forms cyclic compound ... an organic compound (a) c3h8o answers luca’s test within 5-10 mins and...

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K N SUBRAMANI .M.Sc.B.Ed., PGT in Chemistry, VVMHSS, Karamadai –Cbe – 641 104

COMPULSORY PROBLEMS

18. CARBONYL COMPOUNDS {Q.No.70 (c)}

1. Compound (A) of molecular formula C2H4O reduce Tollen’s reagents.(A) ontreatment with HCN gives compound (B).Compound (B) on hydrolysis with acidcompound (C) with molecular formula C3H6O3 which an optically activecompound. Compound (A) on reduction with N2H4/C2H5ONa gives ahydrocarbon (D) of molecular formula C2H6.Identify A,B,C, and D. Explain thereaction. [Mar-2013]

ANSWER: H+/H2OCH3CHO + HCN → CH3CH(OH)CN CH3CH(OH)COOH

(A) (B) (C)

N2H4/C2H5ONaCH3CHO CH3CH3 + N2 + H2O

(A ) 4(H) (D)RESULT:

S.No Compound Formula Name1 A CH3CHO Acetaldehyde2 B CH3CH(OH)CN Acetaldehyde cyanohydrin3 C CH3CH(OH)COOH Lactic acid4 D CH3CH3 Ethane

2. Compound (A) of molecular formula C2H4O reduce Tollen’s reagents.(A) ontreatment with HCN and followed by hydrolysis with acid gives compound (B)with molecular formulaC3H6O3 which an optically active compound. Compound(B) on reaction with Fenton’s reagents forms (C) (C3H4O3).This answersiodoform reaction.Identify A,B, and C.Explain the reaction. [Jun-08,Oct-08,Oct 15]

ANSWER: H+/H2OCH3CHO + HCN → CH3CH(OH)CN CH3CH(OH)COOH

(A) (B)

Fe2+/H2O2

CH3CH(OH)COOH CH3COCOOH(B) (O) (C)

RESULT:S.No Compound Formula Name

1 A CH3CHO Acetaldehyde2 B CH3CH(OH)COOH Lactic acid3 C CH3COCOOH Pyruvic acid

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3. Compound (A) (C2H4O) reduce Tollen’s reagent .(A) on treatment with Zincamalgam and con.HCl gives compound (B) .In the presence of con.H2SO4 (A)forms cyclic compound (C) which is used as hypnotic .Identify A,B and C.Explain the reaction. [Jun-2011]

ANSWER: Zn / Hg – conc.HClCH3CHO CH3CH3 + H2O

(A) 4(H) (B)

(C)RESULT:

S.No Compound Formula Name1 A CH3CHO Acetaldehyde2 B CH3CH3 Ethane

3 C Paraldehyde

4. An organic compound (A) (C2H4O) with HCN gives (B) (C3H5ON). Compound (B)on hydrolysis compound (C) (C3H6O3) which is an optically activecompound.(C) also undergoes iodoform test. Identify A, B and C. Explain thereaction. [Oct-2011]

ANSWER:H+/H2O

CH3CHO + HCN → CH3CH(OH)CN CH3CH(OH)COOH(A) (B) (C)

RESULT:

S.No Compound Formula Name1 A CH3CHO Acetaldehyde2 B CH3CH(OH)CN Acetaldehyde cyanohydrin3 C CH3CH(OH)COOH Lactic acid


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5. Compound (A) with molecular formula C2H4O reduce Tollen’s reagent.(A) ontreatment with HCN gives compound (B) .Compound (B) on hydrolysis with anacid gives compound (C) C3H6O3. Compound (C) is optically active. Compound(C) on treatment with Fenton’s reagent gives compound (D) C3H4O3. Compound(C) & (D) gives effervescence with NaHCO3 solution. Identify the compoundsA,B,Cand D. Explain the reactions. [Mar-2010]

ANSWER:H+/H2O

CH3CHO + HCN → CH3CH(OH)CN CH3CH(OH)COOH(A) (B) (C)

Fe2+/H2O2

CH3CH(OH)COOH CH3COCOOH(C) (O) (D)


1 A CH3CHO Acetaldehyde2 B CH3CH(OH)CN Acetaldehyde cyanohydrin3 C CH3CH(OH)COOH Lactic acid4 D CH3COCOOH Pyruvic acid

6. An organic compound (A) C2H3OCl on treatment with Pd and BaSO4 gives (B)C2H4O which answers iodoform test .(B) when treated with conc.H2SO4

undergoes polymerization to give (C) a cyclic compound. Identify A,B and C.Explain the reaction. [Oct-2009]

ANSWER: Pd / BaSO4

CH3COCl + H2 CH3CHO + HCl(A) (B)


1 A CH3COCl Acetyl chloride2 B CH3CHO Aceytaldehyde3 C (CH3CHO)3 Paraldehyde


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7. An organic compound (A) C2H3N on reduction with SnCl2 / HCl gives (B) C2H4Owhich reduce Tollen’s reagent. Compound (B) on reduction with N2H4/C2H5ONagives (C) C2H6. Identify the compounds A,B and C. Explain the reactions.[Oct-12]

ANSWER:

(B)N2H4/C2H5ONa

CH3CHO CH3CH3 + H2O(B) 4(H) (C)


1 A CH3CN Methyl cyanide2 B CH3CHO Aceytaldehyde3 C CH3CH3 Ethane

8. An organic compound (A) (C2H6O) liberate hydrogen with metallic sodium.(A)on mild oxidation gives (B) C2H4O which answers iodoform test.(B) whentreated with conc.H2SO4 undergoes polymerization to give (C), a cycliccompound. Identify A,B and C. Explain the reaction. [Jun-2006]

(O)

CH3CH2OH CH3CHO + H2O(A) H+/ K2Cr2O7 (B)

(C)


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1 A CH3CH2OH Ethanol2 B CH3CHO Aceytaldehyde3 C (CH3CHO)3 Paraldehyde

9. An organic compound (A) with molecular formula C3H6O undergoes iodoformreaction. Two molecules of compound (A) react with dry HCl to givecompound (B) (C6H10O).Compound (B) react with one more molecule ofcompound (A) to give compound (C) (C9H14O).Identify A,B, and Explain thereaction. [Oct-2010]

ANSWER:


1 A CH3COCH3 Acetone2 B (CH3)2C=CHCOCH3 Mesityl oxide3 C (CH3)2C=CHCOCH=C(CH3)2 Phorone

10. An organic compound (A) C3H8O answers Luca’s Test within 5-10 mins andoxidation form (B) C3H6O. (B) On further oxidation forms (C) C2H4O2 whichgives effervescence with NaHCO3. (B) Also undergoes iodoform test. IdentifyA,B and C. Explain the reaction. [Jun-2009]

ANSWER:(O) (O)

(CH3)2CHOH CH3COCH3 CH3COOH(A) H+/ K2Cr2O7 (B) H+/ K2Cr2O7 (C)


1 A (CH3)2CHOH 2-Propanol2 B CH3COCH3 Acetone3 C CH3COOH Acetic acid


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11. Compound (A) of molecular formula C3H6O does not reduce Tollen’s reagentand Fehling’s solution. Compound (A) undergoes Clemmensen reduction togive compound (B) of molecular formula C3H8. Compound (A) in the presenceof con.H2SO4 condenses to give an aromatic compound (C) of molecularformula C9H12.Identify A,B and C. Explain the reaction. [Mar-2012]

ANSWER: Zn/Hg – conc.HClCH3COCH3 CH3CH2CH3 + H2O

(A) 4(H) (B)

(C)RESULT:

S.No Compound Formula Name1 A CH3COCH3 Acetone2 B CH3CH2CH3 Propane3 C C6H3(CH3)3 Mesitylene

12. Compound (A) C3H6O does not reduce Tollen’s reagent, but under goes haloform reaction. Compound (A) under goes dehydration in the presence of dryHCl to give compound (B) C6H10O.Three molecules of compound (A) undergoes condensation reaction in the presence of conc. H2SO4 to give a cycliccompound (C).Identify the compounds A,B and C. Explain the reaction [Jun-12]

ANSWER:Dry HCl

(CH3 )2C=O + H2CHCOCH3 (CH3)2CHCOCH3

(A) −H2O (B)

(C)RESULT:

S.No Compound Formula Name1 A CH3COCH3 Acetone2 B (CH3)2C=CHCOCH3 Mesityl oxide3 C C6H3(CH3)3 Mesitylene


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13. An organic compound (A) of molecular formula C3H6O on reduction withLiAlH4 gives (B). Compound (B) gives blue colour in Victor Meyer’s test andalso form a Chloride (C) with SOCl2.The Chloride on treatment with alcoholicKOH gives (D). Identify the compounds A,B,C and D. Explain the reactions.[Mar-07]

ANSWER:

LiAlH4

CH3COCH3 (CH3)2 CHOH(A) 2(H) (B)

(CH3)2 CHOH + SOCl2 (CH3)2 CHCl + SO2 + HCl(A) (C)

Alc.KOH(CH3)2 CHCl CH3C=CH2 + KCl + H2O

(B) (D)RESULT:

S.No Compound Formula Name1 A CH3COCH3 Acetone2 B (CH3)2 CHOH 2-Propanol3 C (CH3)2 CHCl Isopropyl chloride4 D CH3C=CH2 Propene

14. Compound A of molecular formula C7H6O reduce Tollen’s reagents and alsogives Cannizzaro reaction. (A) on oxidation gives compound (B) withmolecular formula C7H6O2. Calcium salt of (B) on dry distillation givescompound (C) with molecular formula C13H10O Identify A,B, and C. Explain thereaction. [Mar-2008]

ANSWER: (O)C6H5CHO C6H5COOH

(A) alkaline KMnO4 (B)

(A) (B)


1 A C6H5CHO Benzaldehyde2 B C6H5COOH Benzoic acid3 C C6H5COC6H5 Benzo phenone


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15.An aromatic aldehyde (A) of molecular formula C7H6O which has the smell ofbitter almonds on treatment with (CH3CO)2O and CH3COONa to give compound(B) which is an aromatic unsaturated acid. (A) also react with (A) in thepresence of alc. KCN to give dimer (C).Identify A,B, and C. Explain the reactions.[Oct-2007]

ANSWER:

CH3COONaC6H5CH=O + CH3COOCOCH3 C6H5CH=CHCOOH + CH3COOH

(A) ∆ (B)

alc.KCNC6H5CH=O + H- C- C6H5 C6H5CH(OH)- C- C6H5

|| ||(A) O O

(C)RESULT:

S.No Compound Formula Name1 A C6H5CHO Benzaldehyde2 B C6H5CH=CHCOOH Cinnamic acid3 C C6H5CH(OH)COC6H5 Benzoin

16. An organic compound (A) (C7H6O) form a bisulphate .(A) when treated withalcoholic KCN forms (B) (C14H12O2) and (A) on refluxing with sodium acetateand acetic anhydride forms an acid (C) (C9H8O2).Identify A,B and C. Explain thereaction.[Jun-07]

ANSWER:


|| ||(A) O O

(B)CH3COONa

C6H5CH=O + CH3COOCOCH3 C6H5CH=CHCOOH + CH3COOH(A) ∆ (C)


1 A C6H5CHO Benzaldehyde2 B C6H5CH(OH)COC6H5 Benzoin3 C C6H5CH=CHCOOH Cinnamic acid


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17. An organic compound (A) (C7H6O) reduces Tollen’sreagent. On treating withalkali compound (A) form (B) and (C).Compound (B) treating with soda limeforms benzene (C7H8O) is an antiseptic .Identify A,B and C.Explain thereaction. [Oct-2006]

ANSWER:

NaOHC6H5CHO + C6H5CHO C6H5COOH + C6H5CH2 OH

(A) (B) (C)CaO /NaOH

C6H5COOH C6H6 + CO2

(B) ∆RESULT:

S.No Compound Formula Name1 A C6H5CHO Benzaldehyde3 B C6H5COOH Benzoic acid2 B C6H5CH2OH Benzyl alcohol

18. An organic compound (A) of molecular formula C7H6O is not reduced byFehling’s solution but will undergoes Cannizzaro reaction. Compound (A)react with Aniline to give compound (B) .Compound (A) also react with Cl2 inthe presence of catalyst to give compound (C).Identify A,B and C. Explain thereaction. [Mar-2006]

ANSWER: OH| [- H2O]

C6H5CH=O + C6H5NH2 C6H5CH−NHC6H5 C6H5CH=NC6H5

(A) (B)

(C)


1 A C6H5CHO Benzaldehyde2 B C6H5CH=NC6H5 Schiff’s base

3 C m- chloro benzaldehyde


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19. An aromatic compound (A) with molecular formula C7H6O has the smell ofbitter almonds.(A) react with Cl2 in the absence of catalyst to gives (B) and inthe presence of catalyst to fives compound (C). Identify the compounds A,Band C. Explain the reactions. [Mar-2009, Oct - 2014]

ANSWER:C6H5CHO + Cl2 C6H5COCl + HCl

(A) (B)

(C)


1 A C6H5CHO Benzaldehyde2 B C6H5COCl Benzoyl chloride

3 C m- chloro benzaldehyde

20. An organic compound (A) C7H8,on oxidation at 773K in the presence of V2O5

gives compound (B) of molecular formula C7H6O.(B) reduce Tollen’sreagent.(B) on heating with sodium acetate in the presence of acetic anhydridegives compound (C) of molecular formula C9H8O2.identify A,B and C. Explainthe reaction. [Mar-11, Jun- 13,Oct-13]

ANSWER:

(O)C6H5CH3 C6H5CHO + H2O

(A) air/ V2O5 773K (B)

CH3COONaC6H5CH=O + CH3COOCOCH3 C6H5CH=CHCOOH + CH3COOH

(A) ∆ (C)RESULT:

S.No Compound Formula Name1 A C6H5CH3 Toluene2 B C6H5CHO Benzaldehyde3 C C6H5CH=CHCOOH Cinnamic acid


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21. An organic compound (A) C7H8, on oxidation at 773K in the presence of V2O5

gives compound (B) of molecular formula C7H6O, which has smell of bitteralmonds. Alkaline KMnO4 oxidise compound (B) to (C) of molecular formulaC7H6O2. Compound (B) treatment with N2H4/KOH give back compound (A).Identify A,B and C. Explain the reaction. [Jun-2010]

ANSWER:

(O)C6H5CH3 C6H5CHO + H2O

(A) air/ V2O5 773K (B)

(O)C6H5CHO C6H5COOH

(B) alkaline KMnO4 (C)

N2H4 / KOHC6H5CHO C6H5CH3 + H2O

(B) 4(H) (A)RESULT:

S.No Compound Formula Name1 A C6H5CH3 Toluene2 B C6H5CHO Benzaldehyde3 C C6H5COOH Benzoic acid

22. An organic compound (A) of molecular formula C7H6O is not reduced byFehling’s solution but will undergoes Cannizzaro reaction. Compound (A)react with Aniline to give compound (B) .Compound (A) also react with Cl2 inthe absence of catalyst to give compound (C).Identify A,B and C. Explain thereaction. [Mar-2014]

ANSWER:

OH| [- H2O]

C6H5CH=O + C6H5NH2 C6H5CH−NHC6H5 C6H5CH=NC6H5

(A) (B)

C6H5CHO + Cl2 C6H5COCl + HCl(A) (C)


1 A C6H5CHO Benzaldehyde2 B C6H5CH=NC6H5 Schiff’s base3 C C6H5COCl Benzoyl chloride


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23. An organic compound (A) C7H8, on oxidation at 773K in the presence of V2O5

gives compound (B) of molecular formula C7H6O. (B) on reduction withlithium aluminium hydride to form (C) of molecular formula C7H8O. Identify(A), (B), and (C) and explain the reactions. [Mar – 2015]

ANSWER: (O)C6H5CH3 C6H5CHO + H2O

(A) air/ V2O5 773K (B)

LiAlH4

C6H5CHO C6H5CH2OH(A) [ H⎺ ] (C)


1 A C6H5CH3 Toluene2 B C6H5CHO Benzaldehyde3 C C6H5CH2OH Benzyl alcohol

24. Compound A of molecular formula C7H6O is called oil of bitter almonds. (A)on oxidation gives compound (B) with molecular formula C7H6O2 , which givesbrisk effervescence with NaHCO3 solution ,then, (A) is refluxed with alcoholicKCN compound (C) is formed. Identify A,B and C. Explain the reaction.[Jun-14]

ANSWER: (O)C6H5CHO C6H5COOH

(A) alkaline KMnO4 (B)


(A) || ||O O

(C)RESULT:

S.No Compound Formula Name1 A C6H5CHO Benzaldehyde2 B C6H5COOH Benzoic acid3 C C6H5CH(OH)COC6H5 Benzoin


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IMPORTANT: (From PTA)

1. An organic compound A (C7H6O) has bitter almond smell, with ammonia ‘A’gives ‘B’ (C21H18N2) with aqueous alcoholic KCN ‘A’ gives ‘C’ (C14H12O2). Witharomatic tertiary amine ‘A’ gives ‘D’ (C23H26N2). What are A, B, C and D explainthe reaction.

ANSWER:

RESULT:

S.No Compound Name1 A Benzaldehyde2 B Hydrobenzamide3 C Benzoin4 D Triphenyl methane dye

2. An organic compound ‘A’ (C7H6O) reduce Tollen’s reagent with aceticanhydride in the presence of sodium acetate. ‘A’ gives an α, β – unsaturatedacid ‘B’ (C9H8O2). With acetone in the presence of alkali ‘A’ gives ‘C’ (C10H10O).What are A, B, C and D explains the reactions. [Jun – 2015]


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ANSWER:

CH3COONaC6H5CH=O + CH3COOCOCH3 C6H5CH= CHCOOH + CH3COOH

(A) ∆ (B)

NaOHC6H5CHO + CH3COCH3 C6H5CH=CHCOCH3

(A) (C)RESULT:

S.No Compound Formula Name1 A C6H5CHO Benzaldehyde2 B C6H5CH=CHCOOH Cinnamic acid3 C C6H5CH=CHCOCH3 Bezalacetone

3. An organic compound A (C8H8O) undergoes iodo form test. When reduced withZinc amalgam and HCl it gives B (C8H10). ‘A ’with Br2 in ether at 273K gives ‘C’(C8H7OBr). Identify A, B, and C explain the reactions.

ANSWER: Zn / Hg – conc.HClC6H5COCH3 C6H5CH2CH3 + H2O

(A) 4(H) (B)

C6H5COCH2−H + Br−Br C6H5COCH2Br + HBr(B) Ether (C)


1 A C6H5COCH3 Acetophenone2 B C6H5CH2CH3 Ethyl benzene3 C C6H5COCH2Br Phenaceyl bromide

4. An organic compound ‘A’ (C5H10O) does not reduce Tollen’s reagent. It is alinear compound and undergoes iodoform test. On oxidation ‘A’ gives ‘B’(C2H4O2) and ‘C’ (C3H6O2) as the major product. Identify A, B, and C explain thereactions.

ANSWER:(O)

CH3COCH2CH2CH3 CH3COOH + CH3CH2COOH(A) (B) (C)


1 A CH3COCH2CH2CH3 2-Pentanone2 B CH3COOH Acetic acid3 C CH3CH2COOH Propionic acid


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***All the Best******** Electro Chemistry – I{ Q.No.70 (d) } *****

1. If 50 milli ampere of current is passed through copper coulometer for 60 min,calculate the amount of copper deposited.

Solution:Electrical charge input = I x t coulombs= 50 x10-3 A x60 x60sec= 180 coulombs.The chemical reaction is, Cu2+ + 2e → Cu(s)1 gm atom of copper requires 2F currentAmount of copperDeposted = (63. 5g.mol-1x 180 C )/2 x 96500C= 0.0593 gm.

Log63.5 = 1. 8027Log180 = 2.2552 (+)

4.0579Log 2 = 0.3010Log96500 = 4.9845 (+)

5.28554.0579 – 5.2855 =2.7724Antilog of 2.7724

= 0.059342. 0.1978 g of copper is deposited by a current of 0.2 ampere in 50 minutes. What

is the electro chemical equivalent of copper? (Mar-2009)Solution :Here, t = 50 minutes = 50 x60 seconds ; I = 0.2 ampere.Quantity of electricity used is

Q = I x t = 0.2 x50 x 60 = 600 coulombsAmount of copper deposited by 600 coulombs = 0.1978 gAmt. of copper deposited by1 coulomb = (0.1978 /600) g= 0.0003296 gElectrochemical equivalent of copper = 0.0003296= 3.296 x10–4gc–1= 3.296 x 10–7 kg c–1

Log0.1978=1.2962Log600 = 2.7781

2.5181Ant.log 2.5181

=0.0003297

3. What current strength in amperes will be required to liberate 10 g of iodinefrom potassium iodide solution in one hour? (Mar-2012)

Solution :127 g of iodine (1g eqvt) is liberated by = 96,500 coulomb10 g of iodine is liberated by = (96 500/127) x 10 coulombLet the current strength be = ITime in seconds = 1 x60 x60We know that the quantity of electricity, Q, used is given bythe expressionQ = I x time in secondsCurernt strength, I = Q /t = (96500 x 10) / (127 x 60 x 60)Curernt strength, I= 2.11 ampere.

Log 96500 =4.9845Log 10 =1.0000

5.9845Log127 =2.1078Log 60 =1.7781Log 60 =1.7781

5.66405.9845 – 5.6640

= 0.3205Ant.log of

0.3205=2.11


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4. An electric current is passed through three cells in series containingrespectively solutions of copper sulphate, silver nitrate and potassium iodide.What weights of silver and iodine will be liberated while 1.25 g of copper isbeing deposited? (Mar-2008,Jun-2013)

Solution :

Wt.of Cu / Wt.of Iodine=Eqvt.wt.Cu /Eqvt.wt.of Iodine1.25 / x = 31. 7 / 127x = (1. 25 x 127)/31. 7Hence, Wt. of Iodine ( x ) = 5.0 gAlso,

Wt.of Cu / Wt.ofSilver =Eqvt.wt. of Cu /Eqvt.wt.ofSilver1.25 / y = 31.7 / 108y = (1. 25 x 108)/31. 7

Wt. of silver (y) = 4.26 g

Log 1.25 =0.0969Log 127=2.1038(+)

2.2007Log31.7 =1.5010(-)

0.6997Ant.log of 0.6997

= 5.008Log 1.25 =0.0969Log 108=2.0334(+)

2.1303Log31.7 =1.5010(-)

0.6293Ant.log of 0.6293

= 4.259

5. The resistance of a 0.01 N solution of an electrolyte was found to 210 ohm at25oC using a conductance cell with a cell constant 0.88 cm-1. Calculate thespecific conductance and equivalent conductance of the solution.

Solution : [R = 210 ohm ;l / a =0.88 cm-1]Specific conductance (ĸ) = (l /a) x (1/R)=0.88cm-1/ 210ohm

ĸ = 4.19 x 10–3mho.m–1

Log 0.88 = 1.9444Log 210 = 2.3222 (-)

3.6222Ant.log of 3.6222 = 4.19 x 10-3

Equivalent conductance, λ = k x VV has 1 gram equivalent dissolved given is 0.01 N in 1000 ml.V = (1000 / 0.01)= 1, 00,000 mlλ = 4.19x10-3x1, 00,000λ=419.05 mho.cm2.gm.equiv.-1λ = 4.1905 x 10–2 mho m2 (gm.equiv)–1

6. 0.04 N solution of a weak acid has a specific conductance 4.23 x10-4 mho.cm-1.The degree of dissociation of acid at this dilution is 0.0612. Calculate theequivalent conductance of weak acid at infinite solution. (Oct-2009)

Solution :[Specific conductance k = 4.23 x10-4 mho.cm-1 ; α= 0.0612]λc = (k x 1000) /C = (4.23 x10-4) / 0.04 = (4.23 x10-4) /(4 x 10-2)λc= (4.23/4) x 10-2 = 1.0575 x 10-2 mho.cm2.eq-1.

λc = 105.75 mho.cm2.eq-1.α = λc/ λ∞λ∞ = λc / α = 10. 575 / 0.0612= 172.8 mho.cm2.gm.equiv.-1

Log 10.575 = 1.0242Log 0.0612 = 2.7868 (-)

2.2374Ant.log of 2.2374 = 1.728 x 102


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7. Equivalent conductivity of acetic acid at infinite dilution is 390.7 and for 0.1 Macetic acid the equivalent conductance is 5.2 mho.cm2.gm.equiv.-1. Calculatedegree of dissociation, H+ ion concentration and dissociation constant of theacid. (Jun-2007)

Solution :α = λc/ λ∞= 5.2 / 390.7 = 0.01333 = 1.33%CH3COOH ⇌ CH3COO¯ + H+C (1 – α) Cα Cα[H+] = Cα= 0.1 x0.0133 = 0.00133 MK =α2C /(1- α)= (0.01332 x 0.1) /(1- 0.0133)= (0.0001769 x 0.1)/0.9867 =0.0000179K= 1.79 x 10-5 M

Log 5.2 = 0.7160Log 390.7 = 2.5918 (-)

2.1242Ant.log of 2.1242=1.331 x 10-2

8. Ionic conductances at infinite dilution of Al3+ and SO42–are 189 ohm–1 cm2

gm.equiv.-1 and 160 ohm-1 cm2 gm.equiv.-1. Calculate equivalent and molarconductance of the electrolyte at infinite dilution.(Mar-2010,13)

Solution:

9. Calculate the pH of 0.1 M CH3COOH solution.Dissociation constant of acetic acidis 1.8 x10-5 M.(Jun-2009,Mar-2011)

Solution:For weak acids,[H+] = √KaxC= √1.8x10-5x 0.1 = √1.8x10-6[H+] = 1.34 x 10–3 M

pH = –log [H+] = -log (1.34 x10–3 )=-log1.34 - log10-3 = -log1.34- (-3log10)pH = -0.13 +3 = 2.87

Log1.34 = 0.1271

10. Calculate the pH of 0.02 m Ba(OH)2aqueous solution assuming Ba(OH)2 as astrong electrolyte.

Solution:Ba(OH)2 →Ba2+ + 2OH–

[OH–] = 2 [Ba(OH)2] = 2 x0.02 = 0.04 M= 4 x 10-2 Log4 = 0.6020


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pOH = –log [OH–] = -log (4 x 10-2)= -log4 – (-2)log10 =- 0.6020 + 2=1.398=1.40pH = 14 – pOH ; pH = 14 – 1.4 = 12.60

11. Find the pH of a buffer solution containing 0.20 mole per litre CH3COONa and0.15 mole per litre CH3COOH, Kafor acetic acid is 1.8 x10-5 .(Jun-2006,11,Oct-2006,07,08,11)

Solution :[Ka = 1.8 x10–5 ; [salt] = 0.20 mol /litre ; [Acid] = 0.15 mol /litre]pKa = –log KapKa = –log (1.8 x10–5) = - log 1.8 – log10-5= - log1.8 – (-5log10 ) = -0.2553+5

pKa = 4.7447pH= pKa + log ( [salt] / [Acid] )... Henderson - Hasselbalch equation= 4.7447 + log [0.20] /[0.15]pH = 4.7447 + log (4/3)= 4.7447 + 0.6021 – 0.4771pH = 4.8697

log1.8 = 0.2553

12. The Ka of propionic acid is 1.34 x10-5. What is the pH of a solution containing0.5 M propionic and 0.5 M sodiumpropionate ? What happens to the pH of thesolution when volume is doubled by adding water?(Mar-2006,Jun-2010)

Solution :Ka of propionic acid = 1.34 x10–5pKa = –log Ka = – log (1.34 x10–5)= -log1.34 – log10-5= -log1.34- (-5log10)= -0.1271 + 5 = 4.8729pKa = 4.87By Herderson - Hasselbalch equationpH = pKa + log ( [salt] /[acid])= 4.87 + log (0.5 /0.5)

pH = 4.87

Log1.34 = 0.1271

13. 0.5F of electric current was passed through 5 molar solutions of AgNO3,CuSO4

and AlCl3 connected in series. Find out the concentration of each of theelectrolyte after the electrolysis?

Solution:Amount of Ag deposited by passing 0.5F current = 108 x 0.5 = 54gmConcentration of AgNO3 after the electrolysis = 5 - (54/108) M= 5 –0.5 = 4.5 MAmount of Cu deposited by passing 0.5F current = 31.75 x 0.5 = 15.86gmConcentration of CuSO4 after the electrolysis = 5 - (15.86/63) M= 5 – 0.5 = 4.75 MAmount of Al deposited by passing 0.5F current = 9 x 0.5 = 4.5gmConcentration of AlCl3 after the electrolysis = 5 - (4.5/27) M= 5 – 0.17 = 4.83 M


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14. Specific conductance of 1M KNO3 solution is observed to be5.55 x10-3mho.cm-1. What is the equivalent conductance of KNO3 when one litreof the solution is used?

Solution:λc = (ĸ x 1000) /C= (5.55 x 10-3 x 1000) / 1= 5.55 x 10-3 x 103λc = 5.55 mho.cm2.gm.equiv-1.

15.The equivalent conductances at infinite dilution of HCl, CH3COONa and NaClare 426.16, 91.0 and 126.45 ohm-1 cm2 gm.equivalent-1respectively.Calculatethe λ∞of acetic acid.(Oct-2010, Jun-2008, Oct-2013)

Solution:λ∞ CH3COOH = λ∞ CH3COONa + λ∞ HCl – λ∞ NaClλ∞ CH3COOH = 91.0 + 426.16 – 126.45 = 390.71 ohm-1 cm2 gm.equivalent-1

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Created problem:16. Find the pH of a buffer solution containing 0.30 moles per liter CH3COONa and0.15 moles per litre CH3COOH, Ka for acetic acid is 1.8 x 10-5. (Oct-2008)

Like Problem number 11. { Ans: pKa = 4.7447 & pH = 5.0457 }

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forms cyclic compound ... an organic compound (a) c3h8o answers luca’s test within 5-10 mins and...

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