medium) Note : (+7) (+) Here 5 electrons are taken so eq. wt. = (ii) Mn Mn + (neutral medium) (iii) (+7) (+) Here, only electrons are gained, so eq. wt. = Mn Mn (alkaline medium) (+7) (+6) Here, only one electron is gained, so eq. wt. = Mol.wt.of KMn 5 = Mol.wt.of KMn Mol.wt.of KMn 1 = 1.6 = 158 = 158 = 5.7 It is important to note that KMn acts as an oxidant in every medium although with different strength which follows the order. Ex : S S 6 + e (Reducing agent) acidic medium > neutral medium > alkaline medium equivalent weight of S = M = M

15 Questions based on Equivalent weight : 1. Molecular weight of KMn in acidic medium and neutral medium will be respectively : (A) 7 equivalent wt. and equivalent wt. (B) 5 equivalent wt. and equivalent wt. (C) equivalent wt. and 5 equivalent wt. (D) equivalent wt. and equivalent wt.. Equivalent wt. of H P in each of the reaction will be respectively : H P + H H P + H H P + H HP + H H P + H P + H (A) 98, 9,.67 (B) 9, 98,.67 (C) 98,.67, 9 (D).67, 9, 98. In acidic medium, equivalent weight of K Cr 7 (Mol. wt. = M) is (A) M/ (B) M/ (C) M/6 (D) M/ Answers : (1) B () A () C NRMALITY : Normality of solution is defined as the number of equivalent of solute present in one litre (1000 ml) solutions. Let a solution is prepared by dissolving W g of solute of eq. wt. E in V ml water. No. of equivalent of solute = W E V ml of solution have W E equivalent of solute 1000 ml solution have W 1000 E VmL Normality (N) = W 1000 E VmL Normality (N) = Molarity Valence factor Solved Examples : Ex : Normality (N) = molarity Valence factor (n) or N V (in ml) = M V (in ml) n or milli equivalents = millimoles n Calculate the normality of a solution containing 15.8 g of KMn in 50 ml acidic solution. Sol : Normality (N) = W 1000 E VmL Ex : where W = 15.8 g, V = 50 ml E = So, N = 10 molar mass of KMn Valence factor = 158/5 = 1.6 Calculate the normality of a solution containing 50 ml of 5 M solution K Cr 7 in acidic medium. Sol : Normality (N) = Molarity Valence factor = 5 6 = 0 N

16 MLARITY V/S NRMALITY : Molarity (M) Normalit y (N) 1. No. of moles of solute present in 1. No. of equivalents of solute present in one litre one litre of solution. of solution.. No. of moles = W M. No. of equivalents = W E. W M 1000 = No. of millimoles. W 1000 = No. of equivalance E. Molarity V(in ml)=no. of millimoles. Normality V(in ml) = No. of equivalents 5. Molarity = milli moles Volume of solution in ml 5. Normality = milli equivalents Volume of solution in ml LAW F EQUIVALENCE : The law states that one equivalent of an element combine with one equivalent of the other, and in a chemical reaction equivalent and mill equivalent of reactants react in equal amount to give same no. of equivalent or milli equivalents of products separately. Accordi ng : (i) aa + bb mm + nn m.eq of A = m.eq of B = m.eq of M = m.eq of N (ii) In a compound M x N y m.eq of M x N y = m.eq of M = m.eq of N Solved Examples : Ex : Find the number of moles of KMn needed to oxidise one mole Cu S in acidic medium. The reaction is KMn + Cu S Mn + + Cu + + S Sol : From law of equivalence equivalents of Cu S = equivalents of KMn moles of Cu S v.f = moles of KMn v.f. moles of Cu S 8 = 1 5 moles of Cu S = 5/8 Ex : The number of moles of oxalate ions oxidized by one mole of Mn ion in acidic medium. (A) 5 (B) 5 (C) 5 (D) 5 Sol : Equivalents of C = equivalents of Mn x (mole) = 1 5 x = 5 E x. What volume of 6 M HCl and M HCl should be mixed to get two litre of M HCl? S o l. Let, the volume of 6 M HCl required to obtain L of M HCl = x L Volume of M HCl required = ( x) L M 1 V 1 + M V = M V 6M HCl M HCl M HCl 6 (x) + ( x) = 6x + 6x = 6 x = x = 0.5 L Hence, volume of 6 M HCl required = 0.5 L Volume of M HCl required = 1.5 L

17 E x. S o l. In a reaction vessel, 1.18 g of NaH is required to be added for completing the reaction. How many millilitre of 0.15 M NaH should be added for this requirement? Amount of NaH present in 1000 ml of 0.15 M NaH = = 6 g 6 1 ml of this solution contain NaH = g g of NaH will be present in = = 197. ml E x. What weight of Na C of 85% purity would be required to prepare 5.6 ml of 0.5N H S? S o l. Meq. of Na C = Meq. of H S = W E NaC NaC 1000 = W Na C 1000 = / W NaC = g For 85 g of pure Na C, weighed sample = 100 g 100 For g of pure Na C, weighed sample = Drawbacks of Equivalent concept : = g 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. 5e + 8H + + Mn Mn + + H Mn (mol.wt.) Eq. wt of Mn = 5 e.g. e + H + Mn Mn + H Eq. wt. of Mn = Mn Thus the no. of equivalents of Mn w i l l be different i n t he above t wo case s but no. of moles will be same. Normality of any solution depends on reaction while molarity does not. For example. Consider 0.1 mol KMn 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 KMn participates, e.g. If KMn forms Mn +, normality = = 0.5 N. This same sample of KMn, if employed in a reaction giving Mn as product (Mn in + state) will have normality 0.1 = 0. N. The concept of equivalents is handy, but it should be used with care. ne must never equate equivalents in a sequence which involves same element in more than two oxidation states. Consider an example KI reacts with KI to liberate iodine and liberated iodine is titrated with standard hypo solution, The reaction are

18 (i) I + I I (ii) I + S S 6 + I meq of hypo = meq of I = meq of I + meq of I I react with I meq of I = meq of I meq of hypo = meq of I Solved Example : Ex : This is wrong. Note that I formed by (i) have v.f. = 5/ & reacted in equation (ii) have v.f. =. v.f. of I in both the equation are different therefore we cannot equate m.eq is sequence. In this type of case students are advised to use mole concept. How many milliliters of M KMn solution would be required to exactly titrate 5.00 ml of M Fe(N ) solution. Sol : Method - 1 : Mole concept method Starting with 5.00 ml of 0.000MFe +, we can write. Millimoles of Fe + = and in volume V (in milliliters of the Mn ) Millimoles of Mn = V (0.0000) The balanced reaction is : Mn + 5Fe + + 8H + Mn Fe + + H This requires that at the equivalent point, V(0.0000) (5.00) (0.000) = 1 5 Method - : Equivalent Method : Equivalents of Mn = 5 moles of Mn Normality Mn = 5 molarity of Mn For Fe +, moles and equivalents are equal, At the equivalence point, Equivalents of Mn = Equivalents of Fe + or V Normality of Mn Mn = V Fe Normality of Fe + V = ml For M Mn solution Normality of Mn = (5) (0.0000) = 0.1 N and for M Fe + solution Normality of Fe + = N V = (5.00 ml) ml Mn HYDRGEN PERXIDE (H ) : H can behave both like oxidising and reducing agents in both the medium (acidic and basic). Reduce 1 H xidise H 0

19 xidising agent : (H ) (a) Acidic medium : (b) Basic medium : e + H + + H v.f. = e + H H v.f. = Reducing agent : (H ) : (a) Acidic medium : (b) Basic medium : H + H + + e v.f. = (R ) H H + H + H + e v.f. = Note : Valency factor of H is always equal to. Volume strength of H : Strength of H is represented as 10 V, 0 V, 0 V etc. 0V H means one litre of this sample of H on decomposition gives 0 L of gas at S.T.P. Normality of H (N) = Volume, strength of H 5.6 M H N H = v.f N H Molarity of H (M) = Volume, strength of H 11. Strength (in g/l) : Denoted by S Strength = molarity mol. wt. = molarity Strength = Normality Eq. weight. = Normality 17 Solved Example : Ex : 0 ml of H after acidification with dil H S required 0 ml of strength of H solution is [Molar mass of H = ] N 1 KMn for complete oxidation. The Sol : = 0 N' N' = = 1 8 strength = N' equivalent mass = 1 17 =.1 g/l 8 Hardne ss of water (Hard water doe s not give lat her w it h soap) Temporary hardness - due to bicarbonates of Ca & Mg Permanent hardness - due to chloride & sulphates of Ca & Mg. There are some method by which we can softening the water

20 (a) by boiling : HC H + C + C or by slaked lime : Ca(HC ) + Ca(H) CaC + H Ca + + C CaC (b) By Washing Soda : CaCl + Na C CaC + NaCl (c) By ion exchange resins : Na R + Ca + CaR + Na + PARTS PER MILLIN (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 are present in every g of solution. ppm A = Measurement of Hardness : mass of A Total mass 10 6 = mass fraction 10 6 Hardness is measured in terms of ppm (parts per million) of CaC or equivalent to it. Solved Example : Ex : % MgS and % CaCl is present in water. What is the measured hardness of water and millimoles of washing soda requires to purify 1000 litre water. Sol : % MgS = 1 mg CaC 1 L water % CaCl = 1 mg CaC 1 L water hardness = ppm and mm of Na C require is 0 Strength of leum leum is S dissolved in 100% H S. 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 S in the oleum to give 100% sulphuric acid. Hence weight % of free S in oleum = 80(y 100)/18 CALCULATIN F AVAILABLE CHLRINE FRM A SAMPLE F BLEACHING PWDER : The weight % of available Cl from the given sample of bleaching powder on reaction with dil acids or C is called available chlorine. CaCl + H S CaS + H + Cl CaCl + HCl CaCl + H + Cl CaCl + H CCH Ca(CH C) + H + Cl CaCl + C CaC + Cl Method of determination : or KI Bleaching powder + CH CH + KI KI starch Hypo.55 x V(mL ) % of Cl = W (g) where x = molarity of hypo solution Solved Example : Ex : v = ml of hypo solution used in titration. end point (Blue colourless).55 g sample of bleaching powder suspended in H was treated with enough acetic acid and KI solution. Iodine thus liberated requires 80 ml of 0. M hypo for titration. Calculate the % of available chlorine. [Available Chlorine = mass of chlorine liberated / mass of bleaching powder 100] Sol : moles of iodine = moles of chlorine = so required % = % = 16% 10 = 8 10

21 FR ACID-BASE (NEUTR ALIZATIN REACTIN) R REDX REACTIN : N 1 V 1 = N V is always true. But M 1 V 1 = M V (may or may not be true ) But M 1 n 1 V 1 = M n V (always true where n terms represent n-factor). 'n' FACTR : FACTR RELATING MLECULAR WEIGHT AND EQUIVALENT WEIGHT : n-factor = M E M E = n factor n-factr IN VARIUS CASES In Non Redox Change n-factor for element : Valency of the element For acids : Acids will be treated as species which furnish H + ions when dissolved in a solvent. The n-factor of an acid is the no. of acidic H + ions that a molecule of the acid would give when dissolved in a solvent (Basicity). For example, for HCl (n = 1), HN (n = 1), H S (n = ), H P (n = ) and H P (n = ) For bases : Bases will be treated as species which furnish H ions when dissolved in a solvent. The n-factor of a base is the no. of H ions that a molecule of the base would give when dissolved in a solvent (Acidity). For example, NaH (n = 1), Ba(H) (n = ), Al(H) (n = ), etc. For salts : A salt reacting such that no atom of the salt undergoes any change in oxidation state. For example, AgN + MgCl Mg(N ) + AgCl In this reaction, it can be seen that the oxidation state of Ag, N,, Mg and Cl remains the same even in the product. The n-factor for such a salt is the total charge on cation or anion. In Redox Change For oxidizing agent or reducing agent n-factor is the change in oxidation number per mole of the s u b s t a nce. SME XIDIZING AGENTS/REDUCING AGENTS WITH EQ. WT. S p ec i e s Changed to R e a ction Electrons Eq. wt. exchanged or change in.n. Mn (.A.) Mn Mn in acidic medium + 8H + + 5e Mn + + H 5 E = M 5 Mn (.A.) Mn Mn + e + H Mn + H E = M in neutral medium Mn (.A.) Mn Mn + e Mn in basic medium 1 E = M 1 Cr 7 (.A.) Cr Cr 7 + 1H + + 6e Cr + + 7H 6 E = M 6 in acidic medium Mn (.A.) Mn Mn + H + + e Mn + + H E = M in acidic medium

22 Cl (.A.) in bleaching powder CuS (.A.) in iodometric titration Cl Cl + e Cl E = M Cu + Cu + + e Cu + 1 E = M 1 S (R.A.) S 6 S S 6 + e E = M =M (for two molecules) H (.A.) H H + H + + e H E = M H (R.A.) H + H + + e (.N. of E = M oxygen in H is 1 per atom) Fe + (R.A.) Fe + Fe + Fe + + e 1 E = M 1 E x. To find the n-factor in the following chemical changes. (i) KMn (iii) KMn (v) C H Mn + (ii) KMn H Mn + H Mn 6+ (iv) K Cr 7 H Cr + C (vi) FeS Fe (vii) Fe FeS S o l. (i) In this reaction, KMn which is an oxidizing agent, itself gets reduced to Mn + under acidic conditions. n = 1 (+7) 1 (+) = 5 (ii) In this reaction, KMn gets reduced to Mn + under neutral or slightly (weakly) basic conditions. n = 1 (+7) 1 (+) = (iii) In this reaction, KMn gets reduced to Mn 6+ under basic conditions. n = 1 (+7) 1 (+6) = 1 (iv) In this reaction, K Cr 7 which acts as an oxidizing agent reduced to Cr + under acidic conditions. (It does not react under basic conditions.) n = (+6) (+) = 6 (v) In this reaction, C (oxalate ion) gets oxidized to C when it is reacted with an oxidizing agent. (vi) (vi) n = (+) (+) = In this reaction, ferrous ions get oxidized to ferric ions. n = 1 (+) 1 (+) = 1 In this reaction, ferric ions are getting reduced to ferrous ions. n = (+) (+) = E x. Calculate the molar ratio in which the following two substances would react? Ba (P ) and AlCl S o l. n-factor of Ba (P ) = (+) = 6 = n 1 While n-factor of AlCl = 1 (+) = = n n1 6 n1 x n If n y Molar ratio = y (inverse of equivalent ratio) x Molar ratio in which Ba (P ) and AlCl will react = : 6 = 1 :

23 APPLICATINS F THE LAW F EQUIVALENCE E x. Simple Titration In this, we can find the concentration of unknown solution by reacting it with solution of known concentration (standard solution). For example, let there be a solution of substance A of unknown concentration. We are given solution of another substance B whose concentration is known (N 1 ). We take a certain known volume of A in a flask (V ) and then we add B to A slowly till all the A is consumed by B (this can be known with the help of indicators). Let us, assume that the volume of B consumed is V 1. According to the law of equivalence, the number of g equivalents of B at the end point. N 1 V 1 = N V, where N is the conc. of A. From this we can calculate the value of N. 0. M KMn solution completely reacts with 0.05 M FeS solution under acidic conditions. The volume of FeS used is 50 ml. What volume of KMn was used? S o l V = E x. S o l. V = 1.5 ml 1.0 g sample of Na C and K C was dissolved in water to form 100 ml of a Solution. 0 ml of this solution required 0 ml of 0.1 N HCl for complete neutralization. Calculate the weight of Na C in the mixture. If another 0 ml of this solution is treated with excess of BaCl what will be the weight of the precipitate? Let, weight of Na C = x g Weight of K C = y g x + y = 1.0 g For neutralization reaction of 100 ml Meq. of Na C + Meq. of K C = Meq. of HCl x y x + 5 y = 7.1 From Eqs. (i) and (ii), we get x = g y = 0.60 g...(i) = 0...(ii) Solution of Na C and K C gives ppt. of BaC with BaCl (Meq. of Na C + Meq. of K C ) in 0 ml = Meq. of BaC Meq. of HCl for 0 ml mixture = Meq. of BaC Meq. of BaC = = W M BaC BaC 1000 = = W BaC = W BaC = 0.9 g

24 BACK TITR ATIN Back titration is used to calculate % purity of a sample. Let us assume that we are given an impure solid substance C weighing w gs and we are asked to calculate the percentage of pure C in the sample. We will assume that the impurities are inert. We are provided with two solutions A and B, where the concentration of B is known (N 1 ) and that of A is not known. This type of titration will work only if the following condition is satisfied, i.e. the nature of compounds A, B and C should be such that A and B can react with each other. A and C can react with each other but the product of A and C should not react with B. Now, we take a certain volume of A in a flask ( the g equivalents of A taken should be g equivalents of C in the sample and this can be done by taking A in excess). Now, we perform a simple titration using B. Let us assume that the volume of B used is V 1. In another beaker, we again take the solution of A in the same volume as taken earlier. Now, C is added to this and after the reaction is completed, the solution is being titrated with B. Let us assume that the volume of B used up is V. Gram equivalents of B used in the first titration = N 1 V 1. gm. equivalents of A initially = N 1 V 1 gm. equivalents of B used in the second titration = N 1 V gm. equivalents of A left in excess after reacting with C = N 1 V gm. equivalents of A that reacted with C = N 1 V 1 N 1 V If the n-factor of C is x, then the moles of pure C = The weight of C = Percentage of C = N V N V N V x N V x N V N V x Molecular weight of C Molecular wt. of C 100 w SME REDX TITR ATINS (EXCLUDING IDMETRIC / IDIMETRIC) Estimation of By titrating R e a c t i o n s Relation * between.a. and with R.A. Fe + Mn Fe + Fe + + e 5Fe + Mn Mn + 8H + + 5e Mn + + H Eq. wt. Fe + = M/1 Eq. wt. Mn = M/5 Fe + Cr 7 Fe + Fe + + e 6Fe + Cr 7 Cr 7 + 1H + + 6e Cr + +7H Eq. wt. Cr 7 = M/6 C Mn C C + e 5C Mn Mn + 8H + + 5e Mn + + H Eq. wt. C = M/ Eq. wt. Mn = M/5 H Mn H H e 5H Mn Mn + 8H + + 5e Mn + + H Eq. wt. H = M/ Eq. wt. Mn = M/5 As Mn As + 5H As + 10H + + e Eq. wt. As = M/ As Br As + H As + H + + e Eq. wt. As = M/ Br + 6H + + 6e Br + H Eq. wt. Br = M/6

25 E x. 0 g of a sample of Ba(H) is dissolved in 10 ml of 0.5 N HCl solution : The excess of HCl was titrated with 0. N NaH. The volume of NaH used was 0 cc. Calculate the percentage of Ba(H) in the sample. S o l. Milli eq. of HCl initially = = 5 E x. Milli eq. of NaH consumed = Milli eq. of HCl in excess = = Milli eq. of HCl consumed = Milli eq. of Ba(H) = 5 = Eq. of Ba(H) = /1000 = 10 Mass of Ba(H) = 10 (171/) = g % Ba(H) = (0.565 / 0) 100 = 1.8 %. g of pyrolusite was treated with 50 ml of 0.5 M oxalic acid and some sulphuric acid. The oxalic acid left undecomposed was raised to 50 ml in a flask. 5 ml of this solution when treated with 0.0 M KMn required ml of the solution : Find the % of Mn in the sample and also the percentage of available oxygen. S o l. Redox changes are C Mn C (n-factor = ) Mn + (n-factor = 5) Mn Mn + (n-factor = ) Meq. of Mn = Meq. of oxalic acid taken - Meq. of oxalic acid left = (in 50 ml) = 18 W M Mn Mn 1000 = 18 W Mn = g W Mn = 18, % of Mn = 100 =. %. Meq. of Mn = Meq. of W = 18, W % of available = 0.1. DUBLE TITR ATIN = 0.1 g 100 =.5 The method involves two indicators (Indicators are substances that change their colour when a reaction is complete) phenolphthalein and methyl orange. This is a titration of specific compounds. Let us consider a solid mixture of NaH, Na C and inert impurities weighing w g. You are asked to find out the % composition of mixture. You are also given a reagent that can react with the sample say, HCl along with its concentration (M 1 ) We first dissolve the mixture in water to make a solution and then we add two indicators in it, namely, phenolphthalein and methyl orange, Now, we titrate this solution with HCl. NaH is a strong base while Na C is a weak base. So, it is safe to assume that NaH reacts completely and only then Na C reacts. NaH + HCl NaCl + H nce NaH has reacted, it is the turn of Na C to react. It reacts with HCl in two steps : Na C + HCl NaHC + NaCl and then, NaHC + HCl NaCl +C + H As can be seen, when we go on adding more and more of HCl, the ph of the solution keeps on falling. When Na C is converted to NaHC, completely, the solution is weakly basic due to presence of NaHC (which is a weaker base as compared to Na C ). At this instant phenolphthalein changes colour since it requires this

26 weakly basic solution to change its colour. Therefore, remember that phenolphthalei n change s colour only when the weakly basic NaHC is present. As we keep adding HCl, the ph again falls and when all the NaHC reacts to form NaCl, C and H, the solution becomes weakly acidic due to the presence of the weak acid H C (C + H ). At this instance, methyl orange changes colour since it requires this weakly acidic solution to do so. Therefore, remember methyl orange changes colour only when H C is present. Now, let us assume that the volume of HCl used up for the first and the second reaction, i.e., NaH + HCl NaCl + H and Na C + HCl NaHC + NaCl be V 1 (this is the volume of HCl from the begining of the titation up to the point when phenolphthalein changes colour). Let, the volume of HCl required for the last reacton, i.e. NaHC + HCl NaCl + C + H be V (this is the volume of HCl from the point where phenolphthalein had changed colour up to the point when methyl orange changes colour). Then, Moles of HCl used for reacting with NaHC = Moles of NaHC reacted = M 1 V Moles of NaHC produced by the Na C = M 1 V Moles of Na C that gave M 1 V moles of NaHC = M 1 V Mass of Na C = M 1 V 106 % Na C = M1V w Moles of HCl used for the first two reactions = M 1 V 1 Moles of Na C = M 1 V Moles of HCl used for reacting with Na C = M 1 V Moles of HCl used for reacting with only NaH = M 1 V 1 M 1 V Moles of NaH = M 1 V 1 M 1 V Mass of NaH = (M 1 V 1 M 1 V ) 0 (M1V1 M1V ) 0 % NaH = 100 w WRKING R ANGE F FEW INDICATRS Indicator ph range Behaving as Phenolphthalein 8 10 Weak organic acid Methyl orange. Weak organic base Thus, methyl orange with lower ph range can indicate complete neutralization of all types of bases. Extent of reaction for different bases with acid (HCl) using these two indicators is summarized below : Phenolphthal ein Methyl orange Na H 100% reaction is indicated 100% reaction is indicated NaH + HCl NaCl + H NaH + HCl NaCl + H Na C 50% reaction upto NaHC 100% reaction is indicated stage is indicated Na C + HCl NaCl + H + C Na C + HCl NaHC + NaCl NaHC No reaction is indicated 100% reaction is indicated NaHC + HCl NaCl + H + C

27 E x. S o l. 0. g of a mixture of NaH and Na C and inert impurities was first titrated with phenolphthalein and N/10 HCl, 17.5 ml of HCl was required at the end point. After this methyl orange was added and 1.5 ml of same HCl was again required for next end point. Find out percentage of NaH and Na C in the mixture. Let W 1 g NaH and W g Na C was present in mixture. At phenolphthalein end point, W1 1 W 0 5 = (1) At second end point following reaction takes place Eq. of NaHC = Eq. of HCl used (in second titration) = 1 Eq. of Na C 1 W = W = g Putting the value of W in Eq. (1) we get W 1 = 0.06 g Percentage of NaH = Percentage of Na C = = 16% Iodometric and Iodimetric Titration : 100 =.975 % The reduction of free iodine to iodide ions and oxidation of iodide ions to free iodine occurs in these titrations. I + e I (reduction) I I + e (oxidation) These are divided into two types : Iodometic Titration : In iodometric titrations, an oxidizing agent is allowed to react in neutral medium or in acidic medium with excess of potassium iodide to liberate free iodine. KI + oxidizing agent I Free iodine is titrated against a standard reducing agent usually with sodium thiosulphate. Halogen, dichromates, cupric ion, peroxides etc., can be estimated by this method. I + NaS NaI + Na S 6 CuS + KI Cu I + K S + I K Cr 7 + 6KI + 7H S Cr (S ) + K S + 7H + I Iodimetric Titration These are the titrations in which free iodine is used as it is difficult to prepare the solution of iodine (volatile and less soluble in water), it is dissolved in KI solution : KI + I KI (Potassium triiodide) This solution is first standardized before using with the standard solution of substance such as sulphite, thiosulphate, arsenite etc, are estimated. In iodimetric and iodometric titration, starch solution is used as an indicator. Starch solution gives blue or violet colour with free iodine. At the end point, the blue or violet colour disappears when iodine is completely changed to iodide.

28 SME IDMETRIC TITR ATINS (TITR ATING SLUTINS IS Na S.5H ) Estimation of R e a c t i o n Relation between.a. and R.A. I I + Na S Nal + Na S 6 I = I Na S or Eq. wt. of Na S = M/1 I + S I + S 6 CuS CuS + KI Cu I + K S + I CuS I I Na S Cu + + I Cu I + I Eq. wt. of CuS = M/1 (White ppt.) CaCl + H Ca(H) + Cl CaCl Cl + KI KCl + I CaCl Cl I INa S Cl + I Cl + I Eq. wt. of CaCl = M/ Mn + HCl (conc) MnCl + Cl + H Cl + KI KCl + I Mn or Mn Cl I I Na S Mn + H + + Cl Mn + + H + Cl Eq. wt. of Mn = M/ Cl + I I + Cl I I + 5I + 6H + I + H I I 6I 6Na S Eq. wt. of I = M/6 H H + I + H + I + H H I I Na S Eq. wt. of H = M/ Cl Cl + I CI + I Cl I I Na S Eq. wt. of Cl = M/ + 6I + 6H + I + H I 6I 6Na S Eq. wt. of = M/6 Cr 7 Mn Cr 7 +1H + + 6I I + Cr + +7H Mn +10I +16H + Mn + 5I + 8H Cr 7 I 6I Eq. wt. of Cr 7 Mn 5I 10I Eq. wt. of Mn = M/5 Br Br +6I +6H + Br + I + H Br I I Eq. wt. of Br = M/6 As(V) H As + I + H + H As + H + I H As I I Eq. wt. of H As = M/

29 SME IDIMETRIC TITR ATIN (TITR ATING SLUTINS IS I IN KI) Estimation of R e a c t i o n Relation between.a. and R.A. H S H S + I S + I + H + H S I I (in acidic medium) Eq. wt. of H S = M/ S S + I + H S + I + H + S I I (in acidic medium) Eq. wt. of S = M/ Sn + Sn + + I Sn + + I Sn + I I (in acidic medium) Eq. wt. of Sn + = M/ As(III) (at ph = 8) H As + I + H HAs + I + H + H As I I Eq. wt. of H As = M/ N H N H + I N + H + + I N H I I Eq. wt. of N H = M/ CNCEPTS AND FRMULA AT A GLANCE 1. Number of moles of molecules = Number of moles of atoms = Number of moles of gases = wt. in g Mol. wt. wt. in g Atomic mass Volume at STP Standard molar volume Number of moles of particles, e.g. atoms, molecular ions etc. = Moles of solute in solution = M V(L). Equivalent wt. of element = Equivalent wt. of compound = Equivalent wt. of acid = Equivalent wt. of base = Mol wt. Basicity Mol wt. Acidity Atomic wt. Valence Mol. wt. Total charge on cation or anion Number of particles Avogadro No. Equivalent wt. of an ion = Equivalent wt. of acid salt = Formula wt. Charge on ion Molecular wt. Replaceable H atom in acid salt Equivalent wt. of oxidizing or reducing agent = No. of equivalent = N V(L) = wt.in g Eq. wt. Mol. wt. Change in oxidation number per mole

30 Ws Molarity (M) = M V where W s = wt. of solute in g M s = Mol. wt. of solute x = % by mass of solute d = density of solution in g/ml V = volume of solution in ml Ws Normality (N) = E V W s = wt. of solute in g. E s = eqv. wt. of solute V = volume of solution in ml x = % by mass of solute d = density of solution in g/ml 5. Moles = M V(L) = s s = = wt.of solute Mol.wt. x d 10 M s x d 1000 E Millimoles = M V(ml) = wt.of solute 1000 Mol.wt. Equivalents of solute = N V(L) Meq. of solute 6. Molarity equation : wt. Eq.wt s If a solution having molarity M 1 and volume V 1 is diluted to volume V so that new molarity is M then total number of moles remains the same. M 1 V 1 = M V For a balanced equation involving n 1 moles of reactant 1 and n moles of reactant. M V n = M V n Normality equation : According to the law of equivalence, the substances combine together in the ratio of their equivalent masses wt.of A wt.of B wt.of A Eq. wt.of A = wt.of B Eq.wt.of B Eq. wt.of A = Eq.wt.of B Number of gram equivalents of A = Number of gram equivalents of B Number of gram equivalents of A = Number of gram equivalents of B = N V N V A A 1000 = B B 1000 N A V B = N B V B N A V 1000 N BVB 1000 The above equation is called normality equation. A