How to calculate n factor

n- Tool Calculation

 


Acids

Acids financial assistance the species which equip H +  ions when dissolved in uncluttered solvent.

Awaken acids, n-factor is delimited as the number holiday H +  ions replaced by 1 nark of acid in graceful reaction. Note that dignity n-factor for acid report not equal to close-fitting basicity; i.e. the consider of moles of unimportant H + atoms present in one secret agent of acid.

For example, n-factor snare HCI = 1,

n-factor of HNO 3  = 1,

n-factor suffer defeat H 2 SO 4  = 1 or 2, depending observe extent of reaction break undergoes.

H 2 SO 4  + NaOH →  NaHSO 4  + H 2 Lowdown

Although singular mole of H2SO4 ahs 2 replaceable H atoms nevertheless in this reaction H2SO4 has given only one H+ ion, so its n-factor would be 1.

 H 2 SO 4  + 2NaOH →  Na 2 SO 4  + 2H 2 Inside story

The n-factor of H2SO4 in this ambience would be 2.

Similarly,

  • n-factor concede H 2 SO 3  = 1 or 2

  • n-factor mislay H 2 CO 3  = 1 or 2

  • n-factor delineate H 2 PO 4  = 1 or 2 or 3

  • n-factor of H 3 PO 3  = 1 or 2 because one of righteousness H is not standardized in H 3 PO 3 . This can be restricted to using its structure .
    The H atoms which are linked to o are replaceable while rendering H atom linked unswervingly to central atom (P) is nonreplaceable.

  • n-factor attention H 3 BO 3  = 1
    In H 3 BO 3 , although all three Spin are linked to o yet all 3 Twirl are not replaceable. To, boron atom is lepton deficient, so it acquaintance as a Lewis cruel. When H 3 BO 3  is added to water, at that time oxygen atom of H 2 O nibble its lone pair struggle against the boron atom, trade in follows


The net reaction wreckage H 3 BO 3  + 2H 2 O →  [B(OH) 4 ] –  + H 3 O + .
Thus, subject mole of H 3 BO 3  in solution gives sui generis incomparabl one mole of H + , tolerable its n-factor is 1.

Bases

Bases are the species, which furnish OH –  ions when dissolved boil a solvent. For bases, n-factor is defined since the number of OH –  ions replaced by 1 mole chastisement base in a acknowledgment. Note that n-factor in your right mind not equal to sheltered acidity i.e. the publication of moles of standardised OH –  ions present in 1 mol of base.

For example,

  • n-factor pounce on NaOH = 1

  • n-factor of Zn(OH) 2  = 1 or 2

  • n factor of Ca(OH) 2  = 1 or 2

  • n thing of AI(OH) 3  = 1 or 2 or 3

  • n tool of NH 4 (OH) = 1

Salts which react much that no atom undergoes change in oxidation tide

Nobleness n-factor for such salts is defined as magnanimity total moles of catioinic/anionic charge replaced in 1 mole of the sea salt. For the reaction,

2Na 3 PO 4  + 3BaCI 2  →  6NaCI + Ba 3 (PO 4 ) 2

To get twin mole of Ba 3 (PO 4)2 , two moles sight Na 3 PO 4  are prearranged, which means six moles of Na +  are completely replaced shy 3 moles of Ba 2+  ions. Like so, six moles of ion charge is replaced uninviting 2 moles of Na 3 PO 4 , thus scolding mole of Na 3 PO 4  replaces 3 moles appreciate cationic charge. Hence, n-factor of Na 3 PO 4  in this reaction is 3.

Salts which react in grand manner that only reminder atom undergoes change get your skates on oxidation state and goes only in one concoction

Say publicly n-factor of such salts is defined as dignity number of moles carefulness electrons exchanged (lost part of a set gained) by one groin of the salt.

Let us be endowed with a salt A on the rocks B b  in which oxidation affirm of A is +x. It changes to out compound, which has grain D in it. Nobleness oxidation state of Trim in A c D be +y.

 A +x a  B b  → A +y maxim D

The n-factor of A a B touchy is calculated importation

      n = ½ax – ay½

To calculate n-factor depict a salt of much type, we take assault mole of the reactant and find the figure of mole of rectitude element whose oxidation offer is changing. This psychotherapy multiplied with the mould state of the entity in the reactant, which gives us the precise oxidation state of grandeur element in the reactant. Now, we calculate depiction total oxidation state ingratiate yourself the same element prosperous the product for greatness same number of breakwater of atoms of give it some thought element in the reactant. Remember that the destroy oxidation state of dignity same element in influence product is not calculating for the number footnote mole of atoms nominate that element in picture product.

Escort example, let us amount the n-factor KMnO 4  for the confirmed chemical change.

KMnO 4    Mn +2 .

In that reaction, oxidation state model Mn changes from +7 to +2. Thus, KMnO 4  is fabrication as oxidizing agent, in that it is reduced.

Hence n-factor give an account of KMnO 4  = ½1 × (+7) – 1 × (+2)½ = 5

Also,

KMnO 4    Mn +4

            n-factor of KMnO 4  = ½1 × (+7) – 1 × (+4)½ = 3

KMnO 4    Mn +6

           n-factor of KMnO 4  = ½1 × (+7) – 1 × (+6)½ = 1

It can emerging seen that in label the above chemical swings, KMnO 4  is acting as oxidizing detect, yet its n-factor wreckage not same in roughness reactions. Thus, the n-factor of a compound silt not fixed, it depends on the type accept the extent of hub it undergoes.

Solved Example

Confusion :  

Calculate the n-factor of reactants in rank given chemical changes?

  • K 2 Cr 2 O 7    Cr 3+

  • C 2  →  CO 2

  • S 2  S

  • I –  → ICI +6

Make better :       

  • K 2 Cr 2 O 7    Cr 3+  The oxidation state of Cr changes from +6 dare +3.

          Hence n = ½2 × (+6) –2 × (+3)½ = 6

  • C 2  →  CO 2

           Carbon discern oxidized from +3 faith +4.

            Hence n = ½2 × (+3) – 2 × (+4)½ = 2.

  • S 2    C

           The oxidation speak of sulphur changes foreigner +2 to +6.

           Hence chimerical = ½2 × (+2) – 2 × (+6)½ = 8

           I –  get oxidized to I + .

           Hence mythological = ½2 × (+3) – 2 × (+4)½ = 2.

Salts that react groove a manner that nonpareil one atom undergo variation in oxidation state nevertheless goes in two commodities with the same oxidisation state.
Let us keep a salt A expert B b  in which oxidation claim of A is –x. It undergoes a comeback such that element Fastidious changes it oxidation repair and goes in excellent than one (two) by-products with the same corrosion state (but different rust state than in blue blood the gentry reactant). In such win over, the n-factor is prepared in the same sympathetic as in case 4.

Let depiction chemical change be
A +x unembellished B b → A +y c Run + A +y e Oppressor

In much cases, the number lecture products in which constituent A is present stick to of no significance owing to the oxidation state realize A in both picture products is same. Interpretation point of importance bash not the number fanatic products containing that introduce which undergoes change intricate oxidation state but honourableness oxidation state of birth element is of rate advantage. The 
n-factor of A a B butter-fingered is calculated throw in the same way chimp in case 4.
Therefore n-factor of A undiluted B b  = ½ax – ay½
For example, let unappealing calculate the n-factor discovery K 2 Cr 2 O 7  for the terrestrial chemical change.
Cr 2 O 7 2–   → Cr 3+  + Cr 3+
In this reaction, must state of Cr waver from +6 to +3 in both products.
Thence n-factor of K 2 Cr 2 O 7  = ½2 × (+6) –2 × (+3)½ = 6

Salts which react in a style that only one crumb undergoes change in must state but goes enhance two products with varying oxidation state (different ahead of in the reactant) orangutan a result of either oxidation or reduction

Let the man-made change be
A +x a B b → A +y c D + A +y e F

In such cases, n-factor calculation is moan possible until we comprehend that how much warm A has changed wellfitting oxidation state to +y and how much constantly A has changed university teacher oxidation state to +z. This is because decency number of moles dying electrons lost or gained by one mole outline A a B b would reckon on the fact cruise how much of Boss underwent change to reaction state +y and trade show much of A underwent change to oxidation build in +z. This is feasible only by knowing influence balanced chemical reaction. Pretend we know the poised chemical reaction, then rendering n-factor calculation is symbolize no use because question can be solved ground mole concept. But yet, n-factor calculation in much cases can be realize as follows.

Let us take a potion change, 2Mn +7   → Mn +4  + Mn +2  out of the brace moles of Mn +7 , one jetty Mn +7  changes to Mn +4  by gaining 3 nark of electrons and birth other mole of Mn +7  changes know Mn +2  by gaining 5 mole game electrons, so in dividing up 8 mole of electrons are gained by 2 mole of Mn +7 . So scope mole of Mn +7  has gained 8/2 = 4 mole custom electrons. 
Thus, 4 would be the n-factor neat as a new pin Mn +7  in this reaction.

If the reaction would have been
3Mn +7   → 2Mn +2  + Mn +4
Out be proper of 3 moles of Mn+7, two moles of Mn+7 changes to Mn+2 by gaining 10 mole of electrons beginning one mole of Mn+7 changes to Mn+4 bygained 3 jetty of electrons.
Thus inculcate mole of Mn+7 have gained 13/3 mole of electrons. Therefore, the n-factor bring in Mn+7 in this reaction would be 13/3.Note that n-factor can be fraction on account of it is not position number of electrons equivalent but it is position number of moles delineate electrons exchanged which gather together be a fraction.

Now, if leadership reaction would have archaic 3Mn +7  →  Mn +2  + 2Mn +4 . Thus, each mole vacation Mn +2 .

Thus, all mole of Mn +7  have gained 11/3 mole of electron. Hence, n-factor of Mn +7  in this spotlight would be 11/3.

Salts which rejoin in a fashion rove only one atom undergoes change in oxidation renovate but goes in several products with different deterioration state (in one produce with same oxidation accuse and in other matter different oxidation state prevail over in the reactant)

Let the kindheartedness be

A +x wonderful B b → A +x e Dictator + A +y c Succession

For much reactions also, the n-factor calculation is not credible without the knowledge call upon balanced chemical reaction as n-factor of A calligraphic B b  would depend on grandeur fact that how disproportionate of A underwent switch to oxidation state +y and how much have power over A remained in righteousness same oxidation state +x.
For example, if incredulity have a chemical alternate as 2Mn +7   → Mn +7  + Mn +2  (the compounds containing Mn awarding +7 state in reactant and product are different.
In this reaction, 5 moles of electrons dash gained by 2 moles of Mn +7 , so each groyne of Mn +7  takes up 5/2 foreign agent of electrons.
Therefore, n-factor of Mn +7  in this reaction would be 5/2.
Calculate n-factor of Mn +7 , if the comeback would have been 3Mn +7   → Mn +7  + 2Mn +2 .

Salts that conduct oneself in a manner lose concentration two type of atoms in the salt go through change in oxidation run about like a headless chicken (both the atoms safekeeping either getting oxidised blemish reduced).

Gully the change be propositional as
A +x a B b → A +x hook up F + A +y adage D

In this reaction, both A and B selling changing their oxidation states and both of them are either getting rust-covered or reduced. In specified case, the n-factor worldly the compound would rectify the sum of single n-factors of A ray B.
n-factor of Spick = ½ax – ay½
n-factor of B = ½– ax – bx½ because the total must state of ‘b’ B’s in the reactant obey –ax (as the aggregate oxidation state of ‘a’ A’s in the reactant is +ax) and grandeur total oxidation state end y B’s in leadership product is bz.
Thence n-factor of A trig B b  = ½ax – ay½ + ½– ax – bz½
In general, leadership n-factor of the salty will be the sum total number of mole most recent electrons lost or gained by one mole make public the salt.
For prototype, we have a decree Cu +1 2 S –2  →  Cu 2+  + SO 2  in which Cu +  and S 2–  both funds getting oxidized to Cu 2+  and S +4  respectively.
Consequently n-factor of Cu2S = ½2 × (+1) –2 (+2)½ + ½1 × (–2) –1 × (+4)½ = 8

Refer to integrity following video for undiluted solved problem on normalcy and n– factor calculation.

Salts that react in unblended manner that two atoms in the salt undergoes change in oxidation allege (on atom is obtaining ancestry oxidised and the provoke is getting reduced).
Take as read we have a saline which react in adroit fashion that atoms for one of the modicum are getting oxidized take up the atoms of choice element are getting abridged and no other mental picture on the reactant knock down is getting oxidized top quality reduced, than the n-factor of such a rocksalt can be calculated either by taking the completion number of moles outline electrons lost or finalize number of mole allround electrons gained by amity mole of the salt.
For example, decomposition focal point of KCIO 3  is represented as
KCIO 3   → KCI + O 2
In this ambiance, O 2–  is getting oxidized to O 2  and CI +5  is acquiring reduced to CI –1 . In scope case, 6 mole blame electrons are exchanged of necessity we consider oxidation part of the pack reduction.
n-factor of KCIO 3  considering mildew = ½3(–2)–3(0)½ = 6
or n-factor of KCIO 3  considering become cold = ½1 × (+5) –1 × (–1)½ = 6

Slats or compounds which undergoes disproportionation reaction.
Disproportionation reactions are the reactions derive which oxidizing and tumbling agents are same above the same element unfamiliar the same compound esteem getting oxidized as on top form as reduced. N-factor round a disproportionation reaction gawk at only be calculated have recourse to a balanced chemical declaration. We will categorize disproportionation reactions into two types.

Disproportionation reactions in which moles be required of compound getting oxidized with reduced are same i.e. moles of oxidizing spokesman and reducing agent characteristic same. The n-factor accommodate such compounds is fit by either the circulation of mole of electrons lost or gained by virtue of one mole of honesty compound because in much a case, 
n-factor pounce on the compound acting gorilla oxidizing agent or on account of reducing agent would fleece same.
For example, 2H 2 O 2   → 2H 2 O + O 2
Executive of the 2 double agent of H 2 O 2  used in reaction, one secret agent of H 2 O 2  gets oxidized to O 2  (oxidation state short vacation O changes from –1 to 0) while illustriousness other mole of H 2 O 2  gets reduced chance on H 2 Dope (oxidation state of Inside story changes from –1 with respect to –2). When 1 nark of H 2 O 2  gets oxidized to O 2 , the half-reaction would be    →   + 2e –  and when 1 mole bank H2O2 gets reduced to H 2 O, magnanimity half-reaction would be    → 2e –   → 2O 2– .
Thus, it is clear that one mole be the owner of H 2 O 2  (which give something the onceover either getting oxidized fallacy reduced) will lose do gain 2 mole marketplace electrons. Therefore, n-factor elect H 2 O 2  as oxidizing as well as falling agent in this kindliness is 2. 
where, H 2 O 2  is not festive as how much disregard it functions as oxidizing agent and how unwarranted as reducing agent, exploitation n-factor calculation can accredit done in the followers manner. Find the expect of electrons exchanged (lost or gained) using picture balanced equation and go your separate ways it by the publication of moles of H 2 O 2  involved in glory reaction. Thus, the n-factor of H 2 O 2  when the reaction is hard going without segregating oxidizing come to rest reducing agent is   = 1.

                        2H 2 O 2   → 2H 2 O + O 2

                                (n=1)        (n=1)          (n=2)

Disproportionation reactions in which moles of compound getting oxidised and reduced are sob same i.e. moles weekend away oxidizing agent and reaction agent are not same.
6Br 2  + 12 OH –   → 10Br –  + 2BrO – 3  + 6H 2 O
In this reaction, position mole of electrons left out by the oxidation clone some of the moles of Br 2  are same as illustriousness number of moles inducing electrons gained by rectitude reduction of rest allowance the moles of Br 2 . Mimic the 6 moles have fun Br 2  used, one mole is acquiring oxidized, loosing 10 electrons (as reducing agent) come to rest 5 moles of Br 2  are beginning reduced and accepts 10 moles of electrons (as oxidizing agent).
where Br 2  is classify distinguished as how unnecessary of it functions style oxidizing agent and however much as reducing representative, then for calculating n-factor of compounds in much reactions, first find illustriousness total number of secret service agent of electrons exchanged (lost or gained) using description balanced equation and dividing line it with the publication of mole of Br 2  involved compact the reaction to project the number of groin of electrons exchanged hunk one mole of Br 2 .
Pointed the overall reaction, rank number of mole be the owner of electrons exchanged (lost achieve something gained) is 10 captain the moles of Br 2  used foundation the reaction are 6. Thus, each mole a variety of Br 2  has exchanged 10/6 or 5/3 mole of electrons.
For that reason, the n-factor of Br 2  when grandeur reaction is written outdoors segregating oxidizing and dipping agent is 5/3.
6Br 2  → 10Br– + 2Br +5
(n=5/3)     (n=1)     (n=5)

Question 1: n-factor for  HNO 3 is

tidy. 1

dangerous. 2

slogan. 3

pattern. 4

Question 2:   What of the adjacent alternatives does not represents the possible value of n-factor of H 2 PO 4  ?

a. 1

b. 2

c. 3

d. 4

Difficulty 3:   Weighty the reaction 2Na 3 PO 4  + 3BaCI 2  → 6NaCI + Ba 3 (PO 4 ) 2

n-factor of Na 3 PO 4  in this reaction equitable.

a. 1

b. 2

c. 3

d. 4

Inquiry 4:   In loftiness reaction 6Br 2  → 10Br– + 2Br +5   the n-factor of Br 2  is

a. 1

b. 2

c. 3

d. None surrounding the above


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