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# (a) Mention conjugate base of each of the following:$\text{H}{{\text{S}}^{1-}},\text{ }{{\text{H}}_{3}}{{\text{O}}^{+}},\text{ }{{\text{H}}_{2}}\text{PO}_{4}^{-},\text{ HSO}_{4}^{-}\text{, HF,C}{{\text{H}}_{3}}\text{COOH, }{{\text{C}}_{6}}{{\text{H}}_{5}}\text{OH, HCl}{{\text{O}}_{4}}\text{, NH}_{4}^{+}$(b) Mention conjugate acid of each of the following:$\text{O}{{\text{H}}^{-}}\text{, C}{{\text{H}}_{3}}\text{CO}{{\text{O}}^{-}}\text{, C}{{\text{l}}^{-}}\text{, CO}_{3}^{2-}\text{, }{{\text{H}}_{2}}\text{PO}_{4}^{-}\text{, C}{{\text{H}}_{3}}\text{N}{{\text{H}}_{2}}\text{, C}{{\text{H}}_{3}}\text{COOH, NH}_{2}^{-}$(c) Mention of the following both bases as Bronsted acid as well as Bronsted base:${{\text{H}}_{2}}\text{O, HCO}_{3}^{-}\text{, }{{\text{H}}_{2}}\text{S}{{\text{O}}_{4}}\text{, }{{\text{H}}_{3}}\text{P}{{\text{O}}_{4}}\text{, H}{{\text{S}}^{-}}\text{, N}{{\text{H}}_{3}}$(d) Which is stronger acid in each of the following pairs?$\text{HCl, HI; }{{\text{H}}_{2}}\text{C}{{\text{O}}_{3}}\text{, }{{\text{H}}_{2}}\text{S}{{\text{O}}_{4}}\text{; }{{\text{H}}_{2}}\text{O, HS; }{{\text{C}}_{2}}{{\text{H}}_{5}}\text{OH, }{{\text{C}}_{6}}{{\text{H}}_{5}}\text{OH}$(e) Which is a stronger base in each of the following pairs?$\text{O}{{\text{H}}^{-}}\text{, C}{{\text{l}}^{-}}\text{; O}{{\text{H}}^{-}}\text{, NH}_{2}^{-}\text{; O}{{\text{H}}^{-}}\text{, C}{{\text{H}}_{3}}\text{CO}{{\text{O}}^{-}}\text{; C}{{\text{H}}_{3}}\text{CO}{{\text{O}}^{-}}\text{, C}{{\text{l}}^{-}};\text{ N}{{\text{H}}_{3}}\text{, C}{{\text{H}}_{3}}\text{N}{{\text{H}}_{2}}$(f) Classify the following into acids and bases according to the Lewis concept: ${{\text{S}}^{2-}}\text{, }{{\text{H}}^{+}}\text{, O}{{\text{H}}^{-}}\text{, B}{{\text{F}}_{3}}\text{, N}{{\text{i}}^{2+}}\text{, N}{{\text{F}}_{3}}\text{, AlC}{{\text{l}}_{3}}\text{, SnC}{{\text{l}}_{4}}\text{, N}{{\text{H}}_{3}}\text{, }{{\left( \text{C}{{\text{H}}_{3}} \right)}_{2}}\text{O }$

Answer Verified
Hint: Conjugate base are those molecules in which one H-atom is absent whereas conjugate acid has one extra hydrogen atom. Bronsted acid donates whereas bronsted base accepts a hydrogen ion.

Complete step by step answer:
(a) we have to identify the conjugate base from the given choices.
- As we know that conjugate base contains one less hydrogen atom and also we can identify them by counting the number of H-atoms in the reactant as well the product side.
- If there is any decrease in the number of H-atoms, they will be considered as the conjugate base and vice versa.
- So, to write the conjugate base we will remove one H-atom form each compound.
- Hence, the conjugate is:
${{\text{S}}^{2-}},{{\text{H}}_{2}}\text{O, HPO}_{4}^{-}\text{, SO}_{4}^{-}\text{, }{{\text{F}}^{-}}\text{, C}{{\text{H}}_{3}}\text{CO}{{\text{O}}^{-}}\text{, }{{\text{C}}_{6}}{{\text{H}}_{5}}{{\text{O}}^{-}}\text{, ClO}_{4}^{-}\text{, N}{{\text{H}}_{3}}$

(b) Similarly, for conjugate acid, we will add an extra H-atom to the given compound i.e.
${{\text{H}}_{2}}\text{O, C}{{\text{H}}_{3}}\text{COOH, HCl,HCO}_{3}^{-}\text{,C}{{\text{H}}_{3}}\text{NH}_{3}^{+}\text{, C}{{\text{H}}_{3}}\text{COOH}_{2}^{+}\text{, N}{{\text{H}}_{3}}$

(c) Now, we have to identify those compounds which act as both Bronsted acid as well as base i.e. they can accept as well as donate a pair of electrons.
- So, from the given compound, ${{\text{H}}_{2}}\text{O, HCO}_{3}^{-};\text{ N}{{\text{H}}_{3}},\text{ H}{{\text{S}}^{-}}$can act as both.

Now, we have to identify the substance which is a stronger acid.
- So, we know that strong acids can easily give ${{\text{H}}^{+}}$ i.e. more easily they give ${{\text{H}}^{+}}$ions, more is the acidic strength.
- So, among the given compound the order will be $\text{HI}$˃ $\text{HCl}$; ${{\text{H}}_{2}}\text{S}{{\text{O}}_{4}}$˃${{\text{H}}_{2}}\text{C}{{\text{O}}_{3}}$; ${{\text{H}}_{2}}\text{S}$˃ ${{\text{H}}_{2}}\text{O}$; ${{\text{C}}_{6}}{{\text{H}}_{5}}\text{OH}$˃${{\text{C}}_{2}}{{\text{H}}_{5}}\text{OH}$.
(e) Now, we have to identify the strong base and we know that they can easily dissociate hydroxyl ions.
- So, from the given compounds strong bases are:
$\text{O}{{\text{H}}^{-}}$˃ $\text{C}{{\text{l}}^{-}}$; $\text{NH}_{2}^{-}$˃ $\text{O}{{\text{H}}^{-}}$; $\text{O}{{\text{H}}^{-}}$˃ $\text{C}{{\text{H}}_{3}}\text{CO}{{\text{O}}^{-}}$; $\text{C}{{\text{H}}_{3}}\text{CO}{{\text{O}}^{-}}$˃ $\text{C}{{\text{l}}^{-}}$; $\text{C}{{\text{H}}_{3}}\text{N}{{\text{H}}_{2}}$˃$\text{N}{{\text{H}}_{3}}$

(f) Now, we have to identify the Lewis acid and Lewis base. Those species which have vacant orbitals to accept electrons are Lewis acid where those who can donate electron pairs are Lewis base.
- From the given compound, Lewis acids are ${{\text{H}}^{+}}\text{, B}{{\text{F}}_{3}}\text{, N}{{\text{i}}^{2+}}\text{, AlC}{{\text{l}}_{3}}\text{, SnC}{{\text{l}}_{4}}$ whereas ${{\text{S}}^{2}}\text{, O}{{\text{H}}^{-}}\text{, N}{{\text{F}}_{3}}\text{, N}{{\text{H}}^{3}}\text{, }{{\left( \text{C}{{\text{H}}_{3}} \right)}_{2}}\text{O}$ are Lewis base.

Note: The difference between conjugate acid and acid is that acids can dissociate to hydrogen ions whereas conjugate acids are formed by the acceptance of H-atom.