1.

I
IT–JEESy
ll
abus
Oxi
dati
onnumber
,Bal
anci
ngofRedoxr
eact
ion;chemi
calequi
val
ence,n-
fact
or,l
awofequi
val
ence;
t
it
rat
ions.

2.OXI
DATI
ONNUMBER
Oxidati
onNumber :
Itisthechar gewhichanatom appear
stohav ewhenitisincombinat
ion.I
fallt
hepolarcov
alent
bondsofmol eculeareassumedt obe100% ionicthent
hechargeappearedoncor r
espondi
ngatomsi s
knownast heiroxidati
onstat
es.I
tmaybeawhol enumberorfr
acti
onal
,i
tdependsonnat ur
eofcompound
i
nwhi chitispresent.Ther
earesomeoperati
onalrul
estodet
ermineoxi
dat
ionnumber.

2.
1Cal
cul
ati
onofOxi
dat
ionSt
ate/Oxi
dat
ionNumber
Ther
ear
esev
eral
chemi
cal
react
ionsi
nwhi
choxi
dat
ion-r
educt
iont
akespl
ace.

Oxi
dat
ioni
s

i
) t
hegai
nofoxy
gen
i
i
)thel
ossofhy
drogen
i
i
i)t
hel
ossofel
ect
rons
i
v)t
hei
ncr
easeofO.
N.

Reduct
ioni
s

i
) t
hel
ossofoxy
gen

i
i
)thegai
nofhy
drogen

i
i
i)t
hegai
nofel
ect
ron

i
v)t
hedecr
easei
nO.
N.

Todescr ibethesechanges,t heconceptofoxi dat i

onst at
ebecomesnecessar y.Fori onicspeci es,t he
chargeoneachi oni ssaidtobet heoxidati
onst atef ort hatatom.Forexampl einNaCl ,Naexi stsasNa+
andClexi st
sasCl -.Theref
oretheoxidati
onst ateofNai nNaCli s+1andt hatofCl -i
s–1.Buti ncov alent
molecules,thechar geonanat om wouldbesosmal lthatsomet i
mesi tbecomesi mpossi bl
et ocal culate
theexactchar geoneachat om ofamol ecule.Ther efore,theOxidati
onSt ate( O.
S.)orOxi dationNumber
(O.
N. )i
sdef i
nedast hecharge,anatom wouldhav einamol eculei
fallthebondsassoci atedwi ththisat om
i
nt hemol eculear econsideredtobecompl etelyionic.Forexampl einH2Ot herearetwoO–Hbonds.I fwe
assumebot htheO–Hbondst obecompl etelyionic,theneachHwoul dpossessachar geof+1,whi leO
possessachar geof–2.Thi sisbecauseoxy geni smor eelectr
onegativ
et hanhy dr
ogen.Ont heot herhand,
i
nH2O2 t herear et woO–H bondsandoneO–Obond.Consi deri
ngeachO–H bondt obei oni cbot ht he
oxygenat omsacqui r
eachar geof1andbot ht heH, +1.ThisisbecauseO–Obondcannotbeassumedt o
beionicasbot ht heat omshavethesameel ectronegat iv
ity.
Tocal culatetheoxidati
onstateofanel ementi namol eculeyouneednotal waysknowt hest ructure
ofthemol ecule.Ther earecert
ainsetofrul
esusedt oassi gnoxidati
onst atesi npolyat
omi cmol ecules:

Rul
esForAssi
gni
ngOxi
dat
ionNumber
:

(
i) Oxi
dat
ionnumberoff
reeel
ement
sorat
omsi
szer
o.

(
ii
) Oxi
dat
ionnumberofal
l
otr
opesi
szer
o.

(
ii
i) Oxi
dat
ionnumberofat
omsinhomo-nucl
earmolecul
esi
szer
o.
(
iv) Oxi
dat
ionnumberofmono-
atomi
cionsisequal
totheal
gebr
icchar
geont
hem .
CENTERS:
MUMBAI/DELHI/AKOLA/KOLKATA/LUCKNOW /NASHI
K/GOA/PUNE# 1
 (
v) Oxi
dat
ionnumberofFi
ncompoundsi
s-1.

(
vi) Oxi
dat
ionnumberofHi
nit
scompoundsi
s+1,
excepti
nmet
alhy
dri
deswher
eiti
s-1.

(
vii
) Oxi
dati
onnumberofOis-2init
scompounds,butinF2Oi
tis+2andi
nper
oxi
desi
tis-1
and-0.
5inKO2.
(
vii
i
) Oxi
dati
onnumberofal
kal
imet
alsi
nthei
rcompounds+1.

(
ix) Oxi
dat
ionnumberofal
kal
i
neear
thmet
alsi
nthei
rcompoundsi
s+2.

(
x) Oxi
dat
ionnumberofani
oni
sequal
toi
tschar
ge.

(
xi) Oxi
dat
ionnumberofamol
ecul
easawhol
eiszer
o.

(
xii
) Thesum ofoxi
dat
ionnumberofal
ltheat
omsi
namol
ecul
eshoul
dbezer
oandi
nani
on
equal
toi
tschar
ge.
Quest
ion: Cal
cul
atet
heoxi
dat
ionst
ateoft
heunder
li
nedat
omsi
nthegi
venspeci
es.
(
a) (
b) (
c)
(
d) (
e) (
f)

Sol
uti
on:
(
a) Lett
heoxi
dat
ionst
at nNO2+bex.
eofNi
x+[
2(2)
]=+1
x=+5
Thus,
oxi
dat
ionst
ateofNi
n i
s+5.
b) Letxbet
( heoxi
dat
ionst
ateofNi
n .
x+[3(2)
]=1
x=+5
Thus,
oxidat
ionst
ateofNi
n i
s+5.
(
c) Lettheoxi
dat
ionstateofKMnO4bex.
[1(
+1)]+x+[4(2)
]=0
x=+7 Thus,oxi
dat
ionst
ateofMni
nKMnO4i
s+7.
(
d) Lett
heoxi
dat
ionst
ateofCri
n bex.
2x+[
7(2)]=2
x =+6
Thus,oxi
dati
onstateofCrin is+6.
e) Letxbet
( heoxidati
onstat
eofFei nFe2O3.
2x+[3(2)]=0
x=+3
Thus,oxi
dati
onstateofFeinFe2O3i
s+3.
(
f) Lettheoxi
dati
onstateofFeinFe3O4bex.
3x+[4(2)]=0
x=+8/ 3

Thus,oxi
dationstat
eofFeinFe3O4 i
s+8/ 3.Thi
si st
heaverageoxi
dat
ionst
ateofFei
nFe3O4.
Actual
l
y,Fe3O4ismadeupofequi
molarquanti
tyofFeOandFe2O3.

2.
2Mostcommonoxi
dat
ionst
ates

CENTERS:
MUMBAI/DELHI/AKOLA/KOLKATA/LUCKNOW /NASHI
K/GOA/PUNE# 2
 Gr
oup Out
ershel
lconf
igur
ati
on Commonoxi
d.St
atesexceptzer
oinf
reest
ate

IA ns1 +1

I
IA ns2 +2

I
IIA ns2np1 +3,
+1

I
VA ns2np2 +4,
+3,
+2,
+1,
-1,
-2,
-3,
-4

VA ns2np3 +5,
+3,
+1,
-1,
-3

VIA ns2np4 +6,

+4,
+2,
-2

VI
IA ns2np5 +7,
+5,
+3,
+1,
-1

3.REDOXREACTI
ON
RedoxReact
ions:
Ar
eact
ioni
nwhi
choxi
dat
ion&r
educt
ionoccursi
mul
taneousl
y.
Oxi
disi
ngAgent
s:
Theyoxidi
seothers,t
hemselvesarer educed& gai
nel
ect
rons. eg.O2,O3,HNO3,MnO2,H2O2,
hal
ogens,KMnO4,K2CrO
2 7,KI
O3,FeCl
3,NaOC l
.
Reduci
ngAgent
s:
Theyr
educeot
hers,t
hemselvesgetoxi
dised&l
oseel
ect
rons.egC,
CO,
H2S,
SO2,
SnCl
2,Sodi
um t
hio
Sul
phat
e,Al
,Na,CaH2,NaBH4,Li
AlH4.
Bot
hOxi
disi
ng&Reduci
ngAgent
s:
SO2,
H2O2,
O3,
NO2,et
c.

3.
1Bal
anci
ngofRedoxReact
ions

Redoxreact
ionsi
nvol
veoxidat
ionandreducti
onbot h.Oxi dat
ionmeansl ossofelect
ronsand
reducti
onmeansgai nofel
ect
rons.Thusredoxreacti
onsi nv ol
veelectr
ontransferandthenumber
ofelectr
onslostaresameast henumberofelectronsgai neddur i
ngt hereact
ion.Thi
saspectof
redoxreact
ioncanserv
easthebasisofapatter
nforbalanci ngredoxreacti
ons.
Ther
ear
e2commonandusef
ulmet
hodst
obal
ancer
edoxr
eact
ions.Thesear
e
(
A) Oxi
dat
ionnumbermet
hod
(
B) I
onel
ect
ronmet
hod.

(
A)OXI
DATI
ONNUMBERMETHOD
Forbal
anci
ngaredoxr eact
ionbyoxi
dat
ionnumbermethod,f
oll
ow theorderofst
epsasl
i
sted
bel
ow(ofcour
se,
all
stepsmaynotberequir
edf
orbal
anci
ngsomer eact
ions)
.
(
i) Foreachredoxreacti
on,deducet
heoxi
dat
ionst
ateoft
heel
ement
sthatar
eunder
goi
ng
oxi
dati
onandreduct
ion.
(
ii
) Separat
et her eactant
sandproductsint
ot wohal fr
eacti
onsinvol
vingtheel ement
sthat
changetheiroxidat
ionstat
e.Wr
it
etheskel
etalequat
ionsforeachhalf
react
ion.
(
ii
i) Foreachhal
freacti
on,f
ir
stbal
ancet
henumberofat
omsoft
heel
ementunder
goi
ngchange
i
noxidati
onstate.
(
iv) Now fi
ndthetotalchangeinoxidati
onnumberbydetermi
ningthechangeperatom and
mult
ipl
yi
ngitbyt hetotalnumberofatomsthatunder
goeschange.Also,deci
dewhether
el
ect
ronsarel
ostorgained.Ani
ncreasei
noxi
dati
onstat
eisl
ossofelect
ronsandadecr
ease

CENTERS:
MUMBAI/DELHI/AKOLA/KOLKATA/LUCKNOW /NASHI
K/GOA/PUNE# 3
 i
noxi
dat
ionst
atei
sgai
nofel
ect
rons.
(
v) Addt
heelectronslostorgai
nedtothehal
fequati
on.Lostel
ect
ronsar
epl
acedont
hepr
oduct
si
deandgainedelectr
onsarekeptonther
eactantsi
de.
(
vi) Nowaddboththehal
freact
ionsaf
termul
ti
ply
ingbysui
tabl
eint
eger
stomaket
henumberof
el
ect
ronsl
ostandgai
nedsame.
(
vii
) Tr
ansf
ert
hecoef
fi
cient
sofeachr
eact
antandpr
oductt
othemai
nskel
etonequat
ion.
(
vii
i
)Ifthecoeffi
cientsdevel
opedarenotcor
rect
,thenchangethem byi
nspect
ion.
Suchcoeffi
cientchangesarer
equir
edwhenanel ementfr
om acompoundgoesi n
2diff
erentcompounds, onewit
hthesameoxi dati
onstat
e&t heot
herwithdif
fer
entoxi
dat
ion
st
ate.
(
ix) Countthechar gesonbothsidesoft heequationandbalancet hechargesintheequat i
onby
addingrequisi
teH+orOHt other equir
edside.I
fthereactionoccursinacidicsolut
ion,use
H+andi fitoccursinbasi
csolut
ion,useOH.I ft
hereacti
onoccur si
nneut r
alsolut
ion,useH+
orOHonanyoft hesideasneededi .e.i
naneut r
alsol
ution,i
fnegativechargesareneeded
forbal
ancing,useOHandi fposit
ivechargesareneeded,useH+.
(
x) Balancethehydr
ogensandoxy
gensbyaddi
ngt
heappr
opr
iat
enumberofH2Omol
ecul
eson
therequi
redsi
de.
Quest
ion: Bal
ancet
hef
oll
owi
ngoxi
dat
ionr
educt
ionequat
ion,
KMnO4 +KCl+H2SO4  MnSO4+K2SO4+H2O+Cl
2

Sol
uti
on:
(
i) I
dent
if
ytheoxi
dat
ion&r
educt
ionhal
fequat
ionsi
nioni
cfor
m.
Reduct
ionhal
f:Mn7+  Mn2+

Oxi
dat
ionhal
f:Cl  Cl
2

(
ii
) Bal
ancet
heat
omst
hatunder
gochangei
noxi
dat
ionst
ate.
Reduct
ionhal
f:Mn7+  Mn2+

Oxi
dat
ionhal
f:2Cl  Cl
2

(
ii
i) Addt
heel
ect
ronsl
ostorgai
nedt
oeachhal
fequat
ion.
Reduct
ionhal
f:5e+Mn7+  Mn2+ ……………….
.

Oxi
dat
ionhal
f:2Cl  Cl
2+2e ……………….
.

(
iv) Mul
ti
plyequat
ion wi
th2andequat
ion wi
th5andt
henaddt
het
wohal
freact
ions.
5e +Mn7+  Mn2+ ]2

2Cl  Cl
2 +2e ]5
2Mn7+ +10Cl

 5Cl
2 +2Mn2+
(
v) Tr
ansf
ert
hecoef
fi
ci
ent
stot
hemai
nequat
ion.
2KMnO4 +10KCl
+H2SO4  2MnSO4 +K2SO4+H2O +5Cl
2

(
vi) Bal
anceH&Oat
oms
2KMnO4+10KCl+8H2SO4  5Cl
2+2MnSO4 +8H2O +K2SO4
(
vii
) Fi
nal
l
ybal
ancebyi
nspect
ionmet
hod.

2KMnO4+10KCl+8H2SO4  2MnSO4 +6K2SO4+8H2O+5Cl

2

(
B) I
ON-
ELECTRONMETHOD
Thi
smet
hodofbal
anci
ngr
edoxr
eact
ioni
nvol
vesf
oll
owi
ngst
eps.
(
i) Foreachredoxreacti
on,deducet
heoxi
dat
ionst
ateoft
heel
ement
sthatar
eunder
goi
ng
oxi
dat
ionandreduct
ion.
(
ii
) Separat
et her
eactant
sandproduct
si ntotwohalfr
eacti
onsinvol
vi
ngtheel ement
sthat
changesit
soxi
dati
onstat
e.Wr
it
etheskelet
onequat
ionsforeachhal
fr
eact
ion.
CENTERS:
MUMBAI/DELHI/AKOLA/KOLKATA/LUCKNOW /NASHI
K/GOA/PUNE# 4
 (
ii
i) Bal
anceeachhal
fr
eact
ionsepar
atel
yinv
olv
inggi
venst
eps.
1.Fi
rstbal
ancet
heat
omsoft
heel
ementunder
goi
ngoxi
dat
ionorr
educt
ion.
2.Thenbal
anceat
omsoft
heel
ement
sot
hert
hanhy
drogenandoxy
gen.
3.Forbalanci
ngoxygenatomsi nacidi
corneutr almedium,addsui
tabl
enumberofH2O
moleculestot
hesidedefi
cienti
nOwhi l
einalkali
nemedium,addequalnumberofH2O
moleculesast
heexcessofOont hesidehavi
ngexcessofOat omsandadddoublethe
numberofOHionsontheopposit
esideoftheequation.
4.I
nordert
obalancet
hehy dr
ogenatomsinaci
dicorneutr
almedium,addrequi
rednumber
ofH+t
othesi
dedefi
cienti
nHwhileinal
kal
i
nemedium, addequalnumberofOHionsas
t
heexcessnumberofHatomsonthesi
dehavi
ngexcessHandaddequalnumberofH2O
mol
eculesont
heopposi
tesi
deoft
heequat
ion.
(
iv) Mult
ipl
yeachhal
fr
eact
ionbysuit
ableintegert
omaket henumberofel ect
ronslostand
gai
nedsameandaddbot
hthehal
fequat
ionstogetacomplet
elybal
ancedr
eacti
on.
Quest
ion: Bal
ancet
her
edoxequat
ion,
HNO3+H2S NO+Sbyi
onel
ect
ronmet
hod(
aci
dicmedi
um)
.

Sol
uti
on:
(
i) I
dent
if
ytheoxi
dat
ion&r
educt
ionhal
ves.
Reduct
ionhal
f:HNO3  NO
Oxi
dat
ionhal
f:H2S S
(
ii
) At
omsoft
heel
ementunder
goi
ngoxi
dat
ionandr
educt
ionar
eal
readybal
anced.
(
ii
i) Bal
anci
ngOat
oms,
Reduct
ionhal
f:HNO3  NO+2H2O
Oxi
dat
ionhal
f:H2S S
(
iv) Bal
anci
ngHat
oms,
Reduct
ionhal
f:3H++HNO3  NO+2H2O
Oxi
dat
ionhal
f:H2S S+2H+
(
v) Bal
anci
ngchar
ge,
Reduct
ionhal
f:3e+3H++HNO3  NO+2H2O ……………….
.

Oxi
dat
ionhal
f:H2S S+2H++2e ……………….
.
(
vi) Mul
ti
ply
ingequat
ion by2andequat
ion by3andt
henaddi
ngt
hem.
3e+3H++HNO3 NO+2H2O]2 ……………….
.

H2S S+2H++2e]3 ……………….

.

2HNO3+3H2S 3S+2NO+4H2O
Quest
ion: Bal
ancet
hef
oll
owi
ngr
edoxequat
ion,
FeC2O4 +KMnO4 +H2SO4  Fe2(
SO4)
3+CO2 +MnSO4+K2SO4
usi
ngi
onel
ect
ronmet
hod(
aci
dicmedi
um)
.

Sol
uti
on:
(
i) I
dent
if
ytheoxi
dat
ion&r
educt
ionhal
ves.
Oxi
dat
ionhal
f:KMnO4  MnSO4
Reduct
ionhal
f:FeC2O4  Fe2(
SO4)
3+CO2
(
ii
) I
nbotht hehal
freact
ions,allt
heatoms( ot
hert
henOandH)arenotappearingonbot
hsidesoft
he
r
eacti
on.So,intheoxidationhalf
,H2SO4 i
stobeaddedonthereact
antsidewhil
einr
educti
onhal
f,
H2SO4andK2SO4ar et
obeaddedonr eact
antandpr
oductsi
der
especti
vel
y.
(
ii
i) Bal
anci
ngoft
heat
omsoft
heel
ementunder
goi
ngoxi
dat
ion&r
educt
ion.
Reduct
ionhal
f:H2SO4+2KMnO4  2MnSO4+K2SO4
Oxi
dat
ionhal
f:H2SO4+2FeC2O4  Fe2(
SO4)
3+4CO2

CENTERS:
MUMBAI/DELHI/AKOLA/KOLKATA/LUCKNOW /NASHI
K/GOA/PUNE# 5
 (
iv) Bal
anci
ngoft
heat
omsofel
ement
sot
hert
hanOandH.
Reduct
ionhal
f:3H2SO4+2KMnO4  2MnSO4+K2SO4
Oxi
dat
ionhal
f:3H2SO4+2FeC2O4  Fe2(
SO4)
3+4CO2
(
v) Bal
anci
ngOat
oms,
Reduct
ionhal
f:3H2SO4+2KMnO4  2MnSO4+K2SO4+8H2O
Oxi
dat
ionhal
f:3H2SO4+2FeC2O4  Fe2(
SO4)
3+4CO2
(
vi) Bal
anci
ngHat
oms,
Reduct
ionhal
f:10H++3H2SO4+2KMnO4  2MnSO4+K2SO4+8H2O
Oxi
dat
ionhal
f:3H2SO4+2FeC2O4  Fe2(
SO4)
3+4CO2+6H+
(
vii
) Bal
anci
ngchar
ge,
Reduct
ionhal
f:10e+10H++3H2SO4+2KMnO4 2MnSO4+K2SO4+8H2O…….
.
Oxi
dat
ionhal
f:3H2SO4+2FeC2O4  Fe2(
SO4)
3+4CO2+6H++6e ………….
.
(
vii
i
) Mul
ti
plyequat
ion by5andequat
ion by3andt
henaddi
ngt
hem.
 +
10e +10H +3H2SO4+2KMnO4 2MnSO4+K2SO4+8H2O]3
3H2SO4+2FeC2O4  Fe2(
SO4)
3+4CO2+6H++6e]5
10FeC2O4+6KMnO4+24H2SO4 5Fe2(
SO4)
3+6MnSO4+3K2SO4+20CO2+24H2O

Quest
ion: Bal
ancet
hef
oll
owi
ngr
edoxequat
ion,

usi
ngi
onel
ect
ronmet
hod(
alkal
i
nemedi
um)
.

Sol
uti
on:
(
i) I
dent
if
ytheoxi
dat
ion&r
educt
ionhal
ves.
Reduct
ionhal
f:  MnO2

Oxi
dat
ionhal
f: 
(
ii
) At
omsoft
heel
ementunder
goi
ngoxi
dat
ionandr
educt
ionar
eal
readybal
anced.
(
ii
i) Bal
anci
ngOat
oms,
Reduct
ionhal
f:2H2O+  MnO2+4OH
Oxi
dat
ionhal
f:2OH+  +H2O
(
iv) Bal
anci
ngHat
oms,
Hat
omsar
eal
readybal
ancedi
nbot
hthehal
fr
eact
ions.
(
v) Bal
anci
ngchar
ge,

Reduct
ionhal
f:3e+2H2O+  MnO2+4OH ………….
.

Oxi
dat
ionhal
f:2OH+  +H2O+2e ………….
.

(
vi) Mul
ti
plyequat
ion by3andequat
ion by2andt
henadd and .

3e+2H2O+MnO4  MnO2+4OH]2
2OH+  +H2O+2e]3

+2 +H2O 3 +2MnO2+2OH

Mostl
y,themedium i
nwhi charedoxreact
ionist obebalancedisgivenint
hepr
obl
em buti
f
t
heproblem doesnotstatethemedium expli
cit
ly,t
hent hemedium isdecidedbyl
ooki
ngatthe
r
eact
antsorproduct
s.Ifanacidorbaseisoneofther eact
ant
sorpr oduct
s, t
hent
hemedi
um i
sthe
CENTERS:
MUMBAI/DELHI/AKOLA/KOLKATA/LUCKNOW /NASHI
K/GOA/PUNE# 6
 same.Ifammoniaispresent,t
hesol ut
ionwouldbebasi c,forexampleammonium ionispresent
,it
wouldbeacidi
c.I fmetalswhichf ormsinsolubl
ehy droxidesareshowni ntheirioni
cform,the
sol
uti
onisaci
dic.Alt
hough,theoxidati
onnumbermet hodandi onel
ect
ronmethodbot hl
eadtothe
cor
rectfor
m oft hebalancedredoxr eact
ionbuti onelectr
onmet hodisconsideredsuperi
orto
oxi
dati
onnumbermet hodduet ofoll
owingadvantages:

(
i) Inionel
ectr
onmet hod,all
react ant
sandpr
oduct
sarecompl
etelybalanced
(i
.e.coeff
ici
entisdev el
opedf oreachoneofthem)whethert heyparti
cipat
eint heredox
changeornot ,whi
leinoxidationnumbermethod,t
hebal
ancingcoef f
ici
entisdev
elopedonly
forthespeci
esinvolv
edi nredoxchange.

(
ii
) Thebalanci
ngcoeff
ici
ent
sdev
elopedini
onel
ectr
onmet hodar
ealway
scorrectandneedsno
amendmentwhi l
esuch coef
fi
cient
saret o bechanged somet
imesin oxidat
ion number
method.

COMMONOXI
DATI
ONANDREDUCTI
ONPARTS

OXIDATI
ONPARTS REDUCTI
ONPARTS
Fe2+ Fe3+ Fe3+ Fe2+
Zn Zn2+
X  X2 X2 X
S2  S CrO 2  Cr
2 7
3+

H2O2  O2 NO3  NO
SO32  SO42 MnO4  Mn2+(
neut
ralmed.
)
C2O42  CO2 MnO4  MnO2(
Basi
cmed.
)
S2O32  S4O62 SO42  SO2

I
2
O3
 I MnO2  Mn2+

3.
2Di
spr
opor
ti
onat
ionReact
ions
I
nsameat om i
sgett
ingoxidi
sedaswellasr
educed.Sucht
ypeofr
edoxr
eact
ionsar
ecal
l
ed
Di
spr
opor
ti
onati
onr
eacti
on.Exampl
esar
e

METATHESI
SREACTI
ONS:

CENTERS:
MUMBAI/DELHI/AKOLA/KOLKATA/LUCKNOW /NASHI
K/GOA/PUNE# 7
 I
nt heset
wocompoundsr
eactt
ofor
mtwonewcompoundsandnochangei
noxi
dat
ionnumber
occur.
(
i)Pb(
NO3)
2
+K2Cr
O4 PbCr
O4+2KNO3 (
ii
)HCl+NaOH NaCl+H2O

4.CHEMI
CALEQUI
VALENCE
4.
1Equi
val
entWei
ght
Theequi
val
entwei
ghtofasubst ancei sthenumberofpar
tsbyweightofthesubstancet
hat
combinewi
thordi
spl
acedi
rect
lyofindir
ectl
y1.008par
tsbywei
ghtofhy
drogenor8part
sbyweight
ofoxy
genor35.
5part
sbyweightofchlor
ine.

Ot
her
wise,

Equi
val
ent
smass=

Al
so,
Numberofequi
val
ent
s=

So,

Numberofequi
val
ent
s=(
nfact
or)x(
no.ofmol
es)

4.
2Nor
mal
i
ty(
N)
I
tisdef
inedast
henumberofequi
val
ent
sofasol
utepr
esenti
nonel
i
treofsol
uti
on.Equi
val
enti
s
al
sothet er
m usedf oramountofsubst
anceli
kemolewit
ht hedi
ff
erencethatoneequiv
alentofa
subst
anceindiff
erentreact
ionsmaybedif
fer
entaswel
lastheoneequival
entofeachsubstancei
s
al
sodiff
erent.

Nor
mal
i
ty(
N)=

Lett
hewei
ghtofsol
utebewg,
equi
val
entmassofsol
utebeEg/
eqv
.,mol
ecul
armassbe
Mw g/
mol
eandt
hev
olumeofsol
uti
onbeVl
i
tre.

Numberofequi
val
ent
sofsol
ute= =

 N=

Hence, Nor
mal
i
tyofsol
uti
on=nf
act
ormol
ari
tyofsol
uti
on

4.
3LAW OFCHEMI
CALEQUI
VALENCE
Accor
dingtothelawofequi val
ence,whenev
ertwosubstancesr
eact,t
heequiv
alent
sofonewil
l
beequalt
ot heequi
valentsofotherandtheequi
val
entsofanyproductwi
l
lalsobeequalt
othatoft
he
equi
val
entsofthereactant.
Forar
eact
ionnA+mB pC+qD (
Iti
snotnecessar
ytobal
ancet
heequat
ion)
Equi
val
ent
sofA=Equi
val
ent
sofB=Equi
val
ent
sofC=Equi
val
ent
sofD
CENTERS:
MUMBAI/DELHI/AKOLA/KOLKATA/LUCKNOW /NASHI
K/GOA/PUNE# 8
 4n-
4. fact
orcal
cul
ati
on
(
A) ACI
DS
Aci
dsarethespecieswhichfur shH+ i
ni onswhendissol
vedinasol vent
.Foraci
ds,nfactoris
+
def
inedasthenumberofH i onsrepl
acedby1moleofacidi
nar eact
ion.Notethatt
henfactorfor
aci
disnotequaltoitsbasicit
y;i
.e.t
henumberofmolesofr epl
aceableH+ at
omspr esentinone
moleofaci
d.
Forexampl
e,nf
act
orofHCl
=1,
nf
act
orofHNO3=1,
nf
act
orofH2SO4=1or2,
dependi
nguponext
entofr
eact
ioni
tunder
goes.
H2SO4+NaOH  NaHSO4 +H2O.
Al
thoughonemoleofH2SO4has2r
epl
aceabl
eHat
omsbuti
nthi
sreact
ionH2SO4hasgi
venonl
y
oneH+ion,soi
tsnf
act
orwouldbe1.
H2SO4+2NaOH  Na2SO4+2H2O
Thenf
act
orofH2SO4i
nthi
sreact
ionwoul
dbe2.
Si
mil
arl
y,nf
act
orofH2SO3 =1or2
nf
act
orofH2CO3 =1or2
nf
act
orofH3PO4 =1or2or3
nf
act
orofH3PO3 =1or2becauseoneoft
heHi
snotr
epl
aceabl
einH3PO3.Thi
scan

beseenusi
ngi
tsst
ruct
ure .TheHat
omswhi
char
eli
nkedt
ooxy
genar
erepl
aceabl
e

whi
l
etheHat
om l
i
nkeddi
rect
lyt
ocent
ral
atom (
P)i
snonr
epl
aceabl
e.
nf
act
orofH3BO3=1
I
nH3BO3,al
thoughal lthr
eeHar elinkedt
ooxy gen,y
etal
l3Har enotrepl
aceabl
e.Her
e,boron
at
om isel
ectr
ondef i
cient,soitact
sasaLewi sacid.WhenH3BO3 i
saddedtowater,t
henoxygen
at
om ofH2Othroughit
sl onepairat
tacktheboronatom,asf
oll
ows

 +
Thenetr
eact
ioni
sH3BO3+2H2O [
B(OH)
4] +H3O .

Thus,
onemol
eofH3BO3i
nsol
uti
ongi
vesonl eofH+,
yonemol soi
tsnf
act
ori
s1.

(
B) BASES
Basesarethespeci es,whichfur shOHi
ni onswhendissol
vedi
nasol vent.Forbases,nfactor
i
sdefinedasthenumberofOHi onsrepl
acedby1mol eofbaseinareaction.Notethatnfactoris
notequalt
oitsacidityi
.e.thenumberofmol esofrepl eOHi
aceabl onspresentin1mol eofbase.
Forexample,
nfactorofNaOH=1
nfactorofZn(OH) 2=1o r2
nfactorofCa( OH)2 =1or2

CENTERS:
MUMBAI/DELHI/AKOLA/KOLKATA/LUCKNOW /NASHI
K/GOA/PUNE# 9
 nf
act
orofAl
(OH)3 =1or2or3
nf
act
orofNH4(
OH)=1.

(
C) SALTSWHI
CHREACTSUCHTHATNOATOM UNDERGOESCHANGEI
N

OXI
DATI
ONSTATE

Thenfact
orforsuchsalt
sisdefi
nedast
het
otalmol
esofcat
ioni
c/ani
oni
cchar
ger
epl
acedi
n
1mol
eofthesalt
.Forthereact
ion,
Na3PO4+BaCl
2  NaCl
+Ba3(
PO4)
2

Togetonemol eofBa3(PO4)
2,twomolesofNa3PO4ar erequi
red,whi
chmeanssixmolesofNa+
2+
ar
ecompl etelyreplacedby3mol esofBa i ons.So,si
xmol esofcati
oni
cchar
geisrepl
acedby2
molesofNa3PO4, thuseachmoleofNa3PO4r epl
aces3mol esofcati
oni
cchar
ge.Hence,
nfact
orof
Na3PO4i
nt hisreactioni
s3.

(
D) SALTSWHICH REACTI
N AMANNERTHATONLYONEATOM UNDERGOESCHANGEI
N
OXIDATI
ONSTATEANDGOESI
NONLYONEPRODUCT
Thenfact
orofsuchsal
tsi
sdefi
nedast
henumberofmol
esofel
ect
ronsexchanged
(
lostorgai
ned)byonemoleofthesal
t.
Letushaveasal
tAaBbinwhichoxi
dati
onst
at s+x.I
eofAi tchangest
oacompound,
whi
chhasatom Di
nit.Theoxi
dati
onstat
eofAinAcDbe+y
.

Thenf
act
orofAaBbi
scal
cul
atedas
axay|
n=|
Tocalculatenfact
orofasal tofsucht ype,wetakeonemol eofther eact
antandf i
ndthe
numberofmol eoft heelementwhoseoxi dati
onstateischangi ng.Thisismul ti
pli
edwiththe
oxidati
onstateoftheelementinther eact
ant,whichgi
vesusthet otaloxi
dati
onstateoftheel
ement
i
nt hereactant.Now,wecal cul
atet hetot
aloxidati
onstat
eofthesameel ementintheproductf
or
thesamenumberofmol eofat omsoft hatelementinther eactant.Remembert hatthetotal
oxidati
onst ateofthesameel ementi ntheproductisnotcalculatedforthenumberofmol eof
atomsoft hatelementintheproduct.

Forexampl
e,l
etuscal
cul
atet
henf
act
orKMnO4f
ort
hegi
venchemi
cal
change.

KMnO4 .
Inthi
sreact
ion,oxi
dat
ionst
ateofMnchangesf
rom +7t
o+2.Thus,
KMnO4i
sact
ingasoxi
disi
ng
agent,
sinceiti
sreduced.
 nf
act
orofKMnO4=|
1(
+7)1(
+2)
|=5
Si
mil
arl
y,

(
a) KMnO4 Mn+4
nf
act
orofKMnO4=|
1(
+7)1(
+4)
|=3

(
b)KMnO4 Mn+6
nf
act
orofKMnO4 =|
1(
+7)1(
+6)
|=1

I
tcanbeseenthati
nalltheabovechemicalchanges,KMnO4i
sacti
ngasoxi
disi
ngagent
,yeti
ts
nf
actori
snotsameinall
reacti
ons.Thus,t
henfactorofacompoundi
snotfi
xed,
i
tdependsonthety
peandt heext
entofreacti
onitundergoes.
CENTERS:
MUMBAI/DELHI/AKOLA/KOLKATA/LUCKNOW /NASHI
K/GOA/PUNE# 10
(
E) SALTSTHATREACTI
N AMANNERTHATONLYONEATOM UNDERGOCHANGEI
N OXI
DATI
ON
STATEBUTGOESI
NTWOPRODUCTSWI
THTHESAMEOXI
DATIONSTATE.
Letushav easal
tAaBb i
nwhichoxi
dati
onstat sx.I
eofAi tunder
goesar eacti
onsucht hat
el
ementAchangesi toxidat
ionstat
eandgoesi nmoret
hanone( t
wo)productswiththesame
oxi
dationstat
e(butdi
ff
erentoxi
dati
onstat
ethani nt
hereact
ant)
.Insuchcase,t henfact
ori s
cal
culatedi
nthesamemannerasincase4.

Lett
hechemi
cal
changebe

Insuchcases,thenumberofproductsinwhi
chelementAi spresenti
sofnosi gni
fi
cancesi
nce
theoxidat
ionstat
eofAi nboththeproduct
sissame.Thepointofimpor t
anceisnotthenumberof
productscont
aini
ngthatel
ementwhichundergoeschangeinoxidati
onstatebutt
heoxidati
onst
ate
oftheelementisofimport
ance.Thenfact
orofAaBbi
scalcul
atedinthesamewayasi ncase4.

 nf
act axay
orofAaBb=| |

Forexampl
e,l
etuscal
cul
atet
henf
act
orofK2Cr
2O7f
ort
hegi
venchemi
cal
change.
2 3+ 3+
Cr
2O7  Cr +Cr

I
nthi
sreact
ion,
oxi
dat
ionst
ateofCrchangesf
rom +6t
o+3i
nbot
hpr
oduct
s.

 nf
act
orofK2Cr
2O7=|
2(
+6)2(
+3)
|=6

(
F) SALTSWHICHREACTI
NAMANNERTHATONLYONEATOM UNDERGOESCHANGEI NOXIDATI
ON
STATEBUTGOESINTWOPRODUCTSWI THDI
FFERENTOXI
DATIONSTATE(
DIFFERENTTHANIN
THEREACTANT)ASARESULTOFEI
THEROXI
DATIONORREDUCTION.

Lett
hechemi
cal
changebe

Insuchcases,nf act
orcalculati
onisnotpossi bleuntilweknow t hathow muchofA has
changeditsoxidati
onst at
et o+yandhowmuchofAhaschangedi tsoxidationstat
e.to+z.Thisis
becausethenumberofmol esofelectr
onslostorgainedbyonemol eofAaBbwoul ddependont he
factthathow muchofAunder wentchanget ooxidationstate+yandhow muchofAunder went
changetooxidationstate+z.Thi
si spossi
bleonlybyknowi ngthebalancedchemi calreact
ion.
Ifweknow t hebal ancedchemi calreact
ion,thenthenf actorcalcul
ati
oni sofnousebecause
probl
em canbesol vedusingmol econcept .Butneverthel
ess,nfactorcalculat
ioninsuchcases
canbedoneasf oll
ows.

Letustakeachemi calchange,2Mn+7 Mn+4+Mn+2outoft hetwomolesofMn+7, onemole

+7 +4
Mn changest oMn bygai ning3moleofel
ect
ronsandt heothermoleofMn+7 changestoMn+2
bygaini
ng5mol eofelectr
ons, soi
nall8mol
eofelect
ronsaregainedby2moleofMn+7.
+7
Soeachmol eofMn hasgai ned8/2=4moleofelectr
ons.Thus,4wouldbethenf orofMn+7
act
i
nthisreact
ion.

I
fther
eact
ionwoul
dhav
ebeen

Outof3mol
esof t
womol
esof changest
o bygai
ning10mol
eofel
ect
rons

CENTERS:
MUMBAI/DELHI/AKOLA/KOLKATA/LUCKNOW /NASHI
K/GOA/PUNE# 11
 andonemoleof changesto bygai
ned3moleofel
ect
rons.Thuseachmoleof
havegai
ned13/
3mol
eofel ect
rons.Ther
efor
e,t
henf
act
orof i
nt hi
sreact
ionwoul
dbe13/
3.

Not
ethatnf
act
orcanbeaf r
acti
onbecauseitisnott
henumberofel
ect
ronsexchangedbuti
t
i
sthenumberofmolesofel
ect
ronsexchangedwhichcanbeafr
act
ion.

Now,
ift
her
eact
ionwoul
dhav
ebeen Thus,
eachmol
eof
Thus,eachmoleof hav
egai
ned11/
3mol
eofel
ect
ron.Ther
efor
e,nf
act
orof i
nthi
s
react
ionwoul
dbe11/3.

(G) SALTSWHI CHREACTI NAFASHI

ONTHATONLYONEATOM UNDERGOES
CHANGEINOXIDATIONSTATEBUTGOESINTWOPRODUCTSWITHDIFFERENT
OXI
DATIONSTATE( INONEPRODUCTWI
THSAMEOXIDATI
ONSTATEANDI N
OTHERWITHDI FFERENTOXI
DATI
ONSTATETHANI
NTHEREACTANT)
Lett
her
eact
ionbe

Forsuchreacti
onsalso,t
henf actorcalcul
ati
onisnotpossi
blewit
houttheknowledgeof
bal
ancedchemicalr
eact
ionbecausenfactorofAaBbwoul
ddependonthef
actt
hathowmuchof
Aunderwentchangetooxi
dati
onstat
e+yandhowmuchofAr emainedint
hesameoxidat
ionstat
e
+x.
Forexampl
e,i
fwehav
eachemi
cal
changeas
(
thecompoundscont
aini
ngMni
n+7st
atei
nreact
antandpr
oductar
e
di
ff
erent
.
I
nthi
sreact
ion,
5mol
esofel
ect
ronsar
egai
nedby2mol
esof eofMn+7t
soeachmol akes
up5/
2mol
eofel
ect
rons.Ther
efor
e,nf orofMn+7i
act nthi
sreact
ionwoul
dbe5/2.
(
H) SALTSTHATREACTINAMANNERTHATTWOTYPEOFATOMSINTHESALTUNDERGO
CHANGEINOXIDATI
ONSTATE(
BOTHTHEATOMSAREEITHERGETTI
NGOXI
DISEDOR
REDUCED)
.
Lett
hechangeber
epr
esent
edas

I
nt hi
sr eacti
on,bothAandBarechangingt
heiroxi
dat
ionstat
esandbot
hofthem ar
eeit
her
get
ti
ngoxi di
sedorr educed.I
nsuchcases,thenfact
oroft hecompoundwoul
dbet hesum of
i
ndi
vidualnfactorsofAandB.
nf
act axay
orofA=| |
nfactorofB=|axbz|becausethet
otaloxi
dati
onstat
eof‘b’B’si
nt hereact sax(
anti as
t
hetot
al oxi
dati
onst
ateof‘
a’A’
sint
hereact
antis+ax)andt
hetot
aloxidat
ionstat
eof
yB’
sintheproducti
sbz.
 nf
act axay
orofAaBb=| | axbz|
+|
I
ngeneral
,thenfact
oroft
hesal
twi
l
lbet
het
otalnumberofmol
eofel
ect
ronsl
ostorgai
nedby
onemoleofthesalt
.

Forexampl
e,wehav
ear
eact
ion,
i chCu+andS2bot
nwhi har
eget
ti
ngoxi
di oCu2+andS+4r
sedt espect
ivel
y.
 nf
act
orofCu2S=|
2(
+1)2(
+2)
|+|
1(
2)1(
+4)
|=8
(
I) SALTS THAT REACT I
N A MANNER THAT TWO ATOMS I
N THE SALT UNDERGOES

CENTERS:
MUMBAI/DELHI/AKOLA/KOLKATA/LUCKNOW /NASHI
K/GOA/PUNE# 12
 CHANGEI
N OXIDATI
ON STATE(
ONEATOM I
SGETTI
NG OXI
DISEDANDTHEOTHERI
S
GETTI
NGREDUCED).
Ifwehav easaltwhi chreactinafashi
ont hatatomsofoneoft heelementaregett
ingoxidi
sed
andt heatomsofanot herel
ementar egett
ingr educedandnootherelementonthereactantsi
deis
gett
ingoxidisedorreduced,thanthenfactorofsuchasal tcanbecalculat
edeit
herbyt aki
ngthe
tot
alnumberofmol esofel ectronsl
ostortotalnumberofmol eofelect
ronsgai
nedbyonemol eof
thesalt.
Forexampl
e,decomposi
ti
onr
eact
ionofKCl
O3i
srepr
esent
edas

O2i +5 1
I
nthi
sreact
ion, sget
ti
ngoxi
disedt
oO2andCl isget
ti
ngr
educedt
oCl .I
neachcase,
6mol
e
ofelectr
onsareexchangedwhetherweconsideroxidati
onorreduct
ion.
nfactorofKClO3consi
deri
ngoxi
dation=|3(
2)3(0)|=6
ornfactorofKClO3consi
deri
ngreducti
on=|1( +5)1( 1)|=6

(
J) SALTSORCOMPOUNDSWHI
CHUNDERGOESDI
SPROPORTI
ONATI
ONREACTI
ON.
Dispropor t
ionat i
onr eacti
onsaret hereacti
onsi nwhichoxidisingandr educi
ngagent saresame
orthesameel ementf rom thesamecompoundi sgett
ingoxidi
sedaswel lasreduced.nfactorofa
dispropor t
ionationr eact i
oncanonl ybecal culatedusingabal ancedchemi calreacti
on.Wewi l
l
categor i
zedi spropor t
ionationreacti
onsintotwot ypes.
(
a)Dispropor t
ionat i
onr eact i
onsinwhi chmol esofcompoundget t
ingoxi di
sedandr educedar esame
i
.e.mol esofoxi disi
ngagentandr educi
ngagentar esame.Thenf act
orf orsuchcompoundsi s
calculatedbyei therthenumberofmol eofel ectronslostorgainedbyonemol eofthecompound
becausei nsuchacase,nf actorofthecompoundact i
ngasoxi dizi
ngagentorasr educingagent
woul dbesame.
Forexample,2H2O2 2H2O+O2
Outofthe2mol eofH2O2usedi
nr eact
ion,onemol
eofH2O2get
soxi
disedtoO2
(oxi
dati
onst at
eofO changesfrom 1t o0)whil
etheothermol
eofH2O2 getsreducedtoH2O
(oxi
dati
onst at
eofO changesf rom 1t o2).When1mol eofH2O2 get
soxidi
sedt oO2,t
he
half
react
ionwouldbe

andwhen1mol
eofH2O2get
sreducedt
oH2O,
thehal
fr
eact
ionwoul
dbe
Thus,iti
sevidentthatonemol eofH2O2(
whi
chiseitherget
ti
ngoxidi
sedorreduced)wil
lloseor
gain2mol eofelect
rons.Ther
efor
e,nf
act
orofH2O2asoxidi
zingaswellasr
educingagentinthi
s
react
ioni
s2.Thus,

H2O2 + H2O2  2H2O +O2

Reduci
ngagent Oxi
dizi
ngagent
(n=2) (n=2) (
n=1) (
n=2)

orwhent
her
eact
ioni
swr
it
tenas
2H2O2 2H2O+O2
where,H2O2 i
snotdisti
nguishedashowmuchofi tf uncti
onsasoxi dizi
ngagentandhow
muchasr educi
ngagent ,thennfactorcalcul
ati
oncanbedonei nthef ol
lowingmanner.Findt
he
numberofelect
ronsexchanged(lostorgained)usingt
hebalanced
equat
ionanddivi
deitbyt henumberofmol esofH2O2invol
vedi nthereacti
on.Thus,

CENTERS:
MUMBAI/DELHI/AKOLA/KOLKATA/LUCKNOW /NASHI
K/GOA/PUNE# 13
 t
henf
act
orofH2O2whent
her
eact
ioni
swr
it
tenwi
thoutsegr
egat
ingoxi
disi
ngandr
educi
ngagent
i
s =1.

2H2O2  2H2O +O2

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

(
b)Dispr
oport
ionati
onreacti
onsinwhichmolesofcompoundgett
ingoxi
disedandr
educedar
enot
samei.e.molesofoxi
disi
ngagentandr
educi
ngagentar
enotsame.
Forexample,
2OH  10Br

6Br
2+1 + +6H2O
I
nthi
sreact
ion,
themol
eofel
ect
ronsl
ostbyt
heoxi
dat
ionofsomeoft
hemol
esofBr
2ar
e
sameast henumberofmol eofel
ectronsgai
nedbythereducti
onofrestoft
hemol
esofBr2.Oft
he
6mol esofBr2used,onemolei
sgett
ingoxidi
zed,
l
osi
ng10el ect
rons
(asreducingagent)and5molesofBr 2ar
egetti
ngreducedandaccepts10molesofel
ectr
on(as
oxi
dizingagent
).
+5
Br
2  2Br +10e
0e 10Br

5Br
2+1

 +5
Br
2 + 5Br
2 10Br +2Br

Reduci
ngagent Oxi
dizi
ngagent
(n=10) (n=2) (
n=1) (
n=5)

Thus,nfact
orofBr
2act
ingasoxi
dizi
ngagenti
s2andt
hatBr
2act
ingasr
educi
ngagenthas
nfact
or10.
Orwhent
her
eact
ioni
swr
it
tenas
 +5
6Br
2 10Br +2Br
where,Br2isnotdi st
ingui
shedashow muchofi tfuncti
onsasoxidi
zi
ngagentandhow muchas
reducingagent,t
hen f
orcal
culat
ingnfactorofcompoundinsuchr
eacti
ons,fir
st
fi
ndt hetotalnumberofmol eofel
ectr
onsexchanged( l
ostorgai
ned)usi
ngthebalancedequat
ion
anddi vi
deitwiththenumberofmol eofBr 2i
nvolvedinthereact
iontogett
henumberofmol eof
elect
ronsexchangedbyonemol eofBr2.

I
ntheov
eral
lreact
ion,
thenumberofmol
eofel
ect
ronsexchanged(
lostorgai
ned)i
s
10andthemolesofBr 2usedinthereacti
onare6.Thus,eachmoleofBr2hasexchanged10/6or
5/3 mol
e ofelectr
ons.Ther ef
ore,the nf
actorofBr2 when t
he react
ion i
s wri
tt
en wit
hout
segr
egat
ingoxi
disi
ngandr educi
ngagentis5/3.
 +5
6Br
2  10Br + 2Br
(
n=5/
3) (
n=1) (
n=5)

5.VOLUMETRI
CANALYSI
S
Now,
wehav
edev
elopedenoughpl
atf
ormst
ounder
standt
hel
awofequi
val
ent
sandv
olumet
ri
canal
ysi
s.

Thevolumetr
icanaly
sisisananal
yti
calmet
hodofestimat
ingtheconcentrat
ionofasubst
anceina
sol
uti
onbyaddi ngexact
lysamenumberofequi
val
ent
sofanothersubstancepresenti
nasolut
ionof
knownconcent
rati
on.
Thi
sist
hebasi
cpr
inci
pleoft
it
rat
ion.Vol
umet
ri
canal
ysi
si soknownast
sal it
ri
met
ri
canal
ysi
s.
Thesubstancewhosesol
uti
onisempl oyedtoest
imat
etheconcent
rat
ionofunknownsol
uti
oni
scal
l
ed
ti
tr
antandthesubst
ancewhoseconcentr
ati
onist
obeesti
mat
ediscaledt
l i
tr
ate.

CENTERS:
MUMBAI/DELHI/AKOLA/KOLKATA/LUCKNOW /NASHI
K/GOA/PUNE# 14
 Thev
olumet
ri
canal
ysi
sisdi
vi
dedi
ntof
oll
owi
ngt
ypes:
(
A)Si
mpleti
tr
ations
(
B)Backt
it
rat
ions
(
C)Doubl
eti
tr
ations

5.
1SI
MPLETI
TRATI
ON
Theai
m ofsi
mpletit
rat
ionistofi
ndt heconcent
rat
ionofanunknownsol
uti
onwi
tht
hehel
poft
he
knownconcent
rat
ionofanothersol
uti
on.
Letustakeasoluti
onofasubstance‘A’
ofunknownconcent rati
on( sayN1).Wear eprovi
ded
withsol
uti
onofanothersubst
ance‘B’whoseconcentr
ati
onisknown( N2).Wet akeacertainknown
volume ( V1li
tr
e)of‘
A’ i
nafl
askandst ar
taddi
ng‘B’fr
om burett
et o‘A’slowlyti
llal
lthe‘A’i
s
consumedby‘ B’
.Thi
scanbeknownwi ththeaidofsui
tabl
eindicator,whichshowscol ourchangeafter
thecompleteconsumpti
onof‘A’
.Letthev ol
umeofBconsumedi sV2l i
tr
e.

Accor
dingt
othel
awofequi
val
ent
s,t
henumberofequi
val
ent
sof‘
A’woul
dbeequal
tot
he
numberofequi
val
ent
sof‘
B’.
 N1V1=N2V2, whereN1i st heconcentrat
ionof‘A’.
Thususingt hisequation,thev al
ueofN1canbecal culated.
Forexample,inar edoxtitrati
on,anoxidantisestimatedbyaddi ngr eductantorv
icever
sa.
Forexampl e,Fe2+ionscanbeest imatedbytitrat
ionagainstacidi
fiedKMnO4sol utonwhenFe2+
i
ionsareoxidisedtoFe3+i onsandKMnO4i sr educedtoMn2+i nthepr esenceofacidi
cmedium.
KMnO4f unctionsassel findicat
orasi t
spurplecolourisdischargedatt heequiv
alencepoi
nt.
+8H+ +5e  Mn2+ +4H2O
Fe2+  Fe3+ +e]5
+8H+ +5Fe2+  Mn2+ +5Fe3+ +4H2O
(
n=5) (
n=1)

I
naddi
ti
ont
oaci
dif
iedKMnO4,aci
dif
iedKCr
2O7 canal
sobeempl
oyed.Ot
herr
edoxt
it
rat
ionsar
e
i
odi
met
ry,
iodomet
ryet
c.
(
i)I
ODI
METRY
Thi
sti
tr
ati
oninv
olv
esf r
eeiodi
ne.Suchdir
ectesti
mati
onofi
odi
neiscal
lediodi
metry
.
Thi
sinvol
vesthetit
rati
onofi odi
nesolutionwithknownsodi
um thiosul
phat
esolut
ion,whose
nor
mali
tyi
sN.Letthevolumeofsodium t
hiosul
phat
eusedbeVli
tr
e.

I
2 +2Na2S2O3  2NaI
+Na2S4O6
(
n=2) (
n=1)

Equi
val sofI
ent 2=Equi
val
ent
sofNa2S2O3used=NV

esofI
Mol 2=

Massoff
reeI
2int
hesol
uti
on=

(
ii
)IODOMETRY
Thi
si sanindirectmethodofesti
mationofi
odine.Anoxi di
singagentismadet or
eactwit
hexcess

ofsolidKI.Theoxi di
singagentoxi
disesI t
oI 2.Thi
sl i
beratediodi
neisthenma detoreactwi
th
Na2S2O3solut
ionofnor mal
it
yN.Letthevol
umeoft hi
osul
phat esol
uti
onrequi
redbeVli
tre.
Oxi
disi A)+KI
ngAgent(  I 2 2NaI
+Na2S4O6¬
+
reducedform of
oxidi
zingagent

CENTERS:
MUMBAI/DELHI/AKOLA/KOLKATA/LUCKNOW /NASHI
K/GOA/PUNE# 15
 Equi
val
ent
sof‘
A’=Equi
val sofI
ent 2=Equi
val
ent
sofNa2S2O3used=NV

Equi
val sofI
ent 2li
ber
atedf
rom KI
=NV

Equi
val
ent
sof‘
A’=NV

Lett
henf
act
orof‘
A’i
nit
sreact
ionwi
thKI
bex,
then

Massof‘
A’consumed= (
wher
eMAi
sthemol
armassofA)

Foriodi
metricandiodometri
ctit
rati
ons,st
archsol
uti
oni
susedasi
ndicat
or.Star
chsolut
iongi
ve
bl
ueorv i
oletcolourwit
hf r
eeiodine.Attheendpointthebl
ueorvi
oletcolourdi
sappear
swhen
i
odinei
scompl etel
ychangedtoiodide.

5.
2BACKTI
TRATI
ON
Letusassumet hatwehav eani mpuresoli
dsubst ance‘ C’,weighi
ng‘ w’gandwear e
requir
edt ocalculatethepercentagepurit
yof‘C’ i
nthesample.Wear ealsopr ovi
dedwi th
twosol ut
ions‘A’and‘ B’
,wheret heconcentrat
ionof‘B’i
sknown( N1)andt hatof‘A’i
sunknown.
Forthebackt i
tr
ationtowor k,f
ollowingcondit
ionsaretobesatisfi
ed(a)Compounds‘ A’,‘
B’
and‘C’shoul dbesucht hat‘
A’and‘ B’reactwitheachother.(b)‘A’andpur e‘C’alsor eactwitheach
otherbutt heimpur i
typresentin‘C’doesnotreactwi t
h‘A’
.(c)Alsothepr oductof‘
A’ and‘ C’shoul
dnot
reactwith‘B’.
Nowwet akeoutcer
tai
nvol
umeof‘A’
inafl
ask(theequi
val
entsof‘A’takenshoul
dbeequi
val
entsof
pure‘
C’i
nt hesampl
e)andper
for
m asi
mplet
it
rat
ionusing‘
B’.Letusassumet hatt
hev
olumeof‘
B’used
beV1li
tr
e.
Equi
val
ent
sof‘
B’r
eact
edwi
th‘
A’=N1V1
 Equi
val
ent
sof‘
A’i
nit
ial
l
y=N1V1
I
nanotherfl
ask,weagaintakesamevol
umeof‘
A’butnow ‘C’i
saddedtothisfl
ask.Pur
epar tof‘
C’
r
eact
swi t
h‘A’andexcessof‘
A’i
sbackt
it
rat
edwi
th‘
B’.Lett
hevol
umeof‘B’consumedisV2li
tr
e.
Equi
val
ent
sof‘
B’r
eact
edwi
thexcessof‘
A’=N1V2
 Equi
val
ent
sof‘
A’i
nexcess=N1V2
Equi
val
ent
sof‘
A’r
eact
edwi
thpur
e‘C’
=(N1V1N1V2)
Equi
val
ent
sofpur
e‘C’
=(N1V1N1V2)

Lett
henf
act
orof‘
C’i
nit
sreact
ionwi
th‘
A’bex,
thent
hemol
esofpur
e‘C’
=

 Massofpur
e‘C’
= Mol
armassof‘
C’.

 Per
cent
agepur
it
yof‘
C’=  100

5.
3DOUBLETI
TRATI
ON
Thepurposeofdoubl et itr
ati
onistodeterminetheper
centagecompositi
onofanalkalimixtureoranacid
mixt
ure.Inthepresentcase, wewi l
lfi
ndthepercent
agecomposi ti
onofanal
kalimi
xture.Letusconsidera
sol
idmi xt
ureofNaOH,Na2CO3 andsomei nertimpur
it
ies,weighi
ng‘w’g.Wear erequir
edt of i
ndthe%
compositi
onofthisal kalimixt
ure.Wearealsogivenanacidreagent(HCl
)ofknownconcent rat
ionM1that
canreactwitht
heal kalisample.
Wefi
rstdissolvethi
smi xtur
ei nwat
ertomakeanalkalinesol
uti
onandt henweaddtwoi ndicat
ors,
(I
ndicator
saresubst ancesthatindi
catecol
ourchangeofsoluti
onwhenar eacti
onget
scompleted),
namel y
phenolpht
hal
einandmet hylorangetothesol
uti
on.
Now,wetit
ratethi
sal
kal
inesolut
ionwi
ths
tandar
dHCl .
NaOHi
sast
rongbasewhi
l
eNa2CO3i
saweakbase.Soi
tisobv
ioust
hatNaOHr
eact
sfi
rstwi
thHCl
CENTERS:
MUMBAI/DELHI/AKOLA/KOLKATA/LUCKNOW /NASHI
K/GOA/PUNE# 16
compl
etel
yandNa2CO3r
eact
sonl
yaf
tercompl
eteNaOHi
sneut
ral
i
zed.
NaOH+HCl
 NaCl
+H2O ….
.(
i)
OnceNaOHhasr
eact
edcompl
etel
y,t
henNa2CO3st
art
sreact
ingwi
thHCl
int
wost
eps,
shownas
Na2CO3+HCl
 NaHCO3+NaCl ….
.(
ii
)
NaHCO3+HCl
 NaCl
+CO2+H2O ….
.(
ii
i)
Iti
sclearthatwhenweaddHClt othealkali
nesol ution,alkal
iisneutrali
zedandt hepHoft he
solut
iondecr eases.Ini
ti
all
yt hepH decr easewoul dber apidasst r
ongbase( NaOH)i sneutr
ali
zed
compl etel
y.WhenNa2CO3 isconv ert
edtoNaHCO3 compl etely,thesoluti
onisst il
lweaklybasicdueto
thepr esenceofNaHCO3 ( whichi sweakerascompar edt oNa2CO3) .Att hispoi nt
,phenolpht
halei
n
changescol oursinceitrequirest hi
sweaklybasicsol uti
ont oshow i t
scol ourchange.WhenHCli s
furt
heradded,t hepHagaindecr easesandwhenal ltheNaHCO3r eactstoform NaCl ,CO2andH2Ot he
solut
ionbecomesweakl yaci dicduet othepresenceoft heweakaci d( H2CO3).Att hi
spoint,methyl
orangechangescol ourasitrequiresthi
sweaklyacidi
csol utiontoshowi t
scol ourchange.
Thusingeneral
,phenol
phthal
einshowscol ourchangewhenthesol
utioncontai
nsweakl
ybasic
NaHCO3 alongwithotherneut
ralsubstanceswhilemet hylor
angeshowscolourchangewhensolut
ion
cont
ainsweaklyaci
dicH2CO3al
ongwi t
hot herneut
ralsubst
ances.
Lett
hev olumeofHClusedupf orthef i
rstandthesecondr eacti
onbeV1lit
re(thi
sisthev
olume
ofHClusedf rom thebegi nningoft hetitr
ationupt ot hepointwhenphenol pht
haleinshowscolour
change)andthev ol
umeofHClr equi
redforthet hi
rdreacti
onbeV2l itr
e(thi
sisthevolumeofHClused
fr
om thepointwher ephenolpht hal
einhadchangedcol ouruptothepoi ntwhenmet hylor
angeshows
col
ourchange).Then,
MolesofHCl consumedbyNaHCO3=Mol esofNaHCO3r eact
ed=M1V2
MolesofNaHCO3f ormedf rom Na2CO3=M1V2
MolesofNa2CO3i nt hemixture=M1V2
MassofNa2CO3i nt hemixture=M1V2106

%ofNa2CO3i
nthemi
xtur
e= 100

MolesofHClusedi
nthereact
ion(i
)and(i
i
)=M1V1
MolesofHClusedi
nreact
ion(ii
)=M1V2
MolesofHClusedi
nreact
ion(i)=(
M1V1M1V2)
 MolesofNaOH=(M1V1M1V2)
MassofNaOH=( M1V1M1V2)40

%ofNaOHi
nthemi
xtur
e=

Quest
ion:Asol ut
ioncontai
nsami xt
ureofNa2CO3andNaOH.Usi
ngphenol
phthalei
nasindi
cat
or,25mlof
mixturerequir
ed21mlof1. 1NHClf ort
heendpoi
nt.Wit
hmethylorangeasi
ndicat
or,25mlof
sol
utionr equi
red25mloft hesameHClf ortheendpoint
.Cal
culategramsperli
treofeach
substanceinthemixt
ure.
Sol
uti
on: Since,thevol
umeofHCl r
equiredinti
trat
ionusingmet hy
lorangeisgreatert
hanthev
olumeof
HClr equi
redusi
ngphenolphthal
ein,t
hismeanst hatthetit
rati
oniscarri
edoutsepar
atel
ytwo
ti
mesusi ngphenol
pht
haleinandmet hylorangeindicator
s,respect
ivel
y.
NaOH + HCl NaCl+H2O ……(
i)
Na2CO3+HCl
 NaHCO3+NaCl ……(
ii
)
NaHCO3+HCl
 NaCl+CO2 +H2O ……(
ii
i)
Thus,
thev
olumeofHCl
usedi
nthi
rdr
eact
ion=(
2521)=4.
0ml
 Mol
esofHCl
usedi
nthi
rdr
eact
i 01031.
on=4. 1
Mol
esofNaHCO3r
eact 4103
ed=4.

CENTERS:
MUMBAI/DELHI/AKOLA/KOLKATA/LUCKNOW /NASHI
K/GOA/PUNE# 17
 3
Mol
esofNaHCO3pr
oduced=4.
410
 Mol
esofNa2CO3pr
esenti
n25ml 4103
=4.
 MassofNa2CO3pr
esenti
n1l
i
tr 410310640=18.
e=4. 656g
Mol
esofHCl
react
edi
nsecondr
eact
i 4103
on=4.
Mol
esofHCl
usedi
nfi
rstt
wor
eact
ions=211031. 1103
1=23.
 Mol
esofHCl
usedi
nfi
rstr
eact
ion=(
23. 4)103=18.
14. 7103
Mol
esofNaOHpr
esenti
n25ml 7103
=18.
 MassofNaOHpr
esenti
n1l
i
tr 71034040=29.
e=18. 92g

HOMEASSI
GNMENT-I
1. Oxidati
onstateofnit
rogeni
sincor
rect
lygi
venfor:
Compound Oxi
dat
ionst
ate
(a)[Co(NH3)5Cl
]Cl
2 –3
(b)NH2OH –1
(c)(N2H5)
2SO4 +2
(d)Mg3N2 –3

2. MnO2+zOH  xCl
+y O+

I
nthi
sbalancedequat
ionx,
y,zar
e
(
a)2,
3,6 (b)1,
3,4 (
c)1,
3,
6 (
d)2,
6,
6

3. 1mol
off
err
icoxal
atei
soxi
dizedbyxmol
of andal
so1mol
off
err
ousoxal
atei
soxi
disedby

ymol
of i
naci
dicmedi
um.Ther
ati
o i
s

(
a)2:
1 (
b)1:
2 (
c)3:
1 (
d)1:
3

4. Numberofmol
esofKMnO4thati
sneededt
or eactwi
thonemol
eofFeC2O4i
naci
dicmedi
um i
s
(
a)2/5 (b)3/
5 (c)4/5 (
d)1

5. Theoxidati
onstateofCri nCrO5i
s
(a)+10 (b)+6 (c)+3 (
d)+35
6. Onemol eofN2H4l oses10mol eofel
ect
ronstoform anewcompoundY.Assumingt
hatal
lni
tr
ogen
appearsint henew compound,whati stheoxidati
onstat
eofN inY( t
hereisnochangeinthe
oxidat
ionstateofhydrogen).
(a)3 (b)+3 (c)+5 (
d)+1

7. Numberofmolesofel
ect
ronschangepermol
eofPb(
N3)
2in
Pb(
N3)
2+C o(MnO4)3 C oO+MnO2+Pb3O4+NOis
(
a) (
b) (
c)10 (
d)44

8. Zn+OH–  Zn +H2
I
nt hebal
ancedequat
ion,whatshoul
dbet
hecoef
fi
cientofOH–?
(a)3 (b)4 (
c)2 (
d)1

9. I
nt her
eact
ion,
Zn+NaNO3+NaOH  Na2ZnO2+NH3+H2O
Themolarcoeff
ici
entsofZnandNaNO3ar
e
(a)1and4respecti
vel
y (
b)4and1r
espect
ivel
y
(c)1and8respecti
vely (
d)8and1r
espect
ivel
y

10. Car
boni
sinhi
ghestoxi
dat
ionst
atei
n
CENTERS:
MUMBAI/DELHI/AKOLA/KOLKATA/LUCKNOW /NASHI
K/GOA/PUNE# 18
 (
a)CH3Cl (
b)CCl
4 (
c)CHCl
3 (
d)CH2Cl
2

11. I
nthereacti
on,
N2O5+H2O  2HNO3
t
heoxidati
onstat
eofnitr
ogen
(
a)changesfrom +5to+2. (
b)changesf
rom +2t
o+5.
(
c)changesfrom +10to+5. (
d)donotchange.

12. Theoxidat
ionst
ateof‘
S’i
nMar
shal
l
’saci
d(H2S2O8)i
s
(a)+5 (
b)+3 (c)+6 (
d)+7

13. Amongstt
hef
oll
owi
ng,
ident
if
ythespeci
eswi
thanat
om i
n+6oxi
dat
ionst
ate.
(
a) (
b) (
c) (
d)Cr
O2Cl
2

14. I
nthef ol
lowingreacti
on,
2Na2S2O3+I 2  2NaI+Na2S4O6
Na2S2O3isacti
ngas
(
a)anoxi di
singagent (b)areduci
ngagent
(
c)bot h (d)none

15. Whi
choft
hefoll
owingst
atement
sisi
ncorrect?
(
a)Oxi
dat
ionst
ateofoxy
genis+1inperoxides.
(
b)Oxi
dat
ionst
ateofoxygeni
s+2inOF2.
(
c)Oxi
dat
ionst
ateofoxygeni
s0.5i
nsuper oxides.
(
d)Oxi
dat
ionst
ateofoxygeni
s2inmostofi tscompounds.

16. Whi
chpai
roft hefol
lowingcompoundshasel
ementsi
nt hei
rhi
ghestoxi
dat
ionst
ate?
3 3 
(
a)[
Fe(
CN)6] a nd[Co(CN)6[ (b)[
MnO4]andCr O2Cl
2
 2
(
c)[
MnO4] and[NiF4] (d)MnO2andTiO2

HOMEASSI
GNMENT#2
1. 0.1gofamet
algav
eonreact
ionwi
thdi
laci
datS.T.P.34.
2mlhy
drogengas.Theequi
val
entwei
ght
ofthemet
ali
s:
(A)32.
7 (
B)48.6 (C)64.
2 (D)16.
3

2. 0.5gofmet
alonox
idat
iongav
e0.
79gofi
tsoxi
de.Theequi
val
entweightoft
hemet
ali
s
(A)10 (
B)14 (C)20 (
D)40

3. 74.5goft
hemet
all
i
cchlor
idecont
ains35.
5gofchl
ori
ne.Theequi
valentwei
ghtoft
hemet
ali
s
(A)19.
5 (B)35.5 (
C)39.
0 (D)78.0

4. Thechlor
ideofametal(M)contai
ns65.
5%ofchl or
ine.100ml.ofthevapouroft
hechl
ori
deoft
he
metal
atS.T.P.wei
gh0.72g.Themol
ecul
arf
or mul
aoft hemetalchl
ori
deis
(A)MCl (
B)MCl2 (
C)MCl3 (
D)MCl 4

5. Theequivalentwei
ghtofamet
ali
s4.
5andthemol
ecul
arwei
ghtofit
schl
ori
dei
s80.Theat
omi
c
weightofthe metali
s
(A)18 (
B)9 (
C)4.
5 (D)36

6. Thesulphat
eofanel
ementcont
ains42.
2%el
ement
.Theequi
val
entwei
ghtoft
hemet
alwoul
dbe:
(A)17.
0 (B)35.
0 (
C)51.
0 (
D)68.0

CENTERS:
MUMBAI/DELHI/AKOLA/KOLKATA/LUCKNOW /NASHI
K/GOA/PUNE# 19
7. Theequiv
alentwei
ghtofanel
ement
sis4.I
tschl
ori
dehasav
apourdensi
ty59.
25.Thent
hev
alency
ofthe el
ementsis
(A)4 (
B)3 (
C)2 (D)1

8. Theoxideofanelementpossessesthef
ormul
aM2O3.I
ftheequi
val
entwei
ghtoft
hemet
ali
s9,
then
theat
omi cwei
ghtofthemetalwill
be
(A)9 (B)18 (
C)27 (
D)noneoft
hese.

9. Approximat
eat
omi
cwei
ghtofanel
ementi
s29.89.I
fit
seq.wt
.is8.
9,theexactat
omi
cwt
.is:
(A)26.89 (
B)8.9 (C)17.8 (D)26.7

10. 0.534gMgdi spl

aces1.415gCufr
om t
hesal
tsol
uti
onofCu.Equival
entwei
ghtofMgi
s12.The
equival
entwei
ghtofCuwouldbe:
(A)15.9 (B)47.
7 (
C)31.
8 (D)8.0
11. Inm1 gram ofamet
alAdispl
acesm2 gr am ofanot
hermetalBfr
om it
ssaltsol
uti
onsandi ft
he
equi
valentwei
ght
sar
eE1andE2respect
ivelyt
hentheequi
val
entwei
ghtofAcanbeexpressedby

(
A) (
B) (
C) (
D)

12. Theweightofamet
alofequi
val
entwei
ght12,whi
chwi
ll
giv
e0.
475gofit
schl
ori
de,
is
(A)0.
12g (B)0.
16g (C)0.
18g (D)0.
24g
13. Theequiv
alentwei
ghtofir
oninFe2O3woul
dbe
(A)18.
6 (B)28 (
C)56 (
D)112.
0
14. 1.5gofadi
val
entmet
aldispl
aced4gofcopper(
at.
wt.=64)f
rom asol
uti
onofcoppersul
phat
e.The
atomicwei
ghtoft
hemetalis
(A)12 (
B)24 (C)48 (D)6

15. Equi
val
entwei
ghtofKMnO4wheni
tisconv
ert
edi
ntoMnSO4i
s
(A)M/5 (
B)M/ 6 (
C)M/3 (
D)M/
2

16. Theweightoft
woelement
swhi
chcombi
newit
honeanot herarei
nther
ati
ooft
hei
r
(A)at
omicweight (B)molecularwei
ght
(C)gr
am mole (D)equi
valentweight

17. 0.84gofamet
alhy
dri
decont
ains0.
042gofhydr
ogen.I
tsequi
val
entwei
ghti
s:
(A)80 (
B)40 (C)60 (
D)20

18. Amet
all
i
coxi
decont
ains60%oft
hemet
al.Theequi
val
entwei
ghtoft
hemet
ali
s
(
A)12 (B)24 (
C)40 (D)48

19. Whenamet
ali
sbur
nt,
itswei
ghti
sincr
easedby24per
cent
.Theequi
val
entwei
ghtoft
hemet
alwi
l
l
be
(A)2 (
B)24 (
C)33.
3 (
D)76.

20. 3gofanoxi deofamet alisconv

ert
edt
ochl or
idecompl
etel
yandi tyi
eld5gofchl
ori
de.The
equiv
alentweightoft
hemet al
is
(A)33.25 (B)3.
325 (C)12 (D)20.

21. Abi
val
entmet
alhast
heequi
val
entwei
ghtof12.Themol
ecul
arwei
ghtofi
tsoxi
dewi
l
lbe
(
A)24 (
B)34 (C)36 (
D)40

22. 2H3PO4+3Ca( OH)2 Ca3(

PO4)
2+6 H2O.
Equiv
alentwei
ghtofH3PO4int
hisr
eactioni
s
(A)98 (B)49 (C)32.
66 (
D)24.
5

CENTERS:
MUMBAI/DELHI/AKOLA/KOLKATA/LUCKNOW /NASHI
K/GOA/PUNE# 20
 HOMEASSI
GNMENT-3
1. Whati
stheequi
val
entmassofHCl
int
hegiv
enr
eact
ion:
2KMnO4+16HCl 2KCl +2MnCl
2+5Cl
2+8 H2O

(
a) (
b) (
c) (
d)noneoft
hese

2. Inanexperi
ment50mlof0.1M sol
uti
onofasaltr
eactedwit
h25mlof0.
1M sol
uti
onofsodi
um
sul
phi
te.Thehal
fequat
ionf
ort
heoxi
dati
onofsul
phi
teionis

I
ftheoxi
dat
ionnumberofmet
ali
nthesal
twas3,
whatwoul
dbet
henewoxi
dat
ionnumberofmet
al?
(
a)zero (b)1 (
c)2 (d)4

3. I
nthetit
rat
ionofK2Cr
2O7 andf
err
oussul
phate,f
oll
owingdatai
sobt
ained:V1mlofM1 K2Cr
2O7
r
equi
resV2mlM2FeSO4whichoft
hefol
l
owingrel
ati
onsaretr
ue:
(
a)6M1V1=M2V2 (b)M1V1=6M2V2
(
c)N1V1=2N2V2 (d)M1V1=M2V2

4. Anel
ementAi
nacompoundABDhasoxi
dat
ionnumber .I
tisoxi
disedby i
naci
dic
–3 –3
medium.Intheexper
iment1.
6810 mol eofK2Cr
2O7wer
eusedfor3.
2610 mol
eofABD.The
newoxidat
ionnumberofAafteroxi
dat
ioni
s
(a)3 (b)3–n (c)n–3 (
d)+n

5. Thenumberofmol
esofAs2S3oxi
disedt
oH3AsO4andH2SO4by1mol
eofHNO3(
reducedt
oNO)i
s
(
a) (
b) (
c) (
d)

6. Asoluti
oniscontai
ning2.52gl it
re–1ofareductant.25mLofthi
ssoluti
onrequir
ed20mLof0. 01M
KMnO4inacidmedi um foroxidati
on.Giventhateachofthet
woat omswhichunder gooxi
dat
ionper
molecul
eofreductant,
sufferanincreaseinoxidat
ionstat
ebyoneunit
.Themol .wt.ofr
educt
anti
s-
(a)M =126 (b)M =130 (c)M =128 (d)M =127

7. H2O2isreducedrapi
dlybySn2+toform Sn4+andH2O.100mlofH2O2solut
ioni
sreducedcompl
etel
y
by50ml of0.2M SnCl
2s ol
uti
on.Calcul
atethev ol
umestr
engt
hofH2O2.
(a)11.
2 (b)1.
12 (c)0.
112 (
d)noneofthese

8. I
n10mlofH2O2sol uti
on,excessofaci
dif
iedsoluti
onofKIisadded.Theiodi
nel
i
ber
atedneeds20
mlof0.1M Na2S2O3sol
uti
on.ThevolumestrengthofH2O2i
s
(
a)1.12 (
b)2.24 (c)3.24 (d)4.
44

2 +
O43+Cr+3
9. I
nt hereacti
on,P+Cr
2O7 +H  P +H2Oi
f0.
2mol
esof ar
etakent
henmol
es
of‘P’r
eactedis
(a)0.
34 (b)0.14 (
c)0. 24 (d)0.
04

10. I
nt hefol
l
owingr
eact
ion,
MnO2+4HCl MnCl
2+Cl
2+2H2O
nfact
orofHCli
s
(a)½ (
b)1 (c)2 (
d)4

11. Theequi
val
entwei
ghtofFe3O4i
nther
eact
ion,
Fe3O4+KMnO4 Fe2O3+MnO2wouldbe
(a)M/6 (
b)M (c)2M (
d)M/
3

CENTERS:
MUMBAI/DELHI/AKOLA/KOLKATA/LUCKNOW /NASHI
K/GOA/PUNE# 21
12. I
nthef
oll
owi
ngr
eact
ion,

equi
val
entwei
ghtofAs2S3(
molecul
arwei
ghtM)is
(a)M/2 (b)M/4 (c)M/24 (
d)M/
28

13. I
nar eact
ion,4moleofelect
ronsar
etr
ansf
erredtoonemoleofHNO3 wheni
tact
sasanoxi
dant
.
Thepossi
blereduct
ionpr
oductis
(a)(
1/2)moleN2 (b)(1/
2)moleN2O
(c)1moleofNO2 (d)1moleNH3

HOMEASSI
GNMENT-4
1. 0.5gofimpur eammoni um chl
ori
dewhenheat edwit
hexcessofNaOHgi vesammoniagaswhi
chi
s
absorbedin150mlofN/ 5H2SO4 sol
uti
on.Excessofaci
di scomplet
elyneut
ral
i
sedby20mlof1N
NaOHsol ut
ion.Thepercent
ageofammoni ainammoni um chl
ori
deis
(a)68% (b)34% (
c)48% (d)17%

2. 2.25gm mixt
ureofNa2CO3andNa2SO4aredissolv
edin250mlofsol ut
ion.25mlofabovesol
ution
requi
red20mlof0.1NH2SO4byusingphenol
phthalei
nindi
cat
or.ThemassofNa2CO3inthemixture
i
s
(a)1.
5gm (
b)2.12gm (
c)2.05gm (d)1.
6gm

3. 25mLofasol uti
onofNa2CO3havingaspecifi
cgr av
it 25gmL–1r
yof1. equi
red32.9mLofasol
uti
on
ofHClcontai
ning109.
5goft heacidperli
tr
ef orcompleteneut
ral
i
zati
on.Findthevol
umeof0.
84N
H2SO4thatwi
llbecomplet
elyneut
ral
izedby125gofNa2CO3sol uti
on.
(a)470mL (b)370mL (
c)530mL (d)280mL

4. A solut
ioncontai
ning4.2gofKOH andCa( OH)2i
sneutral
i
zedbyanacid.I
fitconsumes0.
1
equival
entofaci
d,composi
ti
onofsampl
einsol
uti
oniswhi
chofthefol
l
owi
ng?
(a)45%,55% (
b)35%,65% (
c)25%,75% (
d)15%,
85%

5. 100mlof10VH2O2soluti
onisheat
ed.Theevol
vedgasiscompl
etelyr
eact
edwit
hCat oform CaO.
Theaqueoussol
uti
onofCaOisneut
ral
isedby50mlofH2SO4sol
uti
on.Themol
ari
tyofH2SO4i
s
(a)2.
43M (b)1.
78M (
c)1.
55M (d)2.78M

6. Amixt
ureofNaHCO3andNa2CO3i
sneutral
i
sedbyxmlof1M HClbyusi
ngphenol
pht
halei
nindi
cat
or.
Theabovemixt
urei
sneutr
ali
sedbyymlof1M HClbyusingmet hy
lorangei
ndi
cat
or.Themoleof
CO2evol
vedonheat
ingt
heabovemixt
ureatlowert
emper
atur
eis
(
a) (
b) (
c) (
d)

7. Thenor
mal
i
tyof0.
1M H3PO3wheni
tunder
goesf
oll
owi
ngr
eact
ion,
H3PO3+2OH +2H2O
woul
dbe
(
a)0.1 (
b)0.
2 (
c)0.
3 (
d)0.
05

8. Theequi
val
entwei
ghtoft
hecompoundKHC2O4.
H2C2O4.
4H2Ousedi
nneut
ral
i
sat
ioni
s
(
a) (
b)

(
c) (
d)

CENTERS:
MUMBAI/DELHI/AKOLA/KOLKATA/LUCKNOW /NASHI
K/GOA/PUNE# 22
9. Thenormal
i
tyof0.
3M phosphor
ousaci
d(H3PO3)i
s
(a)0.
1 (b)0.9 (
c)0.3 (
d)0.
6

10. 0.52gofadi basi

caci
drequi
red100mlof0.1NNaOHofcompl
eteneutr
ali
sat
ion.Theequi
val
ent
weightofaci
dis
(a)26 (
b)52 (c)104 (d)156

11. Whatwoul
dbet
henor
malit
yofa0.
1M K2Cr
2O7sol
uti
onusedasapreci
pit
ati orPb2+?
ngagentf
(
a)0.1N (b)0.
6N (c)0.
4N (d)0.
2N
12. 3gofanoxi
deofametal i
sconver
tedt
ochl
ori
decompl
etel
yandi
tyi
elded5gchl
ori
de.
Theequi
val
entwei
ghtofmetali
s
(a)33 (b)42 (
c)12 (
d)40

13. Anaqueoussol
uti
onof6.3goxal
icaci
ddi
hydrat
eismadeupt
o250ml.Thev
olumeof0.
1NNaOH
requi
redt
ocomplet
elyneut
ral
i
ze10mloft
hissol
uti
oni
s
(a)40ml (
b)20ml (
c)1ml (d)4ml

14. I
nther
eact
ion,
H3PO4+Ca(OH)2 CaHPO4+2H2O,
theequi
val
entweightofH3PO4i
s
(
a)32.
7 (b)49.
0 (
c)98.
0 (d)196.
0

GetEqui
ppedf
orI
IT-JEE
Onl
yoneopt
ioncor
rect
:
1. MassofKHC2O4 (
pot
assi
um aci
doxalat
e)r
equi
redt
oreduce100mlof0.02M KMnO4 i
naci
dic
2+
medi
um (
toMn )isxg,andtoneut
ral
i
ze100mlof0.
05M Ca(
OH)2isygt
hen
A)x=y
( (B)2x=y C)x=2y
( (
D)nonei
scor
rect
2. I
fequal v
olumesof1M KMnO4and1M K2Cr
2O7sol
uti
onareallowedt
ooxi
dizeFe(
II
)toFe(
II
I)
,then
Fe(II
)oxidi
zedwil
lbe
(A)MorebyKMnO4 (B)MorebyK2Cr2O7
(C)Equalinbot
hcases (D)Dat
aisincomplete

3. 100mlof1M KMnO4oxi
dized100mlofH2O2i
naci
dicmedi
um (
whenMnO i
sr oMn2+)
educedt ;
v
olumeofsameKMnO4 requi
redt
ooxi
dize100mlofH2O2 i
nneut
ralmedi
um (
whenMnO i
s
r
educedt
oMnO2)wi
l
lbe
(
A) ml (
B) ml (
C) ml (
D)100ml
4. 10ml ofH2O2sol
uti
on(vol
umest
rengt
h=x)requi
red10ml
of0.
1NMnO sol
uti
oni
naci
dicmedi
um.
Hence,xis:
(A)0.
56 (
B)5.6 (
C)0.1 (
D)10.
0

5. 20mlofxM HClneutral
izescompletel
y10mlof0.1M NaHCO3andafurt
her5mlof0.
2M Na2CO3
solut
iontomet
hyl
orangeend-poi
nt.Thev ueofxi
al s
(A)0.167M (B)0.133M (C)0.
150M (D)0.
200M

6. 10mlofNaHC2O4soluti
onisneut
ral
i
zedby10mlof0.1M NaOHsol
ution.10mlofsameNaHC2O4
solut
ioni
soxi
dizedby10mlofKMnO4sol
uti
oni
naci
dicmedium.Hence,molar
it
yofKMnO4i
s
(A)0.1M (
B)0.
2M (
C)0.
04M (D)0.02M

7. 1moloff
err
icoxal
atei
soxi
dizedbyxmol
eofMnO andal
so1moloff
err
ousoxal
atei
soxi
dized
byymolofMnO i
naci
dicmedi
um.Ther
ati
ox/yi
s
(A)2:
1 (B)1:2 (C)3:1 (
D)1:
3

8. 40mlof0.
05M sol
uti
onofsesqui
car
bonate(Na2CO3.
NaHCO3.2H2O)i
sti
tr
atedagai
nst0.
05M HCl
.
xmlofHClisusedwhenphenolpht
hal
einistheindi
catorandymlofHCli susedwhenmet hy
l
CENTERS:
MUMBAI/DELHI/AKOLA/KOLKATA/LUCKNOW /NASHI
K/GOA/PUNE# 23
 orangei
sthei
ndi
cat
orint
wosepar
ati
ont
it
rati
on, y–x)i
hence( s
(A)80ml (
B)30ml (
C)120ml (
D)none

9. 3molofamixt
ureofFeSO4andFe2(
SO4)
3requi
red100mlof2M KMnO4sol
uti
oni
naci
dicmedi
um.
Hence,
molf
ract
ionofFeSO4i
nthemixt
ureis
(
A)1/3 (
B)2/3 (
C)2/5 (D)3/
5

10. 5.3gofM2CO3 isdissolv

edin150mlof1M HCl.Unusedaci
drequi
red100mlof0.
5M NaOH.
Hence,equi
val
entweightofM i
s
(A)53 (B)12 (
C)24 (
D)13

11. Equi
val
entwei
ghtofH3PO2wheni
tdi
spr
opor
ti
onat
esi
ntoPH3andH3PO3i
s
(mol
.wt.
ofH3PO2 =M)
(A)M (B)M/2 (
C)M/4 (
D)3M/
4

12. I
2o bt
ainedf
rom 0.
1molofCuSO4r
equi
red100mlof1M hy
posolut
ion,hence,molper
cent
ageof
pureCuSO4i
s:
(A)100 (
B)50 (C)25 (D)40

13. 1.2gofMgi streatedwit

h100mlof1M H2SO4.Molarconcent
rati
onoftheH2SO4 sol
uti
onaf
ter
complet
ereact
ionis:
(A)0.
05M (B)0.
005M (C)0.50M (
D)0.0005M

14. Volumeof18.
0M H2SO4requi
redt
opr
epar
e1.0li
tr
eofa0.
9M sol
uti
onofH2SO4i
s:
(A)50.
0ml (B)10.
0ml (C)500.
0ml (D)5.
0ml

15. Volumeof0.
50M NaOHsoluti
onr
equi
redt
oreactwit
h40.
0ml
of0.
05M H2SO4sol
uti
oni
s:
(A)40.
0ml (
B)80.0ml (C)20.
0ml (D)8.
0ml

16. 150ml of6.

00M H2SO4sol
uti
oni
smi
xedwi
th250mlof3.
00M H2SO4.Resul
ti
ngmol
ari
tyi
s:
(A)4.
125M (
B)8.
250M (C)4.
500M (
D)1.650M

17. Oxal
i
caci
d(H2C2O4)f
ormst
woser
iesofsal
tHC2O andC2O .I
f0.
9gofoxal
i
caci
disi
n100ml
solut
ion,
HC2O andC2O havenormal
i
tyr
espect
ivel
y:
(A)0.1N,0.
1N (B)0.
1N,0.
2N (C)0.
2N, 0.
2N (
D)0.
2N,
0.1N

18. 10gofMnO2 onreact

ionwi
thconc.HCll
i
berated0.1equi
val
entofCl
2(Mn=55)
.Hence,per
cent
puri
tyofMnO2i
s:
(A)87.
0 (
B)21.
75 (C)50.0 (
D)43.
5

19. 0.106gofNa2CO3complet
elyneut
ral
i
zes40.
0mlofH2SO4.Hence,
normali
tyofH2SO4sol
uti
oni
s:
(A)0.05N (
B)0.025N (
C)0.
10N (D)0.
20N

20. Volumeof0.
02M MnO solut
ionr
equi
redt
ooxidi
ze40.
0ml 1M Fe2+sol
of0. uti
oni
s:
(A)200ml (B)100ml (C)40ml (D)20ml

21. A20.0mlsolut
ionofNa2SO3r equi
red30mlof0.
01M K2Cr
2O7sol
uti
onfort
heoxi
dat
iont
oNa2SO3.
Hence,mol
ari
tyofNa2SO3soluti
onis:
(
A)0.015M (
B)0.045M (
C)0.
030M (D)0.
0225M

22. 1.00Lof0.
15M NaOHabsorbed11.
2mmol
ofCO2fr
om ai
r.Hence,newmolar
it
yofNaOHi
s:
(A)0.
1276M (
B)0.1500M (
C)0.
0224M (
D)0.0112M

23. I
fagm ist
hemassofNaHC2O4r
equi
redtoneutr
ali
ze100mLof0.
2M NaOHandbgm t
har
equi
red
t
oreduce100mLof0.
2M KMnO4inaci
dicmedium,then
(
A)a=b (
B)2a=b (C)a=2b (D)Noneoft
hese

24. Ami
xtur
eofNa2C2O4(
A)andKHC2O4.
H2C2O4(
B)r
equi
redequal
vol
umeof0.
1M KMnO4and0.
1M
CENTERS:
MUMBAI/DELHI/AKOLA/KOLKATA/LUCKNOW /NASHI
K/GOA/PUNE# 24
 NaOHsepar
atel
y.Mol
arr
ati
oofAandBint
hemi
xtur
eis
(
A)1:1 (
B)1:
5.5 (C)5.
5:1 (
D)3.
1:1

25. AnexcessofNaOHwasaddedt o100mLofaf er

ricchl
ori
desoluti
on.Thiscausedt
hepr
eci
pit
ati
on
of1.425gofFe(
OH)
3.Cal
culat
e t
he nor
mal
it
y oft
h efer
ri
c chl
ori
desolut
ion
(A)0.20N (
B)0.50N (C)0.25N (D)0.
40N
26. 0.4gofapolybasi
caci
dHnA( al
lthehy dr
ogensareaci
dic)r
equir
es0.5gofNaOHforcompl
ete
neutral
i
zati
on.Thenumberofreplaceabl
ehydrogenat
omsi ntheacidandthemol
ecul
arwei
ghtof
'
A' wouldbe:(Mol
ecul
arweightoftheacidis96gms.)
(A)1,95 (B)2,
94 (C)3,
93 (D)4,92
27. Asol
uti
onofNa2S2O3i
sstandardi
zediodi
metri
call
yagai
nst0.
1262gofKBr O3.Thi
spr
ocess
r
equi
res45.
mLoft heNa2S2O3sol
uti
on.Whatisthest
rengt
hoftheNa2S2O3?
(
A)0.
2M (B)0.1M (C)0.05M (
D)0.1N
28. 25.0gofFeSO4.
7H2Owasdi ssol
vedinwatercontai
ningdi
l
uteH2SO4,andthevolumewasmadeup
to1.0L.25.
0mLoft hi
ssoluti
onrequi
red20mLofanN/ 10KMnO4sol ut
ionforcompl
eteoxi
dat
ion.
Thepercent
ageofFeSO4.
7H2Oi ntheaci
dsoluti
onis
(A)78% (
B)98% (C)89% (D)79%

29. 1.0molofFereactscompletel
ywith0.
65molofO2t
ogi
veami
xtur
eofonl
yFeOandFe2O3.The
molerat
iooff
errousoxidetoferr
icoxi
dei
s
(A)2:
3 (B)4:3 (
C)1:2 (
D)2:7
30. 25mLofasol ut
ioncontainingHCl andH2SO4r equired10mLbfa1NNaOHsol uti
onfor
neutral
izati
on.20mLoft hesameaci dmixtur
eonbei ngtreat
edwi thanexcessofAgNO3gi v
es
0.1435gofAgCl .Thenormal i
tyoftheHCl andt henor mali
tyoftheH2SO4ar erespecti
vel
y
(A)0.40Nand0. 05N (
B)0. 05Nand0. 35N
(C)0.50Nand0. 25N (
D)0. 40Nand0. 5N
31. 0.70gofmi xt
ure(NH4)2SO4wasboi l
edwith100mLof0. 2NNaOHsol ut
ionti
llall
theNH3(g)
evolvedandgetdi ssol
vedi nsolut
ionitsel
f.Ther emainingsoluti
onwasdi lut
edt o250mL.25mLof
thissol
ut i
onwasneut r
alizedusing10mLofa0. 1NH2SO4sol ut
ion.Thepercentagepuri
tyofthe
(NH4)2SO4sampl eis
(A)94.3 (B)50. 8 (
C)47. 4 (
D)79.8
32. Ami xedsolut
ionofpotassi
um hydroxi
deandsodi um carbonat
er equir
ed15mLofanN/20HCl
solut
ionwhent i
tr
atedwithphenoi
phthal
einasani ndi
cator.Butthesameamountofthesol
uti
on
whent i
tr
atedwithmethylorangeasanindicatorr
equired25mLoft hesameaci
d.Theamountof
KOHpr esenti
nthesoluti
oni s
(A)0.014g (
B)0.14g (C)0.
028g (D)1.4g

COMPREHENSI
ONTYPE
Passage1
n–f actorofaci
ds,basesandsal t
sdependupont henatureoft hereacti
ons.Forr edoxreacti
ons,n–factori
s
nothingbuti ti
st henumberofmol esofel ectronslostorgai nedi nagi vennumberofmol esofact iv
e
elementi n1mol esofr educi
ngagentoroxi disingagentr especti
vely.Forredoxr eact
ion,n–factormaybe
fracti
onalquantit
y,buttheno.ofelect r
onwillnotbeaf racti
onalquant i
ty.Therebythen–f act
orofar edox
reactionisnotthenumberofel ectr
onl ostorgai ned.Themol arrati
ooft her eact
ingspecies,aproblemati
c
part,isdeter
minedbyt hehelp
ofn–f act
or.Ther eciprocalofn–f actor
’sr atiooft her eact i
ngspeci esi sthemol arratiooft he
r
eact
ingspecies,andsi
mil
arl
yfort
hepr
oduct
.Thr
ought
hen–f
act
or,wecal
cul
ateequi
val
ent
wei
ghtasunder:

Equi
val
entwt
.=

1. Fort
her
eact
ion,

CENTERS:
MUMBAI/DELHI/AKOLA/KOLKATA/LUCKNOW /NASHI
K/GOA/PUNE# 25
 H3PO4+NaOH NaH2PO4+H2O
Whatwi l
lbetheequi
val
entwt .ofH3PO4.(
H–1, P–31, O–16)
?
(A)96 (B)49 (C)32.66 (D)98
2. Forthereact
ion:Zn+K4[Fe(CN)6] K2Zn3[
Fe(CN)6]
2+K,Whatwil
lbet
heequi
val
entwei
ghtof
K4[Fe(
CN)6]
,i
fthemolecularweightofK4[
Fe( CN)6]i
sM?
(A)M/2 (B)M (C)M/ 3 (D)M/4

3. Forthefoll
owingbalancedredoxr
eact
i aFeS2+bO2
on, x(
Fe2O3)+y
SO2Whati
sthev
alueofa
andbr espect
ivel
y?
(A)5,12 (B)11,4 (
C)5,11 (
D)4,
11

4. I
nt heabov
equest
ionnumber3,whati
stheequi
val
entwei
ghtofFeS2,ift
hemol
ecul
arwei
ghtof
FeS2isM?
(A)M/5 (
B)M/9 (
C)M/15 (D)M/ 11

5. I
ntheabov
equest
ion3,whatwi
l
lbet
heequi
val
entwei
ghtofSO2,
ift
hemolecul
arwei
ghtofSO2i
s
M?
(
A)M/5 (
B)M/ 9 (C)M/15 (
D)M/11

6. Fortheoxi
dat
ionreaction:Fe0.95O Fe2O3,i
fthemol
ecul
arwei
ghtofFe0.95ObeM,t
hent
he
equi
val
entwei
ghtofFe0.95Owil
lbe
(A)M (B)M/ 2 (
C)M/
0.95 (
D)M/
0.85

7. I
nr edoxreacti
on, Ba(MnO4)2
oxidizesK4[Fe(CN)6
]i
ntoK+,Fe+3,
CO3–2 andNO3– i
onsinacidi
cmedium,
whereBa( MnO4)2
i
tsel
freducesi ntoMn ,t
+2
hanhowmanymol esofBa( MnO4)2
wi
l
lreactwi
th1mole
ofK4[Fe(
CN)6
]?
(A)5mol es (B)9mol es (C)8mol es (D)6.1mol es
Passage2

Compoundscont ai
ningi
odi
near
ewi
delyusedi
ntit
rat
ions,commonl
yknownasi
odi
net
it
rat
ions.
I
tisoftwokinds:
(i
)Iodomet
ri
ct i
tr
ati
ons (ii
)Iodi
metr
icti
tr
ations

I
odomet r
ict
it
r at
ions:I
tisnothingbutani ndi
rectmet hodofesti
matingt
hei odi
ne.Inthi
sty
peof
t
it
rati
on,anoxidi
singagenti
smadet or eactwithexcessofKI,
inacidi
cmedium or,basi
cmedi
um i
n
whi –
chI oxi
disesintoI
2.Nowt heli
beratedI2canbet i
trat
edwit
hNa2S2O3soluti
on.
KI I –+NaSO
I
2 24 6
Al
thoughsol
i
dI2isbl
ackandi
nsol
ubl
e i
nwater
,buti
tconv
ert
sint
osol
ubl
eI3i
ons.

Starchisusedasi ndi
catorneartheendpoi ntorequiv
alencepoint.Evensmal
lamountofI 2

mol ecul
es,givesbl
uecolourwithstar
ch.Thecomplet
ionofthereact
ioncanbedet
ectedwhenbl
ue
colourdisappearsattheendpoint
.

I
niodi
met
ri
cti
tr
ati
on,
thest
rengt
hofr
educi
ngagenti
sdet
ermi
nedbyr
eact
ingi
twi
thI
2
.

1. When79.
75gofCuSO4 sampl
econtai
ningi
nertimpur
it
yisreactedwit
hKI ,t
heli
ber
atedI
2
i
sreact
ed
wi
th50mL(1M)Na2S2O3 i
nbasicmedium,whereitoxi
disesintoSO4–2 i
ons,andI
2
r
educesint
oI–
,
t
henwhatwi
l
lbethe%pur i
tyofCuSO4i
nsampl e?
(
A)60% (
B)80% (
C)50% (
D)95%

CENTERS:
MUMBAI/DELHI/AKOLA/KOLKATA/LUCKNOW /NASHI
K/GOA/PUNE# 26
2. When214gofKI O3 r
eact
swi t
hexcessofKIi npresenceofH+ thenitproducesI2
.Now I 2
i
s
complet
elyr
eact
edwi t
h1M Na2S2O3 sol
uti
oninbasicmedi um,whereitconver
tsint
oSO4 i
–2
ons.
Thenwhatvol
umeofNa2S2O3i
sneededtoreacttheendpointoft
hereact
ion?
(A)500mL (
B)800mL (C)1500mL (D)750mL
3. When1.66gofKIi sreactedwi t
hexcessof i
nt hepr esenceof6NHCl,t
henI
Cli
spr
oduced.The
amountofKIO3reactedandt heICl
formedarerespectively.
(A)4×10 mol
–2
e,2×10 mol
–3
e (B)1.5×10 mol
–2
e,5×10–3mol
e
(C)5×10 mol
–2
e,1.5×10 mol
–3
e (D)5×10 mol
–3
e,
1.5×10–2mol
e

4. I
nor dertoest
imat
eO3 intheair,
22.
4Lofai ratNTP, ispassedthr
oughaci
dif
iedKIsol
utionwhere
O2 i
sev ol
vedandI–
i
soxidisedtoI
2
.Theli
b er
atedI2
requi
red200mLo f0.1M Na S O
2 2 3
i
naci
dic
medium, t
henwhatwillbethe%composit
ionofO3byv ol
umei nt
hesample?
(
A)50% (B)10% (C)0.1% (
D)1%

Passage3

H2O2act sasbot hoxidi

singaswel l
asr educingagent.Asoxidisingagent,i
tsproducti
sH2O,
butasr educingagent ,it
spr oductisO2.Thest r
engthofH2O2 i sexpressedinmanymay s
l
ikemol ari
ty,normalit
y,% st rengthandv olumest rengt
h.Butoutofal ltheseform of
strengths, vol
umest rengthhasgr eatsignif
icanceforchemical
reactions.Thedecomposi ti
onofH2O2i sshownasunder :
H2O2( l
)— H2O( l
)+1/ 2O2(g)
x“v ol”strengthofH2O2means1v olume( l
it
reormL)ofH2O2r eleasesxv ol
ume( l
it
reormL)
ofO2atNTP.
Per centagest r
engthofH2O2 i snothingbut,itist
hewei ghtofH2O2i ngm per100mLof
H2O2sol ut i
on.Byvolumest rengthofH2O2, wecanpr edictt
heper centagepuri
tyofthe
subst anceandmanymor ei
nf ormationcanbeknown.

1. Whati
stheper
cent
agestrengt
hof10“
vol
”H2O2?
(
A)11% (B)5% (
C)7% (
D)3.
03%

2. 55gBa( MnO4)
2
samplecontaini
nginer
timpuri
tyi
scompletel
yreact
ingwit
h100mLof56“vol
”of
H2O2,
thenwhatwi l
lbethepercent
agepuri
tyofBa(
MnO4)
2
i
nthesample?(Ba–137,
Mn–55)
(A)40% (
B)25% (
C)10% (D)68.
18%

3. When100mLsol
uti
onofH2O2 r
eact
scompl
etel
ywi
thKI
,theni
toxi
disesKIi
ntoI
2
.Now l
i
ber
ated
i
odiner
equired25mLof1M Na2S2O3 i
nbasi
cmedium,
wher
eitoxi
disesi
nto ,
thenwhati
sthe
v
olumestrengthofH2O2?
(
A)5.6 (B)8.
6 (C)11.
2 (D)None

4. A100mLH2O2 l
i
ber
ates25.4gofI
2
f
rom anaci
dif
iedKIsol
uti
on,t
henwhati
sthev
olumest
rengt
h
ofH2O2?
(A)5 (B)5.
6 (C)10 (
D)11.
2

5. Whatv
olumeofH2O2 sol
uti
onof22.
4“v
ol”st
rengt
hisr
equi
redt
oli
berat
e2240mLofO2 atNTP?
(
A)300mL (
B)200mL (C)500mL (D)100mL

6. A840mLsampl
eofH2O2 sol
uti
oncont
aini
ngy“
vol
”st
rengt
hofH2O2 r
equi
resymLofAl
(MnO4)
3
i
n
CENTERS:
MUMBAI/DELHI/AKOLA/KOLKATA/LUCKNOW /NASHI
K/GOA/PUNE# 27
 acidi
cmedi
um,
thenwhatwi
l
lbet hemol
ari
tyofAl
(MnO4)
3
sol
uti
on?
(A)7M (
B)5M (C)10M (D)15M

7. 100mLofaH2O2 sol
uti
onweredilut
edto400mL,and20mLoft hi
sdil
utedsol
uti
onr
equi
redt
o
reduce10mL0.
1M KMnO4 sol
uti
onthenwhati
sthe“
vol
ume“st
rengt
hofH2O2?
(A)20 (B)11.
2 (C)8.4 (D)5.
6

MATRI
XMATCH
1.Col
umnI Col
umnI
I(React
ion)
(
Eq.Wt.ofReact
ant
)
(A) Eq.Wt=M. Wt
/33 1. WhenCr
I3oxi
desesi
nto and .
(
B) Eq.
Wt=M.
Wt/
27 2. WhenFe(
SCN)
2oxi
disesi
nto , and .
(
C) Eq.
Wt=M.
Wt/
61 3. WhenNH4SCNoxi
disest
o , and .
(
D) Eq.
Wt=M.
Wt/
24 4.When oxi
disesi
ntoK+,
Fe+3, and .

2.Col
umnI ColumnII
(Eq.Wt
.ofunder
li
nedspeci
es)
(
A)  PH3+P4H2
1.

(
B)
2.

(
C)
3.

(
D)
4.

3. Consi
dert
her
eact
ionofBr2wi
thKOHasshownbel
owandmat
cht
hef
oll
owi
ng:
Br2+KOH KBr+KBrO3+H2O

ColumnI(Compound) Col
umnII(
n-Fact
or)
(A) KOH 1. 1
(B) KBr 2. 5
(C) KBrO3 3. 5/3
(D) Br2 4. 5/6
5. 2

ASSI
GNMENT(
SUBJECTI
VE)

1. Cal
cul
atetheoxi dati
onnumber
soft
hef
oll
owi
ng:
i
) Oi nO2F2
i
i) Ni nNH2OH
i
ii
) CinHCN
i
v) FeinK4[ Fe(CN)6]
v) IinHIO3
vi
) Mni nMnO42-
vi
i) CrinCr O2Cl2
vi
ii
) Pi nCa3( PO4)2
i
x) PinP2O74-
x) CrinCr O5
CENTERS:
MUMBAI/DELHI/AKOLA/KOLKATA/LUCKNOW /NASHI
K/GOA/PUNE# 28
 xi
) Fei
nFe(CO)
5
xi
i
) Cri
n[Cr
(NH3)
6]
Cl3

2. Bal
ancet
hefol
l
owingequati
onsbyoxi
dat
ionnumbermet
hod:
-
i
) Cu+HNO3  Cu(NO3)2+NO+H2O
i
i) Zn+HNO3 Zn( NO3)2+N2O+H2O
i
ii
) KMnO4+NH3 KNO3+MnO2+KOH+H2O
i
v) S+HNO3  H2SO4+NO2+H2O

3. Bal
ancethef ol
lowingequat
ionsbyIon-
elect
ronmethodi
naci
dicmedi
um.
2- + 3+
i
) I2 +Cr2O7 +H  C r O3-+H2O
+I
i
i) N2O4 +Br O3-  NO3-+Br-
Cu2O+H +NO3- Cu2++NO+H2O
+
i
ii
)
i
v)  +MnO2
v) KCl O3+H2SO4 KHSO4+HCl O4+Cl O2+H2O

4. Bal
ancethefol
lowi
ngequat i
onsusi
ngbasi
cmedi
um :
i
) C2H5OH+MnO4-  C2H3O-+MnO2 +H2O
i
i) HNO3+AS2S5 H2SO4+H3AsO4+NO2
i
ii
) H2O2 +PbS PbSO4+H2O
i
v) Fe3O4 +MnO4-  Fe2O3+MnO2
v) S+OH-  S2-+S2O32-

5. Gi
vet
henf
act
oroft
heunder
li
nedr
eact
antmol
ecul
esi
nthegi
venr
eact
ions
(
i)
(
ii
)
(
ii
i)
(
iv)
(
v)
6. Acer
tai
namountofmetalwhoseequi
val
entwei
ghti
s2,
displ
aces0.
7li
tr
eofhy
drogenmeasur
edat
NTPfr
om anaci
d.Cal
cul
atethewei
ghtofthemet
al.

7. 9.
44gofmetaloxi
dei
sfor
medbyt
hecombi
nat
ionof5goft
hemet
al.Cal
cul
atet
heequi
val
ent
wei
ghtoft
hemet
al.

8. 14.7gofsulphur
icaci
dwasneededtodissol
ve16.
8gofmet
al.Cal
cul
atet
heequi
val
entwei
ghtof
themetalandthevol
umeofhy
drogenl
iberat
edatNTP.

9. Thesal
tNa2HPO4isformedwhenor
thophosphor
icaci
disr
eact
edwi
thanal
kal
i
.Fi
ndt
heequi
val
ent
wei
ghtofort
hophosphori
caci
d.

10. 0.
501gofsilv
erwasdissolv
edinnit
ri
cacidandHClwasaddedt ot
hissolut
ion,AgClsofor
med
weighed0.
6655g.I
ftheequi
val
entweightofchl
ori
nei
s35.
5,whatwi
llbetheequival
entwei
ghtof
si
l
ver?
11. 5gofzincdi
spl
aced4.846gofcopperf
rom acoppersul
phatesol
uti
on.I
fzi
nchas
equi
val
entwei
ghtof32.5,f
indt
heequiv
alentwei
ghtofcopper?

12. Onegram oft

heacidC6H10O4 r
equir
es0.
768gofKOH f
orcompl
eteneut
ral
i
sat
ion.How many
neut
ral
i
sabl
ehy
drogenat
omsar einthi
smol
ecul
e?

13. 0.
2M KMnO4 sol
uti
oncomplet
elyr
eact
swith0.
05M FeSO4 sol
uti
onunderaci
dic condi
ti
ons.The
vol
umeofFeSO4usedis50ml.Whatv
olumeofKMnO4wasused?

CENTERS:
MUMBAI/DELHI/AKOLA/KOLKATA/LUCKNOW /NASHI
K/GOA/PUNE# 29
14. 20gofasampleofBa(OH)2{
Mol
.Wt=171}isdi
ssol
vedi
n10ml.of0.
5NHClsol
uti
on.Theexcessof
HClwast
it
rat
edwith0.2NNaOH.Thev
olumeofNaOHusedwas10cc.Cal
cul
atetheper
centageof
Ba(
OH)2i
nthesample?

15. Asol
uti
onofKHC2O4.
H2C2O4.
2H2Oi
s0.
2Nasanaci
d.Whati
sit
snor
mal
i
tyasr
educi
ngagent
?

16. Anaqueoussol uti

oncontaini
ng0.10gKI O3 (
formulawt=214.0)wast reat
edwit
hanexcessofKI
sol
uti
on.Thesol uti
onwasaci di
fiedwithHCl .Theliberat
edI2consumed45mloft hi
osulphat
e
sol
uti
on to decolour
ise t
he bl
ue-st
arch-
iodine complex.Cal
cul
atet he mol
ari
tyoft
he sodium
thi
osul
phat
esol ut
ion?

17. A10gsampl eofonl yCuSandCu2Swast reat

edwi t
h100mlof1. 25M K2Cr2O7.Theproduct
s
3+
obt
ainedwereCr ,Cu2+andSO2.Theexcessoxidantwasreact
edwih50mlofFe2+sol
t uti
on.25ml
hesameFe2+sol
oft uti
onr equir
ed0.875M KMnO4underacidi
ccondi
ti
on,
thevol
umeofwhi chused
was20ml .Fi
ndthe%ofCuSi nthesampl
e?

18.100ml sol
uti
onofFeC2O4andFeSO4iscompl
etel
yoxi
dizedby60ml of0.02M KMnO4inaci
dmedium.
Theresul
tingsol
uti
onisthenreducedbyZnanddil.HCl.Ther educedsolut
ionisagainoxi
dized
complet
elyby40ml of0.
02M KMnO4.Cal
cul
atemolari
tyofFeC2O4andFeSO4i nthesol
uti
on?

19. Someamountofcommer cialAgNO3of50%pur i

tyisdissol
vedin50mlofwat er.I
tistr
eatedwith50
mlofKIsol uti
on.Thesilveri
odidethuspreci
pitat
edi sf
il
ter
edoff.ExcessofKIi nthefil
tr
atei s
tit
rat
edwith50mlofM/ 10KIO3 sol
uti
oninpresenceofconc.HClwhenal lt
heiodineisconvert
ed
intoI
Cl.25ml oft
hesamest ocksolut
ionofKIrequi
res30ml ofM/10KIO3undersimil
arcondit
ions.
Calcul
atetheamountofAgNO3t aken?

20. A0. 506gsamplecont

ai ni
ngBa(SCN)2wasdi
ssolvedinabicarbonatesol
uti
on.50mlof0. 1Ni odi
ne
wasaddedandt hemi xtur
ewasal l
owedtostandforfiveminutes.Thesoluti
onwast henacidi
fi
ed
andt heexcessofI2 wa sti
tr
atedwith20mlof0. 1M sodi um thi
osul
phatesoluti
on.Oxidati
on
pr sofSCN– ar
oduct eSO42– andHCN.Calcul
atethepercent
ageofBa( SCN)2inthesample?

21. 100mlofasol uti

onofH2O2 reactswithexcessofKMnO4 solut
ioninacidicmedium andl i
ber
ates
3.294LofoxygenatSTP.Calcul
atethenormali
tyandvolumestrengt
hoft heH2O2sol
ution.Whatis
theamount(i
ng)ofKMnO4t hatreact
swiththeH2O2intheabovereact
ion?

22. Hydrogenperoxidesolut
ion( 20ml)react
squantitat
ivelywit
hasol uti
onofKMnO4 (20ml )acidi
fi
ed
withdilut
eH2SO4.Thesamev ol
umeoft heKMnO4sol uti
onisjustdecolouri
sedby10mlofMnSO4
inneutralmedium simultaneousl
yformingadar kbrownpr ecipi
tateofhy dr
atedMnO2.Thebr own
precipi
tat
eisdissol
vedin10mlof0. 2M sodium oxal
at eunderboil
ingconditi
onsi
nthepresenceof
dil
uteH2SO4.Calcul
atethemol ari
tyofH2O2?

23. 50ml ofasol uti

oncont ai
ning1geachofNa2CO3,NaHCO3andNaOH, wasti
tr
atedwi
th1NHCl
.What
wil
lbet heti
t r
ereadingsif
(
a)onlyphenol pht
haleini
susedasindi
cator
?
(
b)onlymet hylorangeisusedasindi
catorf
rom t
heverybegi
nning?
(
c)met hylorangeisaddedaf t
ert
hefi
rstendpoi
ntwithphenol
phthal
ein?
24. 25.0 cm3 ofa soluti
on contai
ning Na2CO3 and NaOH r equied 10 cm3 of1.
r 0 M H2SO4 for
neutr
alizat
ionwhenphenolpht
haleinwasusedast hei ndi
cator.Whereaswhenmethy
lorangewas
used,itrequi
red15cm3 ofthesameaci df orneut
ral
izati
on.Calcul
atethemassesofNa2CO3 and
3
NaOHperdm oft hesol
uti
on?
25. A soluti
oncont ai
nsNa2CO3 andNaHCO3.10mLoft hi
sr equi
res2.0mLof0. 1M H2SO4 f
or
neut
rali
zati
onusingphenolphthal
einasindicat
or.Methy
lorangeisthenaddedwhenafurt
her2.5
mLof0. 2M H2SO4wasrequir
ed.Calcul
atethestrengt
hofNa2CO3andNaHCO3insol
uti
on?

CENTERS:
MUMBAI/DELHI/AKOLA/KOLKATA/LUCKNOW /NASHI
K/GOA/PUNE# 30
 ANSWERKEY
HOMEASSI
GNMENT-I

1.(C) 2.(C) 3.(A) 4.(B) 5.(B) 6.(B) 7.(B)

8.(C) 9.(B) 10.(B) 11.(D) 12.(C) 13.(D) 14.(B)
15.(A) 16.(B)

HOMEASSI
GNMENT-I
I

1.(A) 2.(B) 3.(C) 4.(C) 5.(B) 6.(B) 7.(B)

8.(C) 9.(D) 10.(C) 11.(A) 12.(A) 13.(A) 14.(B)
15.(A) 16.(D) 17.(D) 18.(A) 19.(C) 20.(A) 21.(D)
22.(C)

HOMEASSI
GNMENT-I
II

1.(
C) 2.(
C) 3.(A) 4.(B) 5.(D) 6.(A) 7.(
B)
8.(
A) 9.(
C) 10.(A) 11.(B) 12.(D) 13.(B)

HOMEASSI
GNMENT-I
V

1.(
B) 2.(
B) 3.(A) 4.(B) 5.(B) 6.(B) 7.(B)
8.(
C) 9.(
D) 10.(B) 11.(D) 12.(A) 13.(A) 14.(B)

GetEqui
ppedf
orI
IT-JEE
ONLYONEOPTIONCORRECT
1.(B) 2.(B) 3.(B) 4.(A) 5.(C) 6.(C) 7.(A)
8.(A) 9.(A) 10.(A) 11.(D) 12.(A) 13.(C) 14.(A)
15.(D) 16.(A) 17.(B) 18.(D) 19.(A) 20.(C) 21.(B)
22.(A) 23.(D) 24.(C) 25.(D) 26.(C) 27.(B) 28.(C)
29.(B) 30.(B) 31.(A) 32.(A)

COMPREHENSI
ONTYPE
1.(
D) 2.(
C) 3.(
D) 4.(
D) 5.(
A) 6.(
D) 7.(
D)
1.(
B) 2.(
D) 3.(
D) 4.(
D)
1.(
D) 2.(
D) 3.(
C) 4.(
D) 5.(
D) 6.(
C) 7.(
D)

MATRI
XMATCH
1.A–2;
B–1;
C–4;
D–3 2.A–4;
B–2;
C–3;
D–1
3.A–4;
B–1;
C–2;
D–3

ASSI
GNMENT(
SUBJECTI
VE)
1. i
) +1 i
i)-1 i
ii
)+2 i
v) +2
v
) +5 vi
)+6 vi
i
)+6 vi
i
i) +5
i
x)+5 x) +6 xi
)0 xi
i) +3

2. i
) 3Cu+8HNO3 3Cu(NO3)2 +2NO+4H2O
i
i
) 4Zn+10HNO3 4Zn(NO3)2 +N2O+5H2O

CENTERS:
MUMBAI/DELHI/AKOLA/KOLKATA/LUCKNOW /NASHI
K/GOA/PUNE# 31
 i
i
i) 8KMnO4+3NH3  3KNO3 +8MnO2+5KOH+2H2O
i
v) S+6HNO3 H2SO4+6NO2+2H2O

2-
4H+ 10Cr3+
3. i
) 3I
2+5 Cr
2O7 +3 O3-+17H2O
+6I
- - -
i
i
) 3N2O4+BrO3 +3H2O 6NO3 +Br +6H+
i
i
i) 3Cu2O+14H++2NO3–  6Cu2++2NO+7H2O
i
v) 3Mn +4H+ 2Mn +MnO2+2H2O
v
) 3KCl
O3+3H2SO4 3KHSO4+HCl
O4+2Cl
O2+H2O

4. i
) 3C2H5OH+2MnO4- +OH- 3C2H3O-+2MnO2+5H2O
i
i
) 40HNO3+As2S5 5H2SO4+2H3AsO4+40NO2+12H2O
i
i
i) 4H2O2+PbS PbSO4+4H2O
i
v) 6Fe3O4+2MnO4-+H2O 9Fe2O3 +2MnO2+2OH-
v
) 4S+6OH- 2S2-+S2O32-+3H2O

5. (
i)1 (
ii
)20 (
ii
i)8 (
iv)8 (
v)1

6. 0.
125g 7. 9.
01g 8. 56g,
3.36L

9. 49g 10. 108g 11. 31.

5g

12. 2 13. 2.
5ml 14. 1.
28%

15. 0.
267M 16. 0.
063M 17. 57.
4%

18. FeC2O4=102M,
FeSO4=3102M

19. 0.
68g

20. 12.
5%

21. N=2.
94,
Vol
umest
rengt
h=16.
464, =9.
29g

22. 0.
1M

23.(
a)34.
4ml (
b)55.
8ml (
c)21.
3ml

24. Na2CO3=42.
4g,
NaOH=16g

25. Na2CO3=4.
24g/
l,NaHCO3=5.
04g/
l

CENTERS:
MUMBAI/DELHI/AKOLA/KOLKATA/LUCKNOW /NASHI
K/GOA/PUNE# 32