Chemistry Compulsory Problem by GEM - WAY TO SUCCESS

5 Mark Compulsory Problems with Solution

1

+2 CHEMISTRY

Q. 70 Compulsory Problems with Solution

Problems are solved in easiest way (As per Government Answer Key)


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SALIENT FEATURES Dear Students ❆

Q.No: 70 is asked as compulsory problem in Govt Exam.

❆

Two problems to be answered out of four problems.

❆

To simplify the problem, hints and expected compounds related to molecular formula, general formula are given in this material.

❆

Problems available in PTA book and Govt exam question paper (upto March 2013) are solved in easiest way.

❆

Repeated practice is enough to get full marks.

❆

Problems are given in the following order 70(a)Hydroxy derivatives (b) d-block elements or (c) Carbonyl compounds (d) Electro chemistry - I


3

CONTENTS

S. No.

Lesson Hydroxy Derivatives

I.

Problems based on primary alcohol

II.

Problems based on secondary alcohol

III.

Problems based on tertiary alcohol

IV.

Problems based on glycol and glycerol

V.

Problems based on phenol

VI.

Problems based on benzyl alcohol d-Block Elements

I.

Problems based on copper

II.

Problems based on chromium

III.

Problems based on zinc

IV.

Problems based on silver

V.

Problems based on gold Carbonyl compounds

I.

Problems based on acetaldehyde and acetone

II.

Problems based on benzedehyde

III.

Problems based on benzophenone & acetophenone Electro Chemistry - I

Page No.


16. Hydroxy Derivatives

8

Q. No. 70 A

Hydroxy derivatives problems are classified into aliphatic group and aromatic group.

The aliphatic problem part is further classified to 1°, 2° and 3° alcohols, glycol and glycerol.

The aromatic problem part is subdivided into phenol and benzyl alcohol.

There is no need of writing equation for hints (e.g. undergoes iodoform test). Equations to be written for conversions. Such as A → B, B → C, A → D.

The number of carbon atoms given in formula is C1 to C5, then the molecules may be aliphatic compound. C2H6O ⇒ C2H5OH.

The number of carbon atoms are C6 (or) greater than C6 in molecular formula, then the molecules may be aromatic compounds C6H6O ⇒ C6H5OH

General formula for saturated aliphatic alcohols is Cn H2n+2 O (Except Glycol, Glycerol)

The general formula for aliphatic aldehydes (or) ketones CnH2nO.


14

II. Problems based on Secondary alcohol 6.

An organic compound (A) C3H8O answers Lucas test within 5-10 minutes and on oxidation forms (B) (C3H6O). This on further oxidation forms (C) (C2H 4O 2) which gives effervescence with Na2CO3. (B) also undergoes iodoform reaction. Identify (A), (B) and (C). Explain the conversion of (A) to (B) and (C). (June-07, 09) (i)

oxidation C3 H8O  → C3 H6 O

(A)

( B)

( ) CH 3 − CH − CH 3 → CH 3 − C− CH 3 O

|

||

OH

O

(A)

(B)

oxidation (ii) C3 H 6 O  → C2 H 4O 2

(C)

( B)

( ) CH 3 − C − CH 3  → CH 3COOH H + K Cr O 0

2

2

7

OB

(C)

( )

Compound

Structure

Name

A

CH 3 − CH − CH 3 | OH

Isopropyl alcohol

B

CH 3 − C − CH3 || O

Acetone

C

CH3COOH

Acetic Acid

5 Mark Compulsory Problems with Solution 7.

15

An organic compound A of molecular formula C3H6O on reduction with LiAlH4 gives B. Compound B gives blue colour in Victor Meyer's test and also forms a chloride C with SOCl2. The chloride on treatment with alcoholic KOH gives B. Identify A, B and C and explain the reactions. (PTA Question Bank, March-07) LiAlH 4 C 3 H 6 O  → C3 H8 O Redcution

(i)

(A)

(B)

CH3

(ii)

C

CH3

( Blue colour in Victor Mayor Test )

LiAlH 4

CH3

CH

O

OH

(A)

(B)

CH3

SOCl2 C3 H 8 O  → C3 H 7 Cl

( B)

CH3

(C)

CH

CH3 + SOCl2

CH 3

CH

OH

Cl

(B)

(C)

CH 3 + SO2 + HCl

Compound

Structure

Name

A

C H 3 − C − CH 3

Acetone

||

O

B

CH 3 − CH − CH 3 | OH

Iso propyl alcohol

C

CH 3 − CH − CH 3 | Cl

Iso propyl chloride


16

Two organic compound A and B have the same molecular formula C2H6O. A react with metalic sodium to give hydrogen where 'B' does not. A on strong oxidation gives C. 'C' gives effervescence with NaHCO3. Identify A, B and C. Explain the reactions. (Model Question Paper-IV) (i)

Compound 'A' (C 2 H 6 O) react with metallic sodium gives hydrogen. So 'A' is ethanol (C2H 5OH). 'B' is dimethyl ether CH3–O–CH3

(ii) C2 H 6 O → ( C ) (gives brisk effervescence with NaHCO3 ) ( 0)

(A)

( ) C 2 H 5 OH  → CH 3COOH H + / K Cr O 0

(A)

2

2

7

(C)

Compound

Structure

Name

A

C2H5OH

Ethanol

B

CH3– O – CH3

Dimethyl ether

C

CH3COOH

Acetic acid


17

Two isomers (A) and (B) have the same molecular formula C4H10O. (A) when heated with copper at 573 K gives an alkene (C) of molecular formula C4H8. (B) on heating with copper at 573 K gives (D) of molecular formula C4H8O which does not reduce Tollen's reagent but answer iodoform test. Identify (A), (B), (C) and (D) and explain the reactions. (March-09) Cu (i) C 4 H10 O  → C4H8 573 K

(A)

(C)

H 3C

CH3 C

H 3C

OH

CH3

Cu 573 K

C

CH2 + H2O

CH3

(A)

(C)

Cu → C4 H8 O (does not reduce Tollens reagent (ii) C 4 H10 O  573 K

(A)

CH 3

(D)

CH 2 (B)

Compounds

CH

CH 3

OH

Structure H3 C H3 C

B

Cu 573 K

CH 3CH 2

CO

CH 3

(D)

Name

CH3 C

A

but answer iodoform test)

3° butyl alcohol

OH

CH 3CH 2 − CH − CH3 |

2° butyl alcohol

OH

CH3

C D

CH3

C

CH2

CH3–CH2–CO–CH3

Isobutylene

Ethyl methyl ketone


18

10. An organic compound 'A' has the formula C3H8O with sodium hypochlorite it gives 'B' (C3H6O). 'B' reacts with chlorine to give 'C' (C3H3Cl3O). 'A' with anhydrous zinc chloride and conc HCl gives turbidity after 5 to 10 minutes. What are A, B and C? Explain the reactions. (PTA, March-06) A + Con.HCl + ZnCl2 ⇒ 

Turbidity after  ⇒ 'A ' is 2° alcohol [ ] 5 − 10 miniutes 

(i)

C3 H8O  → C3 H 6 O (A)

( B)

CH3 CH3

CH OH

CH3

Sodium Hypochlorite (D)

C

CH3

O + H2O

(B)

(A)

→ C3 H3 Cl3 O (ii) C3 H 6 O + Cl2  ( B)

( C)

3Cl2 CH3 – CO – CH3  → CCl3 – CO – CH3 + 3HCl (B)

Compounds A

B

(C)

Structure CH3 CH3

CH OH

CH3 CH3

C

Name

C

O

CCl3 – CO – CH3

Iso propyl alcohol

Acetone

Trichloro acetone


19

11. Compound (A) of molecular formula C3H8O liberates hydrogen with sodium metal. (A) with P/I2 gives (B). Compound (B) on treatment with silver nitrite gives (C) which gives blue colour with nitrous acid. Identify (A), (B) and (C) and explain the reactions. (Sep-09)

→ (B) (i) C3 H8 O  P/I 2

(A)

CH3 CH3

CH

OH

P/I2

CH3 CH

CH3

I

(B)

(A)

(ii) (B) + Silver Nitrite  → (C) CH3 CH3

CH

I

CH3

AgNO 2

NO2 C

CH3

H

(B)

Compounds

(C)

Name

Structure

CH3 A

CH3

CHOH

CH3

B

CH3 CH3

C

CH3

I

CH

C

NO2 H

Isopropyl Alcohol

Isopropyl Iodide

2-Nitro propane


53

Q.No. 70 b

4. d-Block Elements Name of the elements Copper Chromium (Cu) (Cr)

Zinc (Zn)

Silver (Ag)

Gold (Au)

Period

4

4

4

5

6

Group

11

6

12

11

11

Name of the compounds

Formala

Copper sulphate penta hydrate (Blue vitriol)

CuSO4.5H2O

Potassium dichromate

K2Cr2O7

Silver Nitrate (Lunar Caustic)

AgNO3

Zinc carbonate (Calamine)

ZnCO3

Purple of cassius

Colloidal gold (Au)

Hints given in Problem

Name

1. Orange Red orystals

Potassium dichromate

2. Yellow colour compound

Potassium chromate

3. Philosopher's cool

Zinc oxide

4. Compound used in photography

Silver bromide


55

I. Problems based on Copper 1.

An element (A) occupies group number 11 and period number 4. This metal is extracted from its mixed sulphide ore (B). (A) reacts with dil. H2SO4 in presence of air and forms (C) which is colourless. With water (C) gives a blue colour compound D. Identify (A), (B), (C) and D and explain the reactions. (March-07, July-10) (i)

A ⇒ Period 4, Group 11 ⇒ Copper (Cu) B ⇒ Copper pyrite ⇒ CuFeS2

(ii) (A) + dil.H2SO4 + Air  → (C)

2Cu + 2H2 SO4 + O2  → 2CuSO4 + 2H 2 O (A)

(C)

(iii) 'C' + water  → D (Blue colour) 5H 2 O CuSO4  → CuSO45H2O

(C)

(D)

Compounds

Structure

Name

A

Copper

Cu

B

Copper pyrite

CuFeS2

C

Copper sulphate

CuSO4

D

Copper sulphate penta hydrate CuSO45H2O


56

An element (A) belonging to Group No. 11 and period No. 4 is extracted from the pyrite ore. (A) reacts with oxygen at two different temperatures forming compounds (B) and (C). (A) also reacts with conc. HNO3 to give (D) with the evolution of NO2. Find out (A), (B), (C) and (D). Explain the reactions. (Sep-07, March-10, 13) (i)

A is an element ⇒ Period 4 Group 11 ⇒ Copper (Cu) less than 1370 K 2Cu + O 2  → 2CuO B

( )

(A)

Greater than 1370 K 4Cu + O2  → 2Cu 2 O

( A)

(C)

(ii) (A) + conc.HNO3  → (D) + NO2

Cu + 4HNO3  → Cu ( NO3 ) 2 + 2NO2↑ + 2H2O (A)

(D)

Compounds

Structure

Name

A

Copper

Cu

B

Cupric oxide

CuO

C

Cuprous oxide

Cu2O

D

Copper (II) Nitrate

Cu(NO3)2


57

A reddish brown metal 'A' on heating to redness gives 'B' which is Black in colour. 'B' dissolves in dil.H2SO4 to give 'C' which is blue crystal. On heating to 230°C, 'C' gives 'D' which is white in colour, which on further heating to 720°C gives back 'B'. What are A, B, C, and D. Explain the reactions. (Model Question Paper - II) (i)

1370 K Reddish brown metal  → ( B ) Black colour

(A)

1370 K 2Cu + O2  → 2CuO

(A)

( B)

(ii) (B) + dil H2SO4  → (C) Blue colour

CuO + H 2SO4  → CuSO4 + H2 O (A)

CuSO 4 + 5H 2 O  → CuSO 4 ⋅ 5H 2 O (C)

230° C (iii) (C)  → (D) white colour compound

CuSO 4 5H 2 O  → CuSO 4 (C)

( D)

Compounds

Structure

Name

A

Copper

Cu

B

Cupric oxide

CuO

C

Blue vitriol

CuSO45H2O

D

Copper sulphate

CuSO4


58

An element (A) belongs to group number 11 and period number 4. (A) is a reddish brown metal. (A) reacts with HCl in the presence of air and gives compound (B). (A) also reacts with conc. HNO3 to give compound (C) with the liberation of NO2. Identify (A), (B) and (C), Explain the reactions. (Mar-06, July-10)

}

}

Reddish brown Group 11 ⇒ ⇒ Copper 'Cu' metal 'A' Period 4 (i)

( in presence of ) (A) + HCl  → Compound (B) air

2Cu + 4HCl + O2 (air)  → 2CuCl 2 + 2H 2 O (A)

( B)

(ii) (A) + conc.HNO3  → Compound (C) + NO2

→ Cu ( NO 3 )2 + 2NO 2 + 2H 2 O Cu + 4HNO3  ( A)

(C)

Compound

Structure

Name

A

Copper

Cu

B

Copper (II) Chloride

CuCl2

C

Copper Nitrate

Cu(NO3)2


59

Compound (A) also known as blue vitriol can be prepared by dissolving cupric oxide in dil H2SO4. A on heating to 230°C gives compound B which is white in colour. A reacts with excess of NH4OH and gives C which is a complex salt. A also reacts with H2S and gives compound D which is black in colour. Find out A, B, C and D. Explain the reactions. (PTA) A ⇒ Blue vitriol ⇒ CuSO45H2O (i)

230° C (A) → (B) colourless 230° C CuSO 4 i5H 2 O  → CuSO 4 −5H 2 O

(A )

(B)

230° C (ii) (B) + NH4OH → (C) co-ordination compound

CuSO 4 + 4NH 4 OH  →  Cu ( NH 3 )4  SO 4 + 4H 2 O ( B) (C)

(iii) (B) + H2S  → (D) Black colour CuSO 4 + H 2 S  → CuS+ H 2SO 4 ( B)

(D)

Compounds

Structure

Name

A

Blue vitriol

CuSO45H2O

B

Copper sulphate

CuSO4

C

Tetramine Copper (II) Sulphate

 Cu ( NH3 )4  SO4

D

Copper sulphide

CuS


60

Compound (A) is a sulphate compound of group 11 element. This compound is also called as Blue Vitriol. The compound undergoes decomposition at various temperatures. 100°C 230°C 720°C A  → B → C → D

What are (A), (B), (C) and (D). Explain the reactions. (July-09) 100°C 230°C 720°C A  → B → C → D 100° C 230° C 720° C CuSO 4 5H 2 O  → CuSO 4 H 2 O  → CuSO 4  → CuO + SO 3

(A )

( B)

Compounds

( C)

Structure

(D)

Name

A

Copper Sulphate Pentahydrate CuSO45H2O

B

Copper Sulphate Monohydrate CuSO4H2O

C

Copper Sulphate

D

Cupric oxide

CuSO4 CuO


77

Q.No. 70 c

18. Carbonyl Compounds I. Problems based on Acetaldehyde and Acetone

1.

An organic compound A(C2H4O) undergoes iodoform test. With hydrazine and sodium ethoxide 'A' gives 'B' (C2H6), a hydro carbon. 'A' with H2SO4 gives 'C' (C6H12O3). What are A, B and C? Explain the reactions. (PTA) Hydrazine → C2 H 6 (i) C2 H 4 O (undergoes iodoform test)  Sodium ethoxide

( A)

( B)

N 2 H4 CH 3 CHO  → CH 3 CH 3 C 2 H 5 ONa

(A)

( B)

→ C 6 H12 O3 (ii) C 2 H 4 O + conc.H 2SO 4  (A)

(C)

CH3

3CH3CHO

conc.

CH

H2SO4

H3C

O

O

CH

CHCH3 O (C)

Compounds

Structure

Name

A

Acetaldehyde

CH3CHO

B

Ethane

CH3–CH3 CH3 CH

C

Paraldehyde CH3

O

O

CH

CHCH 3 O


78

An organic compound A (C2H4O) with HCN gives B(C3H5ON). B on hydrolysis gives C (C 3 H 6 O 3) which is an optically active compound. A also undergoes iodoform test. What are A, B and C? Explain the reactions. (PTA, Sep-11)

(i) C2 H 4 O (undergoes iodoform test) + HCN  → C2 H5ON (A)

( B)

CH 3 CHO + HCN  → CH 3 CH − CN (A) | OH (B)

Hydrolysis (ii) C3 H5 ON  → C3 H6 O3 (Optically active)

( B)

( C)

H2O CH 3 − CH − CN  → CH 3 − CH − COOH | | OH OH

( B)

(C)

Compounds

Structure

Name

A

CH3CHO

Acetaldehyde

B

CH 3 − CH − CN | OH

Acetal dehyde cyano hydrin

C

CH 3 − CH − COOH | OH

Lactic acid


79

Compound (A) having the molecular formula C2H4O reduces Tollen's reagent. (A) on treatment with HCN followed by hydrolysis gives the compound (B) with molecular formula C3H6O3. Compound B on oxidation by Fenton's reagent gives the compound (C) with the molecular formula C3H4O3. Find (A), (B) and (C). Explain the reactions. (July-08, Oct-08, Mar-10) Hydrolysis → C3 H 6 O 3 (i) C 2 H 4 O + HCN 

(A)

( B)

CH3CHO + HCN

CH3

CH

CN

HOH

CH3

CH

COOH

OH

OH (A)

(B)

Oxidation → C3 H 4 O 3 (ii) ( B ) + Fenton's Reagent 

(C)

(O)

CH3

CH

COOH

2+

Fe / H2O2

CH3COCOOH

OH

(C)

(B)

Compounds

Structure

Name

A

CH3CHO

Acetaldehyde

CH3

B C

CH

COOH

Lactic acid

OH

CH3COCOOH

Pyruvic acid


80

An organic Compound (A) C2H3OCl on treatment with Pd / BaSO4 gives (B) (C2H4O) which answers iodoform test. (B) When treated with conc. H2SO4 undergoes polymerisation to give (C) a cyclic compound. Identify (A), (B) and (C) and explain the reactions. (Sep-09) (i)

C2 H3OCl → C2 H4 O (undergoes iodoform test) pd / BaSO4

( B)

(A)

pd / BaSO4 CH 3 COCl + H 2  → CH3 CHO

(A)

(ii)

( B)

Polymerisation → ( B ) 

(C) Cyclic compound

CH3

3CH3CHO

conc.

CH

H2SO4

H3C

O

O

CH

CHCH 3 O

(C)

Compounds

Structure

Name

A

Acetyl chloride

CH3COCl

B

Acetaldehyde

CH3–CHO CH3 CH

C

Paraldehyde CH3

O

O

CH

CH O

CH3


81

Compound (A) with molecular formula C2H4O reduces Tollen's regent. (A) on treatment with HCN gives compound (B). Compound (B) on hydrolysis with an acid gives compound (C) with molecular formula C3H6O3. Compound (C) is optically active. Compound (C) on treatment with Fenton's reagent gives compound (D) with molecular formula C 3 H 4 O 3 . Compound (C) and (D) give effervesence with explain the reactions. (March-10, Sep-11) (i)

C 2 H 4 O(reduces Tollen's reagent) + HCN  → ( B) (A)

CH 3 CHO + HCN  → CH 3 − CH − CN (A) | OH ( B)

(ii)

Acid hydrolysis → C3 H6 O3 (optically active) ( B ) 

(C)

CH3

(iii)

CH

CN

Hydrolysis

CH3

CH

OH

OH

(B)

(C)

Fenton's Reagent ( C ) →

COOH

C3 H 4 O 3 ( D) 2+

Fe H 2 O 2 CH 3 − CH − COOH  → CH3 − CO − COOH ( 0) (D) | OH

(C)


82

Compounds

Structure

Name

A

CH3CHO

Acetaldehyde

CH3

B C

CH

CN

Acetaldehyde Cyanohydrin

COOH

Lactic acid

OH

CH3

CH OH

D

CH3COCOOH

Pyruvic acid


83

Compound A (C2H4O) reduces Tollen's reagent. A on treatment with zinc amalgam and conc. HCl give compound B. In presence of conc. H2 SO 4. A forms a cyclic structure C which is used as hypnotic. Identify A, B and C. Explain the reactions. (July-11) → ( B) (i) C 2 H 4 O  (A)

Zinc amalgam conc. HCl

Zn/Hg CH 3 CHO  → CH 3 − CH 3 HCl

(A)

(ii)

( B)

→ ( A ) + conc.H 2SO4 

Hypnotic ( C)

CH3

3CH3CHO

conc.

CH

H2SO4

(B)

CH3

O

O

CH

CH

CH3

O (C)

Compounds

Structure

Name

A

CH3CHO

Acetaldehyde

B

CH3CH3

Ethane

CH3 CH

C CH3

O

O

CH

CH O

Paraldehyde CH3


84

An organic compound 'A' (C5H 10O) does not reduce Tollen's reagent. It is a linear compound and undergoes iodoform test on oxidation 'A' gives 'B' (C2H4O2) and 'C' (C3H6O2) as the major product. Identify A, B and C. Explain the reactions. (PTA) → C 2 H 4 O 2 + C3 H 6 O 2 (i) C5 H10 O  Oxidation

(A)

(B)

(C)

( ) CH 3 − C − CH 2 CH 2 CH 3  → CH3 COOH + C2 H5 COOH (B) ( C) O 0

(A)

Compounds

Structure

Name

A

CH 3 − C − CH 2 CH 2 CH 3 || O

Methyl propyl ketone

B

CH3COOH

Acetic acid

C

C2H5COOH

Propionic acid


85

An organic compound A (C2H3N) on reduction with SnCl2/HCl gives B (C2 H 4O) which reduces Tollen's reagent. Compound B on reduction with N 2 H 4 /C 2 H 5 ONa gives C (C 2 H 6 ). Identify the compounds A, B and C. Explain the reactions involved. (Sep-12) SnCl2 /HCl (i) C 2 H 3 N → C 2 H 4 O (reduces Tollen's reagent) reduction

(A)

( B)

SnCl 2 / HCl H2 O CH 3 − CN   → CH 3 CH = NH.HCl  → CH 3 CHO + NH 3

(A)

(B)

→ CH3COO- + 2Ag + 2H2O (ii) CH3CHO + 2Ag+ + 3OH-  N2 H4 (iii) C2 H 4 O  → C2 H 6 C H ONa

( B)

2

5

(C)

N 2 H 4 / C 2 H 5 ONa CH 3 CHO  → CH 3 − CH 3

(C)

Compounds

Structure

Name

A

CH3 – CN

Methyl cyanide

B

CH3CHO

Acetaldehyde

C

CH3 – CH3

Ethane


100

13. Electro Chemistry-I

Q. No. 70 d

Hints

Quantity of current Q = It

Mass of substance liberated by passing current m = ZIt

Equivalent mass =

(Equivalent mass of Cu = 31.77, Ag = 108, I = 127, Al = 9)

Electro chemical equivalent =

1 faraday = 96495 coulomb

Equivalent conductance λC =

Atomic mass Valency

Equivalent mass 96495 κ × 1000 mho.cm2(g.eqiv.)–1 C

(or)

κ × 10 −3 mho.m2 (g.equiv)–1 N

κ × 10 −3 mho.m2.mol–1 M

Molar conductance µC =

Molar conductance = Cell constant × Conductivity (or) Cell constant / Resistance l a

Cell constant =

Degree of dissociation α =

For weak acids  H +  = Cα = K a × C

For weak bases  OH −  = Cα = K b × C

Dissociation constant for weak acid K a =

λα (or) λC

Ka C

α2C 1− α


pH = –log [H+]

pOH = –log [OH–]

pH + pOH = 14

Kw = [H+] [OH–] = 1 × 10–14 at 298 K

pKw = 14

pKa = –log Ka

pKb = –log Kb

Hendersons equation for acid buffer

 Salt  pH = pKa + log    Acid 

Henderson equation for basic buffer

Salt  pOH = pKb + log   Base   

101


102

Exercise Problems 1.

Find the pH a buffer solution containing 0.20 mole per liter sodium acetate and 0.15 mole per litre acetic acid. Ka for acetic acid is 1.8 × 10–5. (June-06, 11, Sep-06, 07, 11) Solution:

[CH3COONa] = 0.20 mole / litre [CH3COOH] = 0.15 mole / litre Ka = 1.8 × 10–5 mole / litre pH of Buffer solution = ? pKa = – log Ka = – log10 1.8 × 10–5 = −  log10 1.8 + log10 10 −5  = – [0.2553 – 5]

= – [–4.7447]

pKa = 4.7447 Henderson equation pH = pK a + log10

[Salt ] [ acid ]

= 4.7447 + log = 4.7447 + log

[ 0.20] [0.15]

20 15

= 4.7447 + log

4 3

= 4.7447 + log4 – log3 = 4.7447 + 0.6021 – 0.4771 = 4.7447 + 0.1250 pH = 4.8697


103

Find the pH of a buffer solution containing 0.3 mole per litre of CH3COONa and 0.15 mole per litre CH3COOH. Ka for acetic acid is 1.8 × 10–5. (Sep-08) Solution:

[CH3COONa] = 0.30 mole / litre–1 [CH3COOH] = 0.15 mole / litre–1 Ka = 1.8 × 10–5 mole / litre–1 pH of buffer solution = ? pKa = – log Ka = – log10 [1.8 × 10–5] pKa = 4.7447 Henderson equation pH = pk a + log10

[Salt ] [ Acid ]

= 4.7447 + log = 4.7447 + log

[ 0.30] [0.15]

30 15

2

= 4.7447 + 0.3010 pH = 5.0457


104

What is the pH of a solution containing 0.5 M propionic acid and 0.5 M sodium propionate? The Ka of propionic acid is 1.34 × 10–5. (March-06, July-10) Solution:

Propionic acid = 0.5 M Sodium propionate = 0.5 M Ka = 1.34 × 10–5 pH of buffer solution = ? pKa = – log Ka = – log10 [1.34 × 10–5] = −  log10 1.34 + log10 10 −5  = − [ log10 0.1217 − 5] pKa = 4.8729 Hendenson Equation pH = pK a + log10

[Salt ] [ Acid ]

= 4.8729 + log

0.5 0.5

= 4.8729 + log1 = 4.8729 + 0 pH = 4.8729 When the volume of buffer solution is doubled, the pH of the solution does not change.

For complete Material Contact

Mrs. Sumathi Gunaseelan ✆ 90802-28421

Chemistry Compulsory Problem by GEM - WAY TO SUCCESS

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