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CHEMISTRY
CET - 2014
C-1
ANSWERS
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1.78 g of an optically active L-amino acid (A) is treated with NaNO2/HCl at 0 °C. 448 cm3 of nitrogen was
at STP is evolved. A sample of protein has 0.25% of this amino acid by mass. The molar mass of the
protein is
1) 34,500 g mol–1
2) 35,600 g mol–1
3) 36,500 g mol–1
4) 35,400 g mol–1
Ans (2)
Solution : Mass of L amino acid = 178g
R
CH
COOH
NH2 + HNO2
1 mol. mass
?
448 cm3  1.78g
22400  ?
N2
1 mole of N2 = 22400 cm3 at STP
448
22400  1.78  102 178

 89
448
2
Sample of proteins contains 0.25% amino acid
Protein
amino acid
100
0.25
?
89
100  89
 35600g / mol
0.25
2.
10 g of a mixture of BaO and CaO requires 100 cm3 of 2.5 M HCl to react completely. The percentage of
calcium oxide in the mixture is approximately (Given : molar mass of BaO = 153)
1) 55.1
2) 47.4
3) 52.6
4) 44.9
Ans (3)
Solution : Consider the mass % of CaO is 55.1% as given in option 1
Then mass of CaO = 5.51 and BaO = 4.49
5.51
28
4.49
No. of g and of BaO =
76.5
No. of g. eq of CaO =
The sum of gram equivalents should be equal to
100  2.5
 0.25
1000
For the option (3) which is correct. Mass of CaO = 5.26g and mass of BaO= 4.74g
i.e. No. of gram equivalent of CaO =
No. of gram equivalent BaO =
5.26
 0.1878
28
4.74
 0.0619
76.5
Total no. of gram equivalent = 0.1878 + 0.0619 = 0.25
1
3.
The ratio of heats liberated at 298 K from the combustion of one kg of coke and by burning water gas
obtained from kg of coke is (Assume coke to be 100% carbon.)
(Given enthalpies of combustion of CO2, CO and H2 as 393.5 kJ, 285 kJ, 285 kJ respectively all at 298 K.)
1) 0.69: 1
2) 0.96: 1
3) 0.79 : 1
4) 0.86 : 1
Ans ( )
Solution : Information given is wrong hence could be consider for grace mark.
4.
Impure copper containing Fe, Au, Ag as impurities is electrolytically refined. A current of 140 A for 482.5
s decreased the mass of the anode by 22.26 g and increased the mass of cathode by 22.011 g. Percentage
of iron in impure copper is (Given molar mass Fe = 55.5 g mol–1 , molar mass Cu = 63.54 g mol–1 )
1) 0.85
2) 0.90
3) 0.95
4) 0.97
Ans (2)
Solution : No. of gram equivalents of copper deposited
22.011
 0.6928
31.77
Same should be the no. of gram equivalents from the anode.
No. of gram equivalents from the current = 140  482.5
Q = I  t = 140  482.5 = 67550C
No. of gram equivalents =
675500
 0.7
96500
Since only Cu and Fe are dissolved from the anode no. of g. equivalents of Fe = 0.7 – 0.6928 = 0.0072
 Mass of Fe = 0.0072  27.75 = 0.1998 g
% Fe =
5.
Mass of Fe
0.1998
 100 
 0.89  0.90%
Mass of impurities
22.26
25 cm3 of oxalic acid completely neutralised 0.054 g of sodium hydroxide. Molarity of the oxalic acid
solution is
1) 0.045
2) 0.032
3) 0.064
4) 0.015
Ans (2)
Solution :
V  N 0.064
25  N 0.064



1000
40
1000
40
N = 0.064
But we know for oxalic acid = Normality 
1
= Molarity = 0.032
2
6.
The statement that is NOT correct is
1) Energies of stationary states in hydrogen like atoms is inversely proportional to the square of the
principal quantum number.
2) The radius of the first orbit of He is half that of the first orbit of hydrogen atom.
3) Angular quantum number signifies the shape of the orbital.
4) Total number of nodes for 3s orbital is three.
Ans (4)
Solution : Total no. of nodes is (n – 1) = 3 – 1 = 2
7.
For the equilibrium :
CaCO3(s)  CaO(s) + CO2(g) ; Kp = 1.64 atm at 1000 K
50 g of CaCO3 in a 10 litre closed vessel is heated to 1000 K. Percentage of CaCO3 that remains unreacted
at equilibrium is (Given R = 0.082 L atm K–1 mol–1)
1) 50
2) 20
3) 40
(4) 60
Ans (4)
Solution : CaCO3  CaO + CO2 Kp = 1.64
50g
= 0.5 moles
K P  PCO
PV = NRT
1.64  10 = n 0.082  1000
n = 0.2
 no. of moles of CO2 formed is 0.2
 No. of moles of unreacted CaCO3 = 0.5 – 0.2 = 0.3
2
Percentage of unreacted CaCO3 =
8.
0.3
 100  60%
0.5
Conversion of oxygen into ozone is non-spontaneous at
1) high temperature
2) low temperature
3) all temperatures
4) room temperature
Ans (3)
Solution : 302  203 Here H = + ve S = – ve
 G = H – TS = + ve at all temperatures.
2
9.
Density of carbon monoxide is maximum at
1) 0.5 atm and 273 K
2) 4 atm and 500 K
3) 2 atm and 600 K
4) 6 atm and 1092 K
Solution : PV = RT
V=
Ans (2)
RT
P
Since no. of moles remain constant.
V
T
P
Lower the value of
T
T
higher is the density. Option (2) has lowest
value.
P
P
10.
The acid strength of active methylene group in
(a) CH3COCH2COOC2H5
(b) CH3COCH2COCH3
(c) C2H5OOCCH2COOC2H5 decreases as
1) a > b > c
2) c > a > b
3) a > c > b
4) b > a > c
Solution : Electron donating group make the hydrogen active.
Ester group is a lesser electron withdrawing group than Carbonyl group.
Ans (4)
11.
A metallic oxide reacts with water to form its hydroxide, hydrogen peroxide and also liberates oxygen.
The metallic oxide could be
1) KO2
(2) Na2O2
3) CaO
4) Li20
Ans (1)
12.
Ozonolysis
X 
Y  Z
(Re ductive)
Y can be obtained by Etard' s reaction, Z undergoes disproportionation reaction with concentrated alkali.
X could be
Ans (1)
3
Solution :
CH
O3
CH2
CHO
+ HCHO
Obtained from
Etard’s reaction
undergoes
disproportionation reaction
(Cannizzaro’s reaction)
13.
Gold Sol is not
1) a lyophobic colloid
2) negatively charged colloid
3) a macro molecular
4) a multimolecular colloid
Solution : Glucose, protein are macromolecular colloid.
Carbocation as an intermediate is likely to be formed in the reaction :

OH
 acetonecyanohydrin
1) Acetone + HCN 
Anhy. AlCl / HCl
2) Hexane 
 2 methyl pentane
hv
 2 chloropropane
3) Propene + Cl2 

 ethyl alcohol
4) Ethylbromide + Aq KOH 
Solution : Pyrolysis
Ans (3)
14.
3
15.
16.
For an ideal binary liquid mixture
1) H(mix) = 0 ; S(mix) < 0
3) S(mix) =0 ; G(mix) = 0
Ans (2)
2) S(mix) > 0 ; G(mix) < 0
4) V(mix) = 0 ; G(mix) > 0
Ans (2)
For hydrogen — oxygen fuel cell at one atm and 298 K
H2(g) +
1
O2(g)  H2O(l) ; G° = —240 kJ
2
E° for the cell is approximately, (Given F = 96,500 C)
1) 1.24 V
2) 1.26 V
3) 2.48 V
o
o
Solution : G = – nFE
– 240 = – 2  96500  Eo
Eo = + 1.24 V
4) 2.5 V
Which one of these is not known ?
1) CuI2
2) CuBr2
3) CuC12
4) CuF2
–
+2
+
Solution : Here I readily transfers electron to Cu and hence Cu and I2 are formed.
Ans (1)
17.
18.
19.
The correct statement is
1) The extent of actinoid contraction is almost the same as lanthanoid contraction.
2) Ce+4 in aqueous solution is not known.
3) The earlier members of lanthanoid series resemble calcium in their chemical properties.
4) In general, lanthanoids and actinoids do not show variable oxidation states.
1.CH 3 MgBr
P 
R
2.H O 
3
1) ethanamine
Ans (1)
Ans (1)
1. dil . NaOH
4—methylpent-3—en-2—one P is
2.

2) ethanal
3) propanone
4) ethanenitrile
Ans (4)
20.
When CH2 = CH — O — CH2 — CH3 reacts with one mole of HI, one of the products formed is
1) ethanol
2) ethanal
3) ethane
4) iodoethene
Ans (2)
Solution : Vinyl alcohol formed undergoes tautomerisation to form ethanal
4
21.
0.44 g of a monohydric alcohol when added to methylmagnesium iodide in ether liberates at S.T.P., 112
cm3 of methane. With PCC the same alcohol forms a carbonyl compound that answers silver mirror test.
The monohydric alcohol is
1) (CH3)3C — CH2OH
2) (CH3)2CH — CH2OH
3)
(4)
Ans (1)
Solution : 0.44g of Monohydric alcohol liberates 112 cm3 of CH4
I
R
OH + CH3MgI
CH4 + MG
1 mole
OR
1 mole
 Mass of Monohydric alcohol which given 22400 cm3 of CH4
22400  0.44
 88
112
Among the alcohols given option 3 and option 4 cannot be the answer as with PCC they give ketoses
which cannot give silver mirror test.
Among option (1) and (2) only option (1) has molecular mass 88.
22.
The IUPAC name of '13' is
1) 2—methylbutan-3—ol
3) 3—methylbutan-2—ol
23.
2) Pentan-2—ol
4) 2—methylbutan-2—ol
For Freundlich isotherm a graph of log
Ans (3)
x
is plotted against log P. The slope of the line and its y-axis
m
intercept, respectively corresponds to
1) log
1
,k
n
2) log
Solution : Freundlich isotherm
1
, log k
n
3)
1
,k
n
1
x
 Pn
m
1
x
 kP n
m
x
1
log  log k  log P
m
n
24.
A plot of
4)
4
x
m
1
, log k
n
Ans (4)
1
n
} log k
log P
1
Vs. k for a reaction gives the slope —1 x 104 K. The energy of activation for the reaction is
T
(Given R = 8.314 J K–1 mol–1)
1) 1.202 kJ mol–1
2) 83.14 kJ mol–1
3) 8314 J mol–1
4) 12.02 J mol–1
Ans ()
Solution : Wrong question since the plot is of logk vs 1/T
In the the question the plot given is of k not log k
Slope =
Ea
2.303R
1  104  2.303  8.314 = Ea
25.
26.
 Ea = 1.9kJ
The IUPAC name of the complex ion formed when gold dissolves in aquaregia is
1) tetrachloridoaurate(I)
2) dichloridoaurate(III)
3) tetrachloridoaurate(III)
4) tetrachloridoaurate(II)
Solution : Product formed is [AuCl4]The correct sequence of reactions to be performed to convert benzene into m-bromoaniline is
1) bromination, nitration, reduction
2) reduction, nitration, bromination
3) nitration, reduction, bromination
4) nitration, bromination, reduction
Ans (3)
Ans (4)
5
Solution :
NO 2
NO 2
Br2
+ HNO 3
NH2
Sn/HCl
Br
Br
27.
1)
2)
3)
4)
Ans (3)
Solution :
O
OH
O
C
+ C 6 H5 COCl
Activated
Benzene
nitration
Deactivated
benzene
O
O
C
NO 2
28.

A (g) 
 P(g)  Q (g)  R (g) follows first order kinetics with a half life of 69.3 s at 500 °C. Starting from the
gas 'A' enclosed in a container at 500 °C and at a pressure of 0.4 atm, the total pressure of the system after
230 s will be
1) 1.32 atm
2) 1.12 atm
3) 1.15 atm
4) 1.22 atm
Ans (2)
Solution : A (g )  P(g) +Q (g) + R (g)
Half life = 69.3s
k=
0.693 0.693

 102 s 1
t1
693
2
[R]o
2.303
log
t
[R]t
2.303
0.4
102 
log
230
Pt
0.4
0.4
1 = log

= 10
Pt
Pt
0.4
 Pt  0.04
10
k=
A  P(g) + Q(g) + R(g)
0.4 – x
= 0.04 0.36 0.36 0.36
 x = 0.36
 Pressure = 0.04 + 0.36 + 0.36 + 0.36 = 1.12
6
29.

MnO2 + HCl 
 A (g)
573
A (g)  F2(excess) 
B(g)
B(l) + U(S)  C(g) + D(g)
The gases A, B, C and D are respectively
1) Cl2, ClF3, UF6, ClF
2) O2, O2F2, U2O3, OF2
3) Cl2, ClF, UF6, ClF3
4) O2, OF2, U2O3, O2F2
Ans (1)
Solution : MnO2 + HCl  MnCl2 + Cl2 + H2O
30.
31.
Acetophenone cannot be prepared easily starting from
1) C6H5CH3
2) C6H6
3) C6H5CH(OH)CH3
4) C6H5C  CH
Ans (1)
One mole of ammonia was completely absorbed in one litre solution each of
(a) 1M HCl, (b) 1M CH3COOH and (c) 1M H2SO4 at 298 K.
The decreasing order for the pH of the resulting solutions is
(Given Kb(NH3) = 4.74)
1) a > b > c
2) c > b > a
3) b > c > a
4) b > a > c
Ans (4)
Solution : NH3 + HCl  NH4Cl
1M NH3 completely reacts with 1M HCl. However the salt formed is acidic salt pH is less than 7
NH3 + CH3COOH  CH3COONH4
1 mole of NH3 completely reacts with 1 mole of CH3COOH. The salt formed in neutral salt and the pH is
very close to 7
2 NH3 + H2SO4  (NH4)2SO4
1 mole of ammonia requires 0.5 mole of H2SO4 and hence H2SO4 remain in the solution in the solution
and hence the pH is very less
 decreasing order of pH is b > a > c
32.
5.5 mg of nitrogen gas dissolves in 180 g of water at 273 K and one atm pressure due to
nitrogen gas. The mole fraction of nitrogen in 180 g of water at 5 atm nitrogen pressure is
approximately
1) 1  10–5
2) 1  10–4
3) 1  10–6
4) 1  10–3
Ans (2)
5.5  103
Solution : 5.5 mg N2 =
moles = 2.36  10–4 moles
28
p N2  k H 
p N2 
n2
n1
2.36  104
10
10  104
kH 
 4.2  104
2.36
1  kH 
n2
n1
2.36  104
     (1)
10
n
5 2
     (2)
10
n 2 / 10
(2) 5
 
(1) 1 2.36  104
10
n2
5
10  10 
 104
10
1
50 cm3 of 0.04 M K2Cr2O7 in acidic medium oxidizes a sample of H2S gas to sulphur.
Volume of 0.03 M KMnO4 required to oxidize the same amount of H2S gas to sulphur, in
acidic medium is
1) 80 cm3
2) 120 cm3
3) 60 cm3
4) 90 cm3
Solution : K2Cl2O4 KMnO4
33.
Ans (1)
V N V N

1000
1000
50  0.24 V  0.15 50  0.24


 80 cm 3
1000
1000
0.15
34.
The compound that reacts the fastest with sodium methoxide is
1)
2)
3)
4)
Ans (3)
7
Solution : Higher the no. of electron withdrawing groups higher is the reactions
• Increasing the number of electron-withdrawing groups increases the reactivity of the aryl halide.
Electron-withdrawing groups stabilize the intermediate carbanion, and by the Hammond postulate, lower
the energy of the transition state that forms it.
35.
The pair of compounds having identical shapes for their molecules is
1) BCl2, ClF3
2) SO2, CO,
3) CH4, SF4
4) XeF2, ZnCl2
Ans (4)
Solution : ZnCl2 linear shape in gaseous state
Conductivity of a saturated solution of a sparingly soluble salt AB at 298 K is 1.85  10–5 m–1
Solubility product of the salt AB at 298 K is
Given om (AB) = 140  10– 4 S m2 mo1–1
(1) 1.32  10–12
2) 1.74  10–12
3) 5.7  10–12
4) 7.5  10–12
o
4
2
Solution : A m  140  10 Sm
K = 1.85  10–5
36.
m 
Ans (2)
K
1.85  105
1.85  105
1.85  10 5

 140  104 

 0.132  10 5  1.32  10 6
1000C 1000  2s
1000  2  140  10 4
28
Ks = s2 = (1.32  10–6)2 = 1.74  10–12
37.
An incorrect statement with respect to SN1 and SN2 mechanisms for alkyl halide is
(1) Competing reaction for an SN2 reaction is rearrangement.
(2) A weak nucleophile and a protic solvent increases the rate or favours SN1 reaction.
(3) A strong nucleophile in an aprotic solvent increases the rate or favours SN2 reaction.
(4) SN1 reactions can be catalysed by some Lewis acids.
Butylated hydroxy toluene as a food additive acts as
1) flavouring agent 2) emulsifier
3) antioxidant
Solution : BHT
Ans (1)
38.
Terylene is NOT a
1) polyester fibre
3) copolymer
Solution : Chain growth polymer
4) colouring agent
Ans (3)
39.
40.
2) step growth polymer
4) chain growth polymer
Ans (4)
The correct statement is
1) One mole each of benzene and hydrogen when reacted gives 1/3 mole of cyclohexane and 2/3 mole
unreacted hydrogen.
2) It is easier to hydrogenate benzene when compared to cyclohexene.
3) Cyclohexadiene and cyclohexene cannot be isolated with ease during controlled hydrogenation of
benzene.
4) Hydrogenation of benzene to cyclohexane is an endothermic process.
Ans (3)
41.
Among the elements from atomic number 1 to 36, the number of elements which have an unpaired
electron in their s subshell is
1) 7
2) 9
3) 4
4) 6
Ans (4)
Solution : H, Li, Na, K, Cr and Cu
42.
The statement that is NOT correct is
1) Van der Waals constant 'a' measures extent of intermolecular attractive forces for real gases.
(2) Boyle point depends on the nature of real gas.
(3) Compressibility factor measures the deviation of real gas from ideal behaviour.
(4) Critical temperature is the lowest temperature at which liquefaction of a gas first occurs.
Ans (4)
43.
The correct arrangement for the ions in the increasing order of their radii is
1) Ca+2 , K+, S-2
2) Cl–, F–, S–2
3) Na+, Cl–, Ca+2
4) Na+, Al+3, Be+2
Ans (1)
8
44.
The correct arrangement of the species in the decreasing order of the bond length between carbon and
oxygen in them is
1) CO2 ,HCO2 ,CO,CO32
2) CO,CO32 ,CO2 ,HCO2
3) CO,CO2 ,HCO2 ,CO32
4) CO32 ,HCO2 ,CO2 ,CO
Ans (4)
Solution : Bond order 
45.
1
Bond length
The species that is not hydrolysed in water is
1) BaO2
2) CaC2
3) P4 O10
(4) Mg3N2
Ans (1)
46.
For the properties mentioned, the correct trend for the differen: ies is in
1) inert pair effect - Al > Ga > In
2) first ionization enthalpy - B > Al > Tl
3) strength as Lewis acid - BCl3 > AlCl3 > GaCl3 4) oxidising property – Al+3 > In+3 > Tl+3
Ans (3)
Solution : Lewis acid strength decreases down the group
First ionization enthalpy is in the order
B > Tl > Al
801 589 577
This is due to poor shielding by f orbitals in the case of T
Tl
Due to the inert pair effect Tl+3 is unstable and hence easily gets reduced to Tl+1. Therefore Tl+3is a
good oxidizing agent
47.
A correct statement is
1) [MnBr4]-2 is tetrahedral.
3) [Co(NH3)6]+2 is paramagnetic.
2) [Ni(NH3)6]+2 is an inner orbital complex.
4) [CoBr2(en)2] – exhibits linkage isomerism.
Ans (1, 3)
Solution : [MnBr4]– 2 is tetrahedral
Br – weak ligand does not allow spin pairing.
[Mn Br4]–2; x – 4 = – 2  x = + 2
48. Iodoform reaction is answered by all, except
1) CH3CHO
3)
2) CH3 — CH2 — CH2OH
4) CH3 — CH2 — OH
Ans (2)
Solution :
O
||
Does not have CH3 – C – group
49. A crystalline solid XY3 has ccp arrangement for its element Y. X occupies
1) 33% of tetrahedral voids
2) 33% of octahedral voids
3) 66% of tetrahedral voids
4) 66% of octahedral voids
Ans (2)
50.
‘R’ is
1) sulphanilamide
3) o-bromo
bromo sulphanilic acid
51.
52.
2) p – bromo sullphanilamide
4) sulphanilic acid
Ans (4)
The statement that is NOT correct is
1) Carbohydrates are optically active.
2) Lactosee has glycosidic linkage between C4 of glucose
lucose and Cl of galactose unit
3) Aldose or ketose sugars in alkaline medium do not isomerise.
4) Penta acetate of glucose does not react with hydroxylamine.
Match the reactant in Column - I with the reaction in Column - II :
I
II
(i) Acetic acid
(a) Stephen
(ii) Sodium phenate
(b) Friedel-Crafts
(iii) Methyl cyanide
(c) HVZ
(iv) Toluene
(d) Kolbe's
(1) i d, b, - c, iv - a (2) i - c, - d, 7 a, iv – b (3) i - c, - a, iii - d, iv – b
Ans (3)
(4) i - b, c, - a, iv – d Ans (2)
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53.
The statement that is NOT correct is
1) In solid state PCl5 exists as [PCl4]+ [PCl6]–
2) Phosphorous acid on heating disproportionates to give metaphosphoric acid and phosphine.
3) Hypophosphorous acid reduces silver nitrate to silver.
4) Pure phosphine is non-inflammable.
Ans (4)
54.
In which one of the pairs of ion given, there is an ion that forms a co-ordination compound with both
aqueous sodium hydroxide and ammonia and another ion that forms a co-ordination compound only with
aqueous sodium hydroxide ?
1) Zn+2, Al+3
2) Al+3, Cu+2
3) Pb+2, Cu+2
4) Cu+2, Zn+2
Ans (1)
55.
A crystalline solid X reacts with dil. HCl to liberate a gas Y. Y decolourises acidified KMnO4. When a gas
'Z' is slowly passed into an aqueous solution of Y, colloidal sulphur is obtained. X and Z could be,
respectively
1) Na2SO4, H2S
2) Na2SO4, SO2
3) Na2S, SO3
4) Na2SO3, H2S
Ans (4)
56.
An aromatic compound 'A' (C7H9N) on reacting with NaNO2/HCl/ at 0 °C forms benzyl alcohol and
nitrogen gas. The number of isomers possible for the compound 'A' is
1) 7
2) 6
3) 5
4) 3
Ans (3)
57.
The statement that is NOT correct is
1) Collectors enhance the wettability of mineral particles during froth flotation.
2) Copper from its low grade ores is extracted by hydrometallurgy.
3) A furnace lined with Haematite is used to convert cast iron to wrought iron.
4) In vapour phase refining, metal should form a volatile compound.
Ans (1)
58.
A solution of 1.25 g of `P' in 50 g of water lowers freezing point by 0.3 °C. Molar mass of ‘P’ is
94.Kf(water) = 1.86 K kg mol–1 . The degree of association of `P' in water is
1) 60 %
2) 75 %
3) 80 %
4) 65 %
Ans (1)
59.
Volume occupied by single CsCl ion pair in a crystal is 7.014  10-23 cm3 . The smallest Cs - Cs
internuclear distance is equal to length of the side of the cube corresponding to volume of one CsCl ion
pair. The smallest Cs to Cs internuclear distance is nearly
0
0
1) 4.3 A
2) 4.5 A
0
3) 4.4 A
0
4) 4 A
Ans (*)
Solution : Since the answer is 4.125 none of the options can be given as answer
For Cr2O72 + 14H+ + 6e–  2Cr+3 + 7H2O ; E° = 1.33 V At Cr2 O 72  = 4.5 millimole, [Cr+3] = 15
millimole, E is 1.067 V. The pH of the solution is nearly equal to
1) 3
2) 4
3) 2
4) 5
Ans (3)
Solution : Cr2 O 72  14H   6e   2Cr 3  7H 2 O
60.
E = E +
0.0591 [Cr2 O72 ][H  ]14
log
6
[Cr 3 ]2
1.067 = 1.33 +

0.0591 
[Cr2 O42 ]
 log[H  ]14 
 log
3 2
6 
[Cr ]


0.0591 
[Cr2 O 22
   log
 log[H  ]14 
1.067 – 1.33 =
3 2
6
[Cr ]


– 0.263 = 
26.3 = log

1 
4.5  103

log
 14 pH 

3 2
100 
(15  10 )

1
+ 14 pH = –1.3010 + 14 pH
20
28 = 14 pH
pH =
28
2
14
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