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)
5 Mark Compulsory Problems with Solution
2
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
5 Mark Compulsory Problems with Solution
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.
5 Mark Compulsory Problems with Solution
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.
5 Mark Compulsory Problems with Solution
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
5 Mark Compulsory Problems with Solution 8.
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
5 Mark Compulsory Problems with Solution 9.
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
5 Mark Compulsory Problems with Solution
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
5 Mark Compulsory Problems with Solution
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
5 Mark Compulsory Problems with Solution
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
5 Mark Compulsory Problems with Solution
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
5 Mark Compulsory Problems with Solution 2.
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
5 Mark Compulsory Problems with Solution 3.
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
5 Mark Compulsory Problems with Solution 4.
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
5 Mark Compulsory Problems with Solution 5.
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
5 Mark Compulsory Problems with Solution 6.
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
5 Mark Compulsory Problems with Solution
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
5 Mark Compulsory Problems with Solution 2.
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
5 Mark Compulsory Problems with Solution 3.
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
5 Mark Compulsory Problems with Solution 4.
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
5 Mark Compulsory Problems with Solution 5.
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)
5 Mark Compulsory Problems with Solution
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
5 Mark Compulsory Problems with Solution 6.
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
5 Mark Compulsory Problems with Solution 7.
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
5 Mark Compulsory Problems with Solution 8.
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
5 Mark Compulsory Problems with Solution
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
5 Mark Compulsory Problems with Solution
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
5 Mark Compulsory Problems with Solution
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
5 Mark Compulsory Problems with Solution 2.
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
5 Mark Compulsory Problems with Solution 3.
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.
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