CONCEPT OF ACIDS & BASES

 

Solution to Subjective Problems 

LEVEL – I

1. To form a conjugate base means removal of a proton 

So answer is 

(i) NH2– (ii) CH3COO–

(iii) H2O (iv) H2NCONH2

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

–Ol, –OlO, – OlO2, –OlO3

The species which have more is 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  –OI, –OIO, – OIO2, –OIO3

1. Volume available for the negative charge is increasing from left to right
2. Charge density is decreasing from left to right.
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 :

HOI < HOIO <  HOIO2  < HOIO3

3. The order in this case is the reverse of that for BX3. π-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

4. NH4+ is slightly stronger acid (Ka for NH4+ = 5.6 × 10-10).

[Larger the value of ionization constant, stronger the acid will be.]

5. Ag+ is stronger Lewis acid due to pseudo inert gas configuration.
6. HF > H2O > NH3 > CH4 > BeH2 > LiH
7. 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. 

8. The pH will shift down below 7. It is clear from the following equation.

CuSO4 + H2O Cu(OH)2 + SO42– + 2H+

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

HIO4 > HI> ICI  

10. 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

 

LEVEL – II

1. I  <  II <  IV <  III

2. CH3 – CH2– CHBr –COOH > CH3 – CHBr – CH2 – COOH > CH2Br – CH2 – CH2 – COOH > CH3 – CH2 – CH2 – COOH
3. Pyrole uses lone pair of electron on nitrogen atom in delocalisation hence less electron are available for protonation, hence less basic than pyridine.
4. CH3CONH2 < (iso-C3H7)3N, C2H5NH2, CH3ONH–Na+
5. Reaction SOCl2+ Na2SO3 ⎯→ 2NaCl + 2SO2

Cation SO++

Anion SO32–

6. NH3 > NH2 NH2 > NH2OH

7. IV > III > I > II

8. 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.

9 o-nitrophenol > phenol > cresol

 

10. Conjugate base of phenol in resonance stabilized while that of alcohol is not.

 

LEVEL – III

1. Due to larger size of iodide ion it is least basic.
2. Since CH4 is weakest acid among the conjugate acids of the given ions therefore CH3– is the most basic. 
3. 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.
4. After the removal of proton we get F–, Cl–, Br– and I–. Due to largest size of I– it is the weakest base therefore its conjugate acid is the strongest acid therefore the order is HF < HCl < HBr < HI.
5. I > V > IV > II > III
6. R2NH > RNH2 > R3N
7. Cl3C–: is less basic than F3C–: because fluorine can disperse charge only by an inductive effect. While Cl (having empty 3d orbitals) disperses charge by inductive effect as well as by pπ – pπ bonding delocalisation. Fluorine is a second period element with no 2d orbital.
8. The conjugate base of CH3OH – CH3O– is stronger than the conjugate base of 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. O–hydroxy benzoate anion is stabilized by chelation due to hydrogen bonding which is absent in meta and para isomers.
10. 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.

⎯⎯→

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.

 

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