Concept of Accid and Bases

CONCEPTS OF ACIDS AND BASES

Strength of acids and bases (qualitative treatment), balancing equations of chemical reaction, (including oxidation-reduction using ion-electron and oxidation number methods)

Introduction

There are several so-called theories of acids and bases, but they are not really theories but merely different definitions of what we choose to call an acid or a base. Since it is only a matter of definition, no theory is more right or wrong than any other, and we use the most convenient theory for a particular chemical situation. So before we talk of strength of acids and bases, we need to know several theories.

Various Theories Regarding Acids & Bases

i) Arrhenius Theory: According to Arrhenius, substances producing H+ ions in solution are acids and those producing OH- ions in solution are bases. Therefore, substances like H2O, HCl, H2SO4, CH3 COOH etc. are acids and the ones like NH4OH, NaOH, KOH, H2O etc. are bases.

ii) Bronsted-Lowry Theory: In 1923, Bronsted and Lowry independently defined acids as proton donors, and bases as proton acceptors. For aqueous solutions the definition does not vary much for acids from the Arrhenius theory but it widens the scope of bases. In this, the bases need not contain OH- ions and simply have to accept protons. So ions like Cl-, CH3COO-, Br- etc. which do not contain OH- ions can be considered as bases under this definition.

Levelling Solvents: Whenever an acid is dissolved in water, it acts as an acid only if the solvent acts as a base. That is, if we dissolve acids like HCl, HNO3 etc in water, their acidic strength is almost the same since water acts as a base for both these acids. Infact, it is known that all strong acids show equal acidic strength when dissolved in water. This is because, water acts as a base to all these acids and thus forces them to donate almost the same amount of protons irrespective of their chemical nature. Since water levels the acidic strength of strong acids, it is referred to as a levelling solvent. In order to measure the strength of strong acids, they are dissolved in glacial acetic acid and the amount of protons measured by conductometry. It is found that the strength of acids varies as

HClO4 > HBr> H2SO4 > HCl > HNO3

Amphiprotic species: Many molecules and ions can behave like water and may either gain or lose a proton under the appropriate conditions. Such species are said to be amphiprotic. eg.

Acid1 Base2 Acid2 Base1

HS- + OH- H2O + S2-

HBr + HS- H2S + Br-

+ CN- HCN +

H3O+ + H2CO3 + H2O

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The hydroxides of metals near the boundary between metals and non-metals in the periodic table, are amphiprotic and so react either as acids or as bases.

[Al(H2O)3 (OH)3] + OH- H2O + [Al(H2O)2(OH)4]-

H3O+ + [Al(H2O)3(OH)3] [Al(H2O)4(OH)2]+ + H2O.

Exercise 1: Identify conjugate acid base pair in the following reactionsC6H5OH + H2O H3O+ + C6H5O–

HCl + OH– H2O + Cl–

iii) Lewis Theory: Lewis developed a definition of acids and bases that did not depend on the presence of protons, nor involve reactions with the solvent. He defined acids as materials which accept electron pairs, and bases as substances which donate electron pairs. Thus a proton is Lewis acid and ammonia is Lewis base since, the lone pair of electrons on the nitrogen atom can be donated to a proton:

H+ : NH3 [H : NH3]+

Conditions to be a Lewis Acid

i) Compounds whose central atoms have an incomplete octet. E.g. BF3, AlCl3, GaCl3

etc.

ii) Compounds in which the central atom has available d-orbital and may acquire more than an octet of valence electrons.

E.g. SiF4 + 2F- SiF62-

Other examples are : PF3, SF4, SeF4, TeCl4.

iii) All simple cations : Na+, Ag+, Cu2+, Al3+, Fe3+ etc.

Conditions to be a Lewis Base

i) All simple negative ions, Cl- , F- etc.

ii) Molecules with unshared pair of electrons, H2O, NH3 etc.

iii) Multiple bonded compounds which form co-ordination compounds with transition-metals . E.g., CO, NO, Ethylene, Acetylene etc.

Exercise 2: In the following reaction, identify each of the reactant as a Lewis acid or Lewis base

i) Cr+3 + 6H2O

ii) BF3 + (C2H5)2O F3B: O(C2H5)2


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Hard and soft acids and bases (Principle of HSAB)

Lewis acids & bases are classified as hard & soft acids & bases. Hardness is defined as the property of retaining valence electrons very strongly.

Thus hard acid is that in which electron-accepting atom is small with a high positive charge and no electron which are easily polarised or removed.

Soft acid is that in which the acceptor atom is large, carries a low positive charge or it has electrons in orbitals which are easily polarised or distorted.

A lewis base which holds its electrons strongly is called hard base eg. OH -, F-, NH3, H2O etc.

A lewis base in which the position of electrons is easily polarised or removed is called a soft base eg. I-, CO, CH3S-, (CH3)3P etc.

A hard acid prefer to bind to hard bases and soft acids prefer to bind to soft bases. The bonding between hard acids & hard bases is ionic & that between soft acids & bases is mainly covalent.

Exericise-3: Identify the hard acid and hard bases among the following

OH–, H2O, Al+3, Cr+3, Sn+4,

Relative Strength of Acids and Bases

i) Predict the relative acidic strength among the following:

CH3 - H, H2N - H, OH - H, F -H

Assume that all these compounds have lost their protons. So we get -CH3 , -NH2, -OH, F-

We shall qualitatively analyse the charge density in each of the species and we shall follow the rule given below:

Larger the volume over which the charge is spread.

Smaller is the charge density.

Smaller is the basic character of the ion to attract a proton.

Larger is the acidity of the conjugate acid.

In the above case one can see that, the size of the central atom over which the negative charge is present is decreasing from left to right. But it can also be seen that 3/4 th of the volume of C is overlapped by hydrogens in -CH3, 2/3 of the volume of N is overlapped by hydrogens in -NH2, 1/2 the volume of oxygen is overlapped by hydrogens in -OH where as, the entire volume of F is available for the charge in -F SO, actually the space available for the charge is increasing in the order -CH3 < -NH2 < -OH < -F. Therefore, the following conclusion may be arrived at.

CH4, NH3, H2O, HF.


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1. Volume available for the negative charge is increasing in the conjugate bases from left to right.

2. Charge density of the conjugate bases is decreasing from left to right.

3. Basicity of the conjugate bases is decreasing from left to right.

4. Acidity of the acids is increasing from left to right.

Therefore the increasing acidic character is CH4 < NH3 <H2O <HF.

ii) Predict the relative acidic strength among the following :

HF, HCl, HBr, HI.

Assume that each has lost a proton

We get F- , Cl- , Br- , I-

1. Volume available for the charge is increasing from left to right

2. Charge density is decreasing from left to right.

3. Basicity of the ions is decreasing from left to right.

4. Acidity of the conjugate acids is increasing from left to right.

Therefore increasing acidic strength is : HF <HCl <HBr <HI

iii) Predict the relative acidic strength among the following :

HOCl, HOClO, HOClO2 , HOClO3 .

Assume that each has lost a proton. So we get,

-OCl, -OClO, - OClO2, -OClO3

The species which have more than one oxygen atoms show resonance.

That is:

It can be clearly seen that more the no. of oxygen atoms, more the resonance and more is the volume available for the negative charge.

Therefore -OCl, -OClO, - OClO2, -OClO3

1. Volume available for the negative charge is increasing from left to right



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3. Basicity of the anion is decreasing from left to right.

4. Acidity of the conjugate acid is increasing from left to right.

Therefore, the order of acidic strength is :

HOCl < HOClO < HOClO2 < HOClO3

iv) Predict the relative acidic strength among the following :

H2O, H2S, H2Se, H2Te.

Assume that each has lost a proton : So, we get : HO-, HS- , HSe- , HTe-

It can be easily seen that the volume available for the negative charge is increasing from HO- to HTe-, Therefore :

1. Volume available for the negative charge is increasing from left to right.


3. Basiciy is decreasing from left to right.

4. Acidity of conjugate acids is increasing from left to right.

v) In general, for oxo-acids with different central atoms of same oxidation states, acidity increases with the decrease in size of the central atom.

This is because, smaller the size of the central atom of an oxo- acid of the type HOY, larger is the electro negativity of Y, more is the pull of the electrons of O-Y bond towards Y. This would result in a more positive character on O which would consequently help the H-O electrons to be pulled by O and release H as H+.

Therefore,

(a) HOI < HOBr <HOCl (b) HIO4 < HBrO4 <HClO4

(c) HPO3 < HNO3 (c) H3 AsO4 < H3 PO4.

For oxo - acids having the same central atom at different oxidation states, the rule to be followed is :

For oxo- acids of the general formula, (HO)m ZOn , where Z is the central atom, m is the no. of oxygen atoms attached to Z and attached to H, while n is the no of oxygen atoms attached to Z but not attached to H. It is these oxygen atoms which help in resonance and which determine the acidic strength.

For n= 0 Acid is very weak HOCl, (HO)3Br

n = 1 Acid is weak HOClO, HONO

n = 2 Acid is strong HOClO2, HONO2

n = 3 Acid is very strong HOClO3, HOIO3


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Origin of Acidity and Basicity in Organic Compounds

i) Among hydrocarbons, acidity increases with the % `s' character . This is because, higher the `s' character, closer are the electrons to the nucleus of C and farther the electrons go from H, easily can H be removed as H+.

Therefore CH3 - CH3 < CH2 = CH2 < CH = CH

ii) Alcohols are more acidic than C because oxygen is more electro negative than C and consequently O - H bond easily give H+ than C - H bond

H3C - H < CH3 - O - H

iii) Carboxylic acids are more acidic than alcohols because carboxylate anion is more resonance stabilised than carboxylic acid as compared to alkoxide anion with respect to alcohol.

Therefore CH3O - H

v) Phenols are more acidic than alcohols because phenoxide ion is more resonance stabilised as compared to phenol than alkoxide as compared to alcohol.

Therefore CH3 - O - H< C6H5 - O - H

vi) Phenols are less acidic than carboxylic acids. This is because in phenoxide ion some of the resonance structures are of higher energy and this makes phenoxide ion not very much stable as compared to phenol.

Therefore C6H5 O- H

vii) In simple aliphatic acids, more the no. of CH3 added, less is the acidity. This is due to the +I (inductive) effect of CH3. This effect of adding CH3 decreases with increase in distance between CH3 added to the substituent, which releases H+.

CH3 -COOH > CH3 - CH2 - COOH

(viii) Increase in `s' character increases acidity of carboxylic acids.

(CH3)2 - COOH < CH2 = CHCOOH <HC = C - COOH

ix) Increase in electron withdrawing substituents into simple aliphatic acids increases acidity.

CH3 - COOH < I - CH2 - COOH <Br - CH2 - COOH , F - CH2 - COOH


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x) Nearer Cl is placed to COOH group, more is acidity.

xi) Benzoic acid is more acidic than carboxylic acids because benzoate anion shows/has more resonance.

CH3 COOH < C6H5 COOH

Exception to this is formic acid.

There is no destabilising factors for formate anion. Therefore, it is highly stabilised compared to formic acid because of resonance. Where as, Benzoic acid itself is resonance stabilised due to benzene ring. Therefore, benzoate ion is not very stable compared to benzoic acid. Thus C6H5 COOH < HCOOH

xii) More the no. of methyl or ethyl groups in NH3, more is its basicity. This is because of +I effect of methyl or ethyl which helps nitrogen to donate its lone pair of electrons.

NH3 < CH3 - NH2 < (CH3)2 NHNH3 < EtNH2 < (Et)2 NH

An exception to the above trend is the trisubstituted derivative. It is seen that if we introduce an alkyl group in a secondary amine, the basic strength of amines actually decreases. This is due to the fact that, the basic strength of an amine in water is determined not only by electron - availability on the nitrogen atom, but also by the extent to which the cation, formed by uptake of a proton, can undergo solvation, and so become stabilised. The more hydrogen atoms attached to nitrogen in the cation, the greater the possibilities of powerful solvation via hydrogen bonding between these and water which increases stabilisation by solvation.


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Thus, on going along the series NH3 RNH2 R2NH R3N, the inductive effect will tend to increase the basicity, but progressively less stabilisation of the cation by hydration will occur, which will tend to decrease the basicity. The net effect of introducing successive alkyl groups thus becomes progressively smaller, and an actual change over takes place, on going from secondary to a tertiary amine. That is why, (CH3)3 N < (CH3)2 NH

If this is the explanation, no such changeover should be observed, if measurements of basicity are made in solvent in which hydrogen- bonding cannot take place; it has, indeed, been found that, in chlorobenzene the order of basicity of the butylamines is

BuH2 < Bu2 NH<Bu3N

xiii) Aniline is less basic than NH3. This is because in aniline the nitrogen atom is again bonded to an sp2 hybridised carbon atom but, more significantly, the unshared electron pair on nitrogen can interact with the delocalised orbital of the nucleus:

If aniline is protonated, any such interaction, with resultant stabilisation in the anilinium cation is prohibited, as the electron pair on N is no longer available :

The aniline molecule is thus stabilised with respect to the anilinium cation, and it is therefore energetically unprofitable for aniline to take up a proton ; it thus functions as a base with almost reluctance.

xiv) Introduction of alkyl, e.g. Me groups on the nitrogen atom of aniline results in small increase in pKa due to the +I effect of alkyl groups.

C6H5NH2 < C6H5 NHMe < C6H5NMe2

xv) Introduction of phenyl groups on N lowers basicity because the substituted amine becomes much more stable than the ion.

ph2NH < phNH2


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Answer to Exercises

Exercise 1: Conjugate acid conjugate baseC6H5OH C6H5O–

H3O+ H2OHCl Cl–

H2O OH–

Exercise 2: i) Lewis acid Cr+3

Lewis base H2Oii) Lewis acid BF3

Lewis base (C2H5)2O

Exercise 3: Hard acids: Al+3, Cr+3, Sn+4

Hard bases: OH–, H2O,


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Solved Problems

Subjective

Problem 1: On the basis of H-bonding explain that the second ionization constant K2 for fumaric acid is greater than for maleic acid.

Solution: We know that H-bonding involving acidic H has an acid weakening effect and H-bonding in conjugate base has an acid strengthening effect.

Both dicarboxylic acids have two ionisable hydrogen atoms. Considering second ionization step.

Since the second ionisable H of the Maleate participates in H-bonding more energy is needed to remove this H because the H-bond must be broken. The maleate mono anion is, therefore, the weaker acid.

Problem 2: Arrange o-nitro, p-nitro, m-nitrophenol in decreasing acid strength

Solution: p-nitrophenol o-nitrophenol m-nitrophenol

The –NO2 is electron – withdrawing and acid – strengthening. Its resonance effect, which occurs only from para and ortho positions, predominates over its inductive effect, with occurs also from the meta position between ortho and para derivative, para derivative is more acidic because incase of ortho the hydrogen of phenolic group is attached through hydrogen bonding with oxygen of NO2 group and hence not possible to remove easily.

Problem 3: Among N,N dimethyl aniline and N,N,2,6 – tetramethyl aniline which one is a stronger base and why.

Solution:

In N,N-2,6 – tetramethyl aniline the methyl groups on nitrogen and the ortho position are very close to each other resulting in a steric crowding. Now to


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avoid steric crowding the C—N bond rotates and becomes perpendicular to the benzene ring. In this process the lone pair on nitrogen becomes perpendicular to the p-orbitals of benzene ring thereby inhibiting resonance. But in N,N dimethyl aniline there is no steric hindrance, so the lone pair is in the same plane as the benzene ring and undergoes resonance. Therefore the lone pair on the tetramethyl derivative is more available and hence it is more basic.

Problem 4: Which is the stronger base towards a proton or and why?

Solution: Bond energy (N—H P—H) ionisation suggests that will be stronger

base. This is constituent with the relative strengths of the respective conjugate acids: NH3 PH3.

Problem5: Predict the nature of change, if any on the acidity of the solution consequent to the addition (i) HgO to an aqueous solution of KI. (ii) CuSO 4 to an aq. solution of (NH)2SO4, (iii) Al(OH)3 to aqueous caustic soda solution.

Solution: i) Due to the formation of the stable complex species, the

concentration of OH– increase according to the reactionHgO + 4KI + H2O HgI4 + 2OH– + 4K+

Hence acidity decreases

ii) Due to the formation of Cu(NH3)4+2 complex ion increases H+ ion

concentration. Hence the acidity increases acid to the reaction

Cu2+(aq) + 4NH4+(aq) + 4H+(aq)

iii) Formation of the complex anion consumes. OH– ions. Hence

acidity increases.

Al(OH)3(s) + OH– (aq)


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Objective

Problem 1: The conjugate base of OH– ion is(A) H2O (B) O2

2–

(C) H3O+ (D) O–

Solution:

(B)

Problem 2: Which salt undergoes hydrolysis(A) CH3COONa (B) KNO3

(C) NaCl (D)K2SO4

Solution: Salt of strong acid and strong base does not undergo hydrolysis. (A)

Problem 3: The compound HCl behaves as …. in the reactionCl + HF H2

+Cl + F–

(A) Strong acid (B) Strong base(C) Weak acid (D) Weak base

Solution: HCl is accepting proton in HF medium and acts as weak base. (D)

Problem 4: The conjugate acid of is

(A) H3PO4 (B) H2

(C) (D)

Solution: Conjugate acid and base differs by one proton hence the conjugate acid of PO4

–3 is HPO3–2

(D)

Problem 5: The anhydride of H3PO4 is(A) P2O5 (B) P2O3

(C) PO2 (D) None

Solution: P has +5 oxidation number in P2O5 and H3PO4

(A)

Problem 6: The conjugate base of hydrazoic acid is(A) HN3

– (B) N3H(C) N3– (D) N2

–

Solution: N3H N3–

(C)

Problem 7: Which of the following is a strong acid(A) HClO4 (B) HBrO4

(C) HIO4 (D) HNO3


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Solution: The acidic character of oxy acids decreases down the group and increases along the period. Also acidity increases with increase in oxidation number of central atom. (A)

Problem 8: H3BO3 is ….acid(A) Monobasic (B) Dibasic(C) Tribasic (D) None

Solution: H3BO3 is monobasic acidH3BO3 + H2O B(OH)4

–

(A)

Problem 9: Which is not Lewis base(A) Ag+ (B) Cl–

(C) CN– (D) H2O

Solution: metal cations are Lewis acids (A)

Problem 10: Tribasic acid furnishes…. Type of anion(A) 2 (B) 1(C) 3 (D) 4

Solution: H3PO4 furnish H2PO4–, HPO4

2– and PO43– anions

(C)


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Assignment

Subjective

LEVEL - I

1. The ionization constant of HCN is given as 4 10-10. The ionization constant of NH3 is 1.8 10-5. Decide which is the stronger acid, HCN or NH4

+.

2. Write down the conjugate base of the following (i) NH4

+, (ii) HCOOH (iii) H3O+, (iv) H2NCONH3+

3. Predict the relative acidic strength among the following: HI, HIO4, ICl.

4. Predict the relative basic strength among the following :HOCl, HOClO,HOClO2,HOClO3

5. Compare the strength of formic acid and acetic acid

6. Phenol is acidic but why ethanol is neutral

7. Among CH4, H2S, HI which one is most acidic? Justify your answer

8. Which one of the two i.e., CH3SH and CH2OH more strong acid and why?

9. Which of the following oxide is most acidic Ag2O, V2O5, CO, N2O5

10. Identify soft and hard bases among the followingF–, I–, RS–, RSH


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LEVEL - II

1. Which way would the pH shift when CuSO4 is added to pure water. Write a net ionic equation to support your answer.

2. What is the order of increasing strength of -chloro butanoic acid, -chloro butanoic acid, -chloro butanoic acid and n – butanoic acid?

3. Write the cation, anion and the neutralisation reaction if thionyl chloride reacts with sodium sulphate.

4. Among different hydroxy benzoic acids o-hydroxybenzoic acid is the strongest compared to m-and p-isomers considerably. Why?

5. Why diphenyl amine is less basic than aniline?

6. Arrange the following in increasing order of basicityPhenol, o-nitrophenol, cresol

7. Arrange the following in the increasing order of basicity. Justify answer R3—N—, —R2NH, RNH2

8. N-ethyl aniline is more basic than N methyl aniline why?

9. Arrange the following in the order of their basic strength. Justify your answerNH2NH2,NH3,NH2OH

10. Arrange the following in the order of their decreasing acidic strength CH4 , LiH, BeH2, H2O, HF, NH3


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LEVEL - III

1. Why CHCl3 is more acidic than CHF3?

2. Why F–– is most basic among the halogen anion

3. Arrange the following in the increasing order of acidic strength.

4. Why pyrole is less basic than pyridine

5. Arrange the following compounds in the increasing order of their basic strength.CH3NH–Na+, C2H5NH2, (iso-C3H7)3N and CH3CONH2

6. What should be the order of acidic strength in the series H3PO4, H3PO3 and H3PO2?

7. Arrange the following in correct order of acidity

8. In dilute benzene solutions, equimolar additions of (C4H9)3N and HCl produce a substance with a dipole moment. In the same solvent, equimolar additions of (C4H9)3N and SO3 produce a substance having an almost identical dipole moment. What is the nature of the polar substances formed and what is the unifying feature of HCl and SO3?

9. Arrange according to increasing Lewis acid character, B(n–Bu)3, B(t–Bu)3

10. Arrange according to increasing Lewis acid character, SiF4, SiCl4, SiBr4, Sil4


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Objective

LEVEL - I

1. Which one of following is the strongest acid (A) ClO3(OH) (B) ClO2(OH)(C) SO(OH)2 (D) SO2(OH)2

2. Amongst the following the most basic compound is (A) Benzyl amine (B) Aniline(C) Acetanilide (D) p –nitro aniline

3. Amongst the trihalides of nitrogen, which one is least basic (A) NF3 (B) NCl3(C) NBr3 (D) NI3

4. What is the decreasing order of strength of bases(A) CH3 – CH2

NH2 > H – C C > OH

(B) H – C C > CH3- CH2 > NH2

> OH

(C) OH > NH2 > H- C C > CH3CH2

(D) NH2

> H – C C > OH > CH3 -CH2

5. H2O is capable to act as an acid with (A) NH3 (B) H2SO4

(C) C6H6 (D) HCl

6. Which of the following is Lewis acid (A) PCl3 (B) AlCl3(C) NCl3 (D) AsCl3

7. The strongest bronsted base in the following anions is (A) ClO (B) ClO2

(C) ClO3 (D) ClO4

8. With reference to protonic acid which of the following statement is correct ?(A) PH3 is more basic than NH3 (B) PH3 is less basic than NH3 (C) PH3 is equally basic than NH3 (D) PH3 is amphoteric while NH3 is basic

9. Which of the following is a hard base (A) F– (B) Li+

(C) I– (D) Mg+

10. The conjugate acid of NH2– is

(A) NH3 (B) NH2OH(C) NH4

+ (D) N2H4

11. The conjugate acid of S2O8–2

(A) H2S2O8 (B) H2SO4

(C) HSO4– (D) HS2O8

–


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12. The conjugate acid of is

(A) (B)

(C) H3PO4 (D) H3PO3

13. Which of the following are amphiprotic in nature

(A) OH– (B)

(C) (D) HF

14. Among the following compounds, the strongest acid is (A) HC CH (B) C6H6

(C) C2H6 (D) CH3OH [IIT-JEE ’98]

15. The following acids have been arranged in order of decreasing acid strength. Identify the correct order.ClOH (I) BrOH (II) IOH (III)(A) I > II > III (B) II > I > III(C) III > II > I (D) I > III > II [IIT-JEE ’96]


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LEVEL – II

1. The conjugate acid of HF is

(A) F– (B)

(C) (D) none of these

2. Which of the following species is an acid and also a conjugate base of another acid

(A) (B) H2SO4

(C) OH– (D) H3O+

3. Which of the following is the strongest acid (A) H3PO4 (B) H3PO3

(C) H3PO2 (D) All have same strength

4. In the equilibrium CH3COOH + HF + F–

(A) F– is the conjugate acid of CH3COOH(B) F– is the conjugate base of HF

(C) CH3COOH is the conjugate acid of

(D) is the conjugate base of CH3COOH.

5. The conjugate base of is

(A) (B)

(C) (D)

6. Which of the following is a hard acid (A) CO2 (B) Br2

(C) Fe3+ (D) Cd2+

7. Pyrophosphoric acid is a (A) mononbasic acid (B) dibasic acid (C) tribasic acid (D) tetrabasic acid

8. Which among the following is the strongest base (A) aniline (B) N,N dimethyl amine(C) para nitro aniline (D) meta nitro aniline

9. Which of the following statements is/are correct (A) I– is a weaker base than F–

(B) (C) HONH2 is a weaker base than NH3

(D) F3C:– is a stronger base than Cl3C:–

10. Which of the following orders regarding acid strength is correct?(A) HCOOH < CH3COOH < PhCOOH(B) HCOOH > PhCOOH > CH3COOH(C) HCOOH > CH3COOH > PhCOOH


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(D) CH3COOH > HCOOH > PhCOOH11. Which of the following statements is/are correct?

(A) Maleic acid is a stronger acid than fumaric acid but maleate monoanion is a weaker acid than fumarate monoanion

(B) Maleic acid is a stronger acid than fumaric acid and also maleate monoanion is a stronger acid than fumarate monoanion

(C) Maleic acid is a weaker acid than fumaric acid but maleate monoanion is a stronger acid than fumarate monoanion

(D) Maleic acid is a weaker acid than Fumaric acid and also maleate monoanion is a weaker acid than fumarate monoanion.

12. Of the following orders regarding basicity, which one is correct (A) CH3CH2NH2 > PhCONH2 > CH3CONH2

(B) CH3CH2NH2 > PhCONH2 > CH3CONH2

(C) CH3CH2NH2 > PhCONH2 < CH3CONH2

(D) CH3CH2NH2 < PhCONH2 < CH3CONH2

13. In the following compounds:

(A) III > IV > I > II (B) I > IV > III > II(C) II > I > III > IV (D) IV > III > I > II [IIT-JEE ’96]

14. In the following compounds, the correct order of basicity is

(A) IV > I > III > II (B) III > I > IV > II(C) II > I > III > IV (D) I > III > II > IV [IIT-JEE ’97]

15. Arrange the following ions in increasing order of acidity.[Na (H2O)n]+, [Al (H2O)6]3+,[Mn (H2O)6]2+, [Ni (H2O)6]2+

(I) (II) (III) (IV)(A) I < II < III < IV (B) I < III < IV < II(C) I < III < II < IV (D) II < IV < III < II


SP-CH-CAB-20

Answers

Subjective

LEVEL - I

1. NH4+ is slightly stronger acid (Ka for NH4

+ = 5.6 10-10).[Larger the value of ionization constant, stronger the acid will be.]

2. To form a conjugate base means removal of a proton So answer is (i) NH3 (ii) HCOO–

(iii) H2O (iv) H2NCONH2

3. IO4- is more stable than I- and I+ is unstable. Hence HIO4 will be strongest acid.

HIO4 HI> ICI

4. HOCl < HOClO < HOClO2 < HOClO3

5. The –CH3 group which is present in acetic acid decrease the acidic character of the–COOH group.

6. Conjugate base of phenol in resonance stabilized white that of alcohol is not.

7. Acidic natureCH4 H2S HISize of the central atoms is greatest in HI hence it is most acidic.

8. The conjugate base of CH3OH – CH3O– is stronger than the conjugate base of CH3SH i.e CH3SH i.e. CH3S–. Since S is large, the negative charge in CH3S– is dispersed over a large. Hence CH3S– is the weaker base.

9. The oxide with the highest positive oxidation state on the element other than O should be most acidic. Oxidation states of V in V2O5 and N in N2O5 are both +5. But the electronegativity of N is higher, making N2O5 the most acidic oxide.

10. Hard bases F–

Soft bases I–, RSH, RS–


SP-CH-CAB-21

LEVEL - II

1. The pH will shift down below 7. It is clear from the following equation.CuSO4 + H2O Cu(OH)2 + SO4

2– + 2H+

2. CH3 – CH2 – CHCl –COOH CH3 – CHCl – CH2 – COOH CH2Cl – CH2 – CH2 – COOH CH3 – CH2 – CH2 – COOH

3. Reaction SOCl2 + Na2SO3 2NaCl + 2SO2

Cation SO++

Anion SO32–

4. O–hydroxy benzoate anion is stabilized by chelation due to hydrogen bonding which is absent in meta and para isomers.

5. Diphenyl amine is less basic than aniline because lone pair of electron on N-atom is used by two benzene rings in case of diphenyl amine.

6. o-nitrophenol phenol cresol

7. R2NH RNH2 R3N

8. N-alkylated anilines are stronger bases than aniline because of steric effects which decrease the resonance of the lone pair of electron on nitrogen and hence makes it more available for protonation. Ethyl group is bigger than the methyl group so N-ethyl aniline is stronger base than N-methyl aniline.

9. NH3 NH2 NH2 NH2OH

10. HF H2O NH3 CH4 BeH2 LiH


SP-CH-CAB-22

LEVEL - III

1. Cl3C–: is less basic than F3C–: because fluroine can disperse charge only by an inductive effect. White Cl (having empty 3d orbitals) disperses charge by inductive effect as well as by p - p bonding delocalisatioin. Fluorine is a second period element with no 2d orbital.

2. Among the halogens, fluorine has the smallest size hence it has availability of electron most.

3. I V IV II III

4. Pyrole uses lone pair of electron on nitrogen atom in delocalisation hence less electron are available for protonation, hence less basic than pyridine.

5. CH3CONH2 (iso-C3H7)3N, C2H5NH2, CH3ONH–Na+

6. It can be seen that hydrogens in these molecules are not all bonded to oxygens. It is clear from their structures,

that the number of terminal oxygen atoms is 1 in all three acids. The electronegativities of P and H are almost the same. Thus no much difference in acidity is expected.

7. IV III I II

8. Both HCl and SO3 are Lewis acids and can react with amine base to form polar substances which could presumably undergo ionic dissociation in a solvent sufficiently more polar than benzene. The reactioins may be represented as follows (R = C4H9):

Sulphur is a third element which can expand its octet due to availability of vacant d-orbitals. Thus sulphur expands its number of valance electron by attaching to the lone pair on the nitrogen. The N – S bond will be polar because of the big difference in electronegativity between N and S.


SP-CH-CAB-23

Here the proton of the HCl attaches to the lone pair on the N. The connection between N – H and Cl is designated by (----) symbolizing an electron pair on the Cl connected to nitrogen by a hydrogen bond.

9. The highly branched tertiary butyl group involve appreciable back – strain (B-strain) when the boron atom changes to pyramidal environment on adduct formation. This destabilizes the adduct. Hence the order is B(t–Bu)3 < B(n–Bu)3

10. The order in this case is the reverse of that for BX3. -conjugation from the halogen p-orbital to the Si-d orbital is not as intense as in the case of BX3 and the order of acidity follows the increase in electron withdrawing power of the halogen from I to F. Hence the order is SiI4 < SiBr4 < SiCl4 < SiF4


SP-CH-CAB-24

Objective

LEVEL - I

1. A 2. A

3. A 4. A

5. A 6. B

7. A 8. B

9. A 10. A

11. D 12. A

13. A 14. A

15. A

LEVEL - II

1. C 2. A

3. C 4. B

5. D 6. A,C

7. D 8. B

9. A, C, D 10. B

11. A 12. A

13. D 14. D

15. B

6


SP-CH-CAB-25

Concept of Accid and Bases

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Transcript of Concept of Accid and Bases