i cho 2005 solutions

7/28/2019 i Cho 2005 Solutions

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61

Answer 1: Separation and Identification of Ions

1-1 For example

1-2 For example


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62

1-3-1 BY 2 = B2+

+ 2 Y-

Ksp = (S 1)(2S 1)2

= 3.20 10-8

S 1 2S 1

4S 13 =3.20 10

-8, S 1(solubility of BY 2) = 2.0 10

-3 M

1-3-2 CY 2 = C2+

+ 2 Y-

Ksp = (S 2)(2S 2)2

= 2.56 10-13

S 2 2S 2

4S 23 =2.56 10

-13, S 2(solubility of CY 2) = 4.0 10

-5 M

1-4-1 Plot of Absorbance (A) vs Volume (V L) of L added as follows:

From the volume of L at Break point B (all B2+

ions form complex with L) in the plot, n canbe calculated:

n/1 = (#of moles of L) / (#of moles of B2+

)

= (5.1 mL 1.0 10-2

) / (2.0 mL 8.2 10-3

)

3

It means that B2+

forms BL 32+

complex with L.

1-4-2 (1) Calculation of Molar Absorption Coefficient

At Break point, A = 0.66 = 1 (concentration of BL 32+

)

And = 0.66 / (2.0 mL 8.2 10-3

/ 50 mL) =2.01 103

(2) Choose a point in the curve of the plot, for example:

At Point P (2.0 mL of L added): A = 0.26

A = 0.26 = 1 [BL32+

]


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[BL32+

] = 0.26 / = 0.26 / (2.01 103) =1.29 10

-4 M

[B2+

] = (2.0 mL 8.2 10-3

50 mL 1.29 10-4 M) / 50 mL

[B2+

] = 1.99 10-4 M

[ L ] = (2.0 mL 1.0 10-2

3 50 mL 1.29 10-4 M) / 50 mL

[ L ] = 1.3 10-5 M

[Calculation of formation constant]

So K f = [BL 32+

] / ([B2+

][ L ]3) =(1.29 10

-4 ) / ((1.99 10-4

)(1.3 10-5

)3)

K f = 8.74 1011

1-5-1 For CY 2: Ksp = [C2+] [Y -]2 =2.56 10 -13

[Y-] =((2.56 10

-13) / 0.05)

1/2= 2.26 10

-6 M When CY 2 begins to form

For BY 2: Ksp = [B2+

] [Y-]2

=3.20 10-8

[Y-] =((3.20 10

-8) / 0.05)

1/2= 5.66 10

-4 M When BY 2 begins to form

CY2 forms first

1-5-2 The precipitation of C2+

as CY 2 considered to be completed at [C2+

] = 10-6

M

Thus Ksp = [C2+

] [Y-]2

= 2.56 10-13

and [Y-] =((2.56 10

-13) / 10

-6)

1/2=5.06 10

-4 M

It means that [Y-] =5.06 10

-4 M, when CY 2 precipitates completely.

When [Y-] = 5.06 10

-4 M, for BY 2 :

[B2+

] [Y-]2

= (0.1) (5.06 10-4

)2

=2.56 10-8


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Answer 2: Preparation and Applications of Radioisotopes

2-1

2-2-1 A = N, A o = N o and N =N o e- t

A / A o = N / N o = e- t

and A = A o e - t

2-2-2 N =N o e- t

If N =1/2 N o, t = t1/2

then 1/2 N o = N o e- t1/2

= 2.303 log 0.5 / t 1/2 = 0.693 / t 1/2

For C-14, = 0.693 / 5730 = 1.2 x 10-4

Also A =A o e- t

10.2 = 16.5 e -1.2 x 10-4

t

and t =4008 years

2-3-1 A = Rp Rd = N - N ( e- t

) = N (1- e- t

)

and =0.693 / (14.3 x 24 x 60 x 60) =5.61 x 10-7

N =[(10 x 10-3

) / 98] x 6 x 1023

=6.12 x 1019

A = N (1- e- t

) =(6.12 x 1019

)(1.00 x 1013

)(0.9 x 10-24

)(1-e-5.61 x 10 -7x60x60

)

and A =1.11 x 106

cps = 1.11 x 106 / (3.7 x 10

10) Ci =3 x 10

-5Ci = 0.03 mCi

2-3-2 Total amount of P-32 is constant after and before the injection,


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so, V o Ao = V x Ax (V = volume, A = Activity, x for pool water)

2.0 x 1.0 =V x (12.4 / (3.7 x 1010

)

and V x(pool water) =5.97 x 109

mL = 5.97 x 106

L

Answer 3: Ion Exchangers

3-1-1 2 RNa +Ca2+

= (R) 2Ca +2 Na+

or 2 RNa +CaCl 2 = (R) 2Ca +2NaCl

3-1-2(a) The tap water contains trace HCl after the adsorption of Ca2+

by the ion exchanger R-H

+

and the chemical equation of the adsorption is:

2 RH +Ca2+

= (R) 2Ca + 2H+

or 2 RH +CaCl 2 = (R) 2Ca +2HCl

3-1-2(b) RNa is suitable for drinking purpose. Because the product of the adsorption of Ca 2+ byRNa is Na

+or NaCl while the product is H

+or HCl after the adsorption of Ca

2+by RH.

3-2-1 The removal of H+

can be achieved by using the anionic ion exchanger R+OH

-with the

equation:

ROH + HCl = R+Cl

- + H 2O

3-2-2 Firstly, the anionic ion exchanger R+OH

-is used to adsorb the SO 4

2-ion with the

equation :

2 ROH + SO 42-

= (R +)2SO 42- +2 OH

-

Secondly, a standard solution of HCl can be used for the titration of the OH-

in thesolution after adsorption of SO 42- by the anionic ion exchanger R +OH -.

H+

+ OH-

= H 2O (Acid-Base Titration)

3-3 RH +M+

= RM + H+, Kc =[RM][H

+] / ([M

+][RH]) (3-3-1)

K d = [RM] / [M+

] (3-3-2)

S = ([RM] + [RH]) 10-3

(3-3-3)

We substitute Equations (3-3-1) and (3-3-2) into Equation (3-3-3) and obtain:


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66

S = (Kd[M+

] + [RM][H+

] / Kc[M+

]) 10-3

= (Kd[M+

] + Kd[M+

][H+

] / Kc[M+

]) 10-3

= (Kd[M+

] +Kd[H+

] / Kc ) 10-3

S Kc 103

= Kd Kc [M+] +Kd[H

+]

1 / Kd = [M +] / (S(103)) + [H +] / (S Kc (10

3)) (3-3-4)

3-4-1 N 1 = 16 (t 1 / 1)2 = 16 (10 / 1.0)

2=1600

N2 = 16 (t 2 / 2)2 = 16 (14 / 1.5)2

=1394

N =(N 1 + N 2) / 2 = (1600 + 1394) / 2 = 1497

3-4-2 H = L / N = 30 / 1497 = 0.021 cm

3-4-3 R =2 (t 2 - t 1) / ( 1 + 2) = 2 (14 10) / (1.0 +1.5) = 3.2

3-4-4 = (t 2 - t 0) / (t 1 - t 0) = (14 1) / (10 - 1) = 1.44

3-5-1 Z-Na+

+ Ca2+

= Z-Ca2+

+ Na+

3-5-2 Z-Na+

+ K +

= Z-K + +Na

+

Answer 4: Determination of Calcium Ion by Precipitation Followed by Redox Titration

4-1 (NH 2)2CO + H 2O

CO 2 +2NH 3

4-2 titration reaction 5 H 2C2O4 + 2 MnO 4-

+6 H+ 10 CO 2 + 2 Mn

2++8 H 2O

[Ca2+

] = 2.5 x 10-3

M x 0.02741 L x 25 / 0.025 L

=6.85 x 10 -3 M

4-3 Mass-balance: [Ca2+

] =[C 2O42-

] + [HC 2O4-

] + [H 2C2O4]


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= [C 2O42-

] (1 + [H+] / K 2 +[H

+]2 / K 1K 2)

[C 2O42-

] = [Ca2+

] / (1 + [H+] / K 2 + [H

+]2

/ K 1K 2) (1)

Substituting (1) into [Ca2+

] [C 2O42-

] =K sp

[Ca2+

] = 1.92 x 10-4 M

4-4 C Ca = [Ca2+

] + [CaC 2O4(aq) ] + [Ca(C 2O4)22-

]

= K sp (][

12

42OC

+ K f1 +K f1 K f2 [C 2O42-

])

][ 242OCd

dC Ca =0 =- K sp22

42 ][1

OC + K sp K f1 K f2

[C 2O42-

] =1.0 x 10-2

M

[Ca2+

] = K sp / [C 2O42-

] =1.3 x 10-6

M

4-5 Charge balance: 2[Ca2+

] + [H+] = 2[C 2O4

2-] + [HC 2O4

-] + [OH

-] (1)

Mass balance: [Ca2+

] = [C 2O42-

] + [HC 2O4-] + [H 2C2O4] (2)

Because K b2 is too small, [H 2C2O4] can be neglected.

Comparing (1), (2), [HC 2O4-] = K w / [H

+] [H

+] (3)

[C 2O42-

] = (K 2 K w) / [H+]2

K 2 (4)

[Ca 2+] = K sp / [C 2O42-] = K sp [H+]2 / (K 2K w K 2[H+]2) (5)

Substituting (3), (4), (5) into (2)

K 2 [H+]5

+ (K 22

- K sp ) [H+]4

2 K 2 K w [H+]3

2 K 22

K w [H+]2

+ K 2 K w2

[H+] + K 2

2K w

2= 0

Solving [H+], [H

+] = 5.5x10

-8M (or pH = 7.26)

Substituting [H+] into (5), [Ca

2+] = 1.04 x 10

-4M


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Answer 5: Nitrogen in Wastewater

5-1-1 NH 4+

5-1-2 NH 3

5-1-3 H 3BO3 + NH 3 NH4+ + H 2BO3

5-1-4 H 2BO3 + H 3O+ H3BO3 + H 2O

5-1-5 Methyl orange. At the equivalence point, the solution contains boric acid and ammonium

ion, therefore, an indicator with an acidic transition interval is required.

5-2-1 2 MnO 4 (aq) +5 HNO 2 + H+

(aq) 2 Mn2+

+5 NO 3 +3 H 2O

5-2-2 2 MnO 4 + 5 H 2C2O4 + 6 H+ 2 Mn

2++10 CO 2 + 8 H 2O

5-2-3 A = [ 5 ( B C) 2 ( D E ) ] 7 / F

Answer 6: Use of Isotopes in Mass Spectrometry

6-1 0.77

6-2 Corrected signal = (m / z 136 signal) - 1.03 (m / z 138 signal)


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69

Answer 7: Atomic Orbitals

7-1

1s: 0, 2s: 1 and 3s: 2.

7-2

2p z , 3p z

There is one angular node for 2p z; one angular node and one spherical node for 3p z.

7-3 (0, 2, 4, 1, 3)

Answer 8: Intermolecular Forces

8-1-1

Na+

OH 2

OH 2

OH 2H2O

OH 2

H2O

8-1-2


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70

OO

O

OO

ONa+

8-2-1

FH

F

H

FH

FH

F

H

FH

8-2-2

CH 3 CO

OC CH 3

O

O

H

H

8-3

N

N N

N

N

N

H

O

OR

H

H3C NH

H

H

R

HThymine

Adenine

N

N N

N

N

NN H

H

OR

H

H

H

O

NH

H

R

H

Cytosine

Guanine


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Answer 9: Crystal Packing

9-1 Simple cubic: 6, body-centered cubic: 8 and face-centered cubic: 12

9-2

For simple cubic, a = 2r, %4.52a

34

f 3

3

v==

r

For body-centered cubic, ra 43 = , %68a34

2f 3

3

v=

=r

For face-centered cubic, ra 42 = , %74a34

4f 3

3

v=

=r

9-3 ra 42 = , pmra 40722 ==

33

23g/cm6.10

)407(1002.69.107

4=

=pm

d

9-4

pm22917.34sin2012

sin2o ==

=

d

Answer 10: Applications of Transition Metals

10-1-1

2 CrO 42 (aq ) + 2 H + (aq ) Cr2O7

2 (aq ) + H 2O ( l)

10-1-2 CrO 42

: + 6, Cr 2O72

: +6.

10-1-3 This is not a redox reaction because the oxidation state in each metal center does not

change.

10-1-4 Hydrogen ion concentration is the main factor to control the equilibrium position.


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10-1-5

Cr

O

OCr

O

Cr

OO

OO

O

O

O

O

2 2

10-2-1

+ + 12e +

+ + 4e O2

Cr2O72 (aq ) 2 H + (aq )+ H2O ( l) 3 O 2 (g)+ 2 Cr ( s) +

Cathode

Anode

Overall

Cr2O72 14 H + 2 Cr 7 H 2O

2 H 2O 4 H +

10-2-2 1.5 moles of oxygen gas will evolve.

+ + 12e +

+ + 4e O2

Cr2O72 (aq ) 2 H + (aq )+ H2O ( l) 3 O 2 (g)+ 2 Cr ( s) +

Cathode

Anode

Overall

( ) 3x

52 g Cr x1 mol Cr

52 g

x3 mol O2

2 mol Cr

1.5 mol O 2=

Cr2O72 14 H + 2 Cr 7 H 2O

2 H 2O 4 H +

10-2-3 16 h

52 g Cr x1 ol Cr

52 gx

6 F

mol Cr16 h=x

96485 C1 F

x1 sec

10 C

1 min60 sec

x x1 h

60 min

10-2-4 Chromium readily forms a thin, adherent, transparent coating of Cr 2O3 in air, making the

metal extremely useful as an attractive protective coating on easily corroded metals.

Answer 11: Electrochemistry of Inorganic Compounds

11-1 For the concentration cell: Mn (s) | Mn2+

(aq) (1M) || Mn2+

(aq) / MnCO 3 | Mn (s) ,


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E cell = Eo

(0.0592 / 2) log ([Mn2+

]right / [Mn2+

]left)

Ksp = 1.810-11

= [Mn2+

][CO 32-

]

[Mn2+

]right = 1.010-8

M and [Mn2+

]left= 1.0 M with Eo

= 0.0 V (both are Mn)

E cell = 0.0 - (0.0592 / 2) log (1.010-8

M / 1.0 M) = 0.237 V

11-2

Reduction of O 2 to H 2O is obtained as (0.70V+1.76V) / 2 = 1.23 V,

for O 2 + 4H+

+ 4e- 2 H 2O E

o= 1.23V

The Eo

value could be obtained directly from the diagram by dividing the differences (2.46)

of O 2 and H 2O by the differences of the oxidation number (2).

For H 2O2 O2 + H 2O Eo

= 1. 06 > 0.0

The disproportionation reaction is spontaneous.

11-3 The number of electron pair should be 5 (trigonal bipyramidal) with three electron pairs in

the equatorial plane, thus the molecular geometry of XeF 2 is linear.

2 H 2O O 2 + 4H+

+ 4e-

Eo

= -1.23V

XeF 2(aq) + 2H+

(aq) + 2e- Xe (g) + 2HF (aq) E = 2.32V

2 XeF 2(aq) + 2H+

(aq) + 2 H 2O 2 Xe (g) + O 2 + 4HF (aq) E = 1.09 V

The decomposition of XeF 2 in aqueous solution is favored in acidic solution.

Answer 12: Metal Carbonyl Compounds

12-1 Compound A is anionic, the absorption bands attributed to CO stretching appear at lower

: 2H2

: 00

: 2H


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74

frequency because of stronger back donation of the anionic charge to the anti bonding

orbital of CO thus weakening the CO bond. For the neutral species B , absorption bands

appear at the higher frequency.

12-2

WOC

CO CO

WOCCOOC

WOC

CO CO

CCH 2

WOC

CO CO

WOC

CO CO

W(CO) 6NaC 5H5 Na

HCCCH 2Br

FeSO 4

Na/Hg

metal

migration

A B

C D

12-3

CCH 2

WOC

COCO

WOC

CO CO

D

D

metal migration

C D


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Answer 13: Carbocat ion and Aromat ici ty

13-1 (CH 3)3C+

SbF 6-

13-2 Spectrum I: (CH 3)3CF in SbF 5

13-3 6 electrons

13-4 yes

13-5 (a) A singlet at 9.17

13-6

O

OH

O

OH

O

OH

O

OH

O

OH

O

OH

O

OH

OOH

or

13-7 D

O

OFe

3

Answer 14: Photochemical Ring Closure and Opening

14-1 (2 E ,4Z ,6Z)-octatriene

14-2 E

14-3 F


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96

H3C

HO

H3CH3C

CH3H3C

H

14-4 I

O

O

O

14-5 No.

Answ er 15: Stereochemis try

15-1 ( 2S,3S )

15-2 X=CH 3, Y = PPh 2

15-3 36%

15-4

15-5

O

OH

HOHO OH

HO

15-6

H OHCH

HO H

OHHCH2OH

OHH

O

15-7 none


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97

15-8 99:1

15-9 0

Answ er 16: Organic Synthes is

16-1

I

A =

HOB =

C =

16-2

BrD =

BrO

E =

BrF =

FF

C8H17 OG =


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Answ er 17: Spectroscopy and Polymer Chemist ry

17-1 C 4H6O2

17-2 C=O group

17-3 A

O CH 3

OHH

H

17-4 B

O O OO O O

In n

In: initiator

17-5 Organic reactions that could transform acetate to alcohol such as acid or base hydrolysis,

alcoholysis, or LiAlH 4 reduction.

17-6 There are 100 units/molecule. However, the last one does not contain chiral center,

therefore, there are 99 chiral centers and each of which would have R or S configuration.

Totally there will be 299 stereoisomers, including enantiomers and diastereomers.

Therefore, the number of pairs of enantiomers is 299

/2 = 2 98 .

17-7 C

OCH

3

HH

H O

17-8 E : CO 2 F: (CH 3)2C=CH 2

G: H: (CH3)2CBr-CH 2Br

OH

n

17-9 I: (d)


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Answer 18: Crown Ether and Molecular Recognition

18-1 B

O

OO

O

OO

18-2 (c) To remove the tetrahydropyran group

18-3 C D

Cl

O O

Cl

OO

O O

OO

NH

OO

NH

E F

O O

OO

NHNH

O O

N

OO

N

O OO O

18-4 (b) A high dilution condition is employed in order to inhibit polymer formation.

18-5 Curve I to I; Curve II to G ; Curve III to H


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Answ er 19: Works in Thermodynamics

19-1 Isothermal reversible expansion

We have 100/22.41=4.461 moles, and the final volume is

l1001

1010V

2

112 =

==PVP

(1)

The work done by gas is

joules

VV

nRTqw

2329010ln2.273341.8461.4

ln1

2

==

==

(2)

19-2 Adiabatic reversible expansion

Notice that

352

3C =

+==

R

RR

CVP

(3)

Thus

l8.39

10)10()( 53

121

2

12

=

== VPPV

(4)

and the final temperature is obtained from

K nRVP

T

o8.10808205.0461.481.39122

2

=

==(5)

For adiabatic processes,

0=q and wwqE =+=

ie

joules TvCnEw 9141=== (6)

19-3 Irreversible adiabatic expansion

Since q=0, we have

)( 12 T TCnwE v == (7)


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102

)( 122 VVPw = (8)

and

)10

2.273

1()2.273(

2

3 22

= nRnRT TnR (9)

It follows that

K T o8.1742 = (10)

and

joules

nRwE

5474

)2.2738.174(23

=

==(11)

Answ er 20: Kinet ic s At mosphere Chemis try

20-1

kt

PP

kPdt

dP

NONO

NONO

+=

=

0

2

22

2

2

11

where 02NO

P devotes the initial pressure of NO 2

20-2 At 2/1tt = , 0 22 21

NONO PP =

or

tPk

NO 2/10

2

1=

422.0

7606003

1 =

=k l / atm min

Answer 21: Kineti cs and Thermodynamics

21-1 In the beginning 4 min of the reaction,


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103

.. (1)

.. (2)

(1) divided by (2) gives

21-2 When the reaction is complete, the system reaches thermal equilibrium.

21-3 A C thermodynamic- controlled reaction process is favored when temperature increases.

The system will reach thermal equilibrium more rapidly.

Answer 22: Phase Diagram

22-1 A: solid; B: solid, liquid, and gas states coexist; C: liquid and gas states coexist.

22-2 The negative slope of the solid/liquid line indicates the liquid state of water is denser than

its solid state. Therefore, ice may not sink in its own liquid.

22-3 Clapeyron equation is expressed as

V T

H

dT

dP

= ,

where H is molar enthalpy of water and V is volume change. The phase diagramshows that the slope of dP/dT for the liquid-solid coexistence region is negative, indicating

the volume expands when water freezes.

22-4 As pressure is lowered, liquid phase transforms directly to gas phase at the same

temperature. Thus water may vaporize. At the same time, the process of water

evaporation is endothermic as to make the surrounding cooled. The left water becomes

frozen. The solid state will sublime until none is left, if the pump is left on.

22-5 The ice surface, exerted by a pressure more than one atm, turns to liquid state at 0 oC.

][][ 1 AkdtBd =

][][ 2 Akdt

Cd =

2

1

][][

kk

CdBd = 10

1.01

2

1 ===kk

CB

1

1

][

][

=k

k

A

B

2

2

][

][

=k

k

A

C

21

200100

0005.01.001.01

][][

22

11 ====

kkkk

CB


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104

Answ er 23: Standard Deviation in One-Dimensional Quantum Mechanics

23-1 average speed :

=

0)( dvvvF dvv

RT

Mv

RT

M 32

0

2/3

2exp

24

=

=MRT

8=

032.014.330031.88

xxx

= 4.45 x 102 ms

-1

standard deviation v:

=

0)( dvvvF dvvRT

MvRT

M 42

0

2/3

2exp

24

=

=032.0

30031.833 xxMRT = =2.33 x 10 5 m 2s -2

v = 22 >< vv = 225 )1045.4(1033.2 xx = 1.87 x 102

ms-1

23-2 average position :

=

dxx * = dx

xx )

2exp(

21 2

= 0

standard deviation x:

=

dxx 2* = dx

xx )

2exp(

21 22

= 1

x = 22 >< xx =1.

23-3

=

dx

xh

i

)2

(* =

dxihxe x

24

2/2

=0

=

dxh

)

4( 22

22* = 2

2

16 h

p= 4

22 hpp =><

23-4 xp= 4h


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105

Answ er 24: A Part ic le in 2-D Box Quantum Mechanics

24-1 E 1,1 = 2 E 0E 1,2 = E 2,1 = 5 E 0

E 2,2 = 8 E 0E 1,3 = E 3,1 = 10 E 0E 2,3 = E 3,2 = 13 E 0E 1,4 = E 4,1 = 17 E 0

E 3,3 = 18 E 0E 2,4 = E 4,2 = 20 E 0E 3,4 = E 4,3 = 25 E 0E 1,5 = E 5,1 = 26 E 0

where E 0 = h2

/ 8 mL2

24-2 The total number of electrons in the highest occupied energy level is 4.

24-3 Ground state is diamagnetic.

24-4 The longest-wavelength excitation energy is E = (25-20) E 0, where

E 0= (6.63 10-34

Js)2

/ [8 9.11 10-31

kg (1 10-9

m)2] = 6.02 10

-20J (1)

E = (25-20) E 0 = 3.01 10-19

J (2)

The wavelength is

= hc / E = [6.63 10-34

Js 3 108] / 3.01 10

-19= 660 nm (3)

Answer 25: Spect rum Analysis

25-1 = 355.00 / 2 sin60.00 = 204.96

DFDL = 2 1.40 204.96 = 573.89 nm

Answer 26: Time-of-Fl ight Mass Spectr ometer

26-1 (a)

= [(2 1 1.602210-19

C 20000 V) / (12362 1.660510-27

kg)]

=17669.5 m/s

26-2 (c)

t = 1.00 m / 17669.5 m/s = 56.59 s


25/25

Answer 27: Enzyme Catalys is

27-1 A = bC; C = A/ b; mol = C x V (volume)

[0.1 / ((27.7-9.2) x 103)] x 5 x 10

-3= 2.7 x 10

-8mol/sec

27-2 Four electrons are needed to reduce one molecule of oxygen, therefore, the oxygen

consuming rate is 2.7 x 10-8

/ 4 = 6.75 x 10-9

mol/sec

27-3 By definition, the turnover number equals 6.75 x 10-9

(mol/sec) / (2.7 x 10-9

M x 5 x 10-3 L).

Therefore, oxidase has a turnover number of 500.

Practical Problems Answer 30: Ident if ication of Unknown Solu tions II

1. Use the indicator to find out NaOH, HCl, and H 2SO 4 (confirmed by Pb2+)

2. Find out the Na 2S by the odor, and use it to find Cd2+ and Zn 2+ (by precipitation. and color).

3. By electrolysis of the four solutions remained, KI solution can be found by the trace of yellowishbrown (I 2) formed in the anode.

4. The color of I 2 will be disappeared by Na 2S 2O3 solution. 5. The concentration of unknown solution is about 0.5 M (mol/L)

i cho 2005 solutions

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