Chapter 7: Equilibrium

Ans: Conjugate acid and conjugate base can be shown as-

$\text{HCl}\left( \text{aq} \right)\text{ + }{{\text{H}}_{\text{2}}}\text{O }\left( \text{l } \right)\text{ }\rightleftharpoons \text{ }{{\text{H}}_{\text{3}}}{{\text{O}}^{\text{+}}}\text{ }\left( \text{aq} \right)\text{ + C}{{\text{l}}^{\text{--}}}\text{ }\left( \text{aq} \right)$ 

 acid             base        conjugate acid    conjugate base

Ans: Conductance of a solution depends upon the number of ions present in the solution since sodium chloride is an ionic compound when mixed in water it ionizes completely forming an aqueous electrolytic solution. Hence, the solution conducts electricity while sugar being a non-electrolyte does not ionize in water and when concentration of NaCl is increased, more $\text{N}{{\text{a}}^{\text{+}}}$ and $\text{C}{{\text{l}}^{\text{-}}}$ ions will be produced. Hence, conductance increases. 

Ans: $\text{B}{{\text{F}}_{\text{3}}}$ does not have a proton but still it acts as a Lewis acid as it is an electron pair acceptor and reacts with $\text{N}{{\text{H}}_{\text{3}}}$by accepting its lone pair of electrons. The reaction can be represented by Coordinate bond is formed between $\text{B}{{\text{F}}_{\text{3}}}\text{ and N}{{\text{H}}_{\text{3}}}$. It can be represented as-

$\text{ }\!\![\!\!\text{ B}{{\text{F}}_{\text{3}}}\leftarrow \mathbf{:}\text{N}{{\text{H}}_{\text{3}}}]$

Nitrogen acts as lone pair donator in this combination. 

Ans: Higher the value of ${{\text{K}}_{\text{b}}}$ stronger will be the base. 

${{\text{K}}_{\text{b}}}$-  $\text{5}\text{.4 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-4}}}$>$\text{1}\text{.77 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-5}}}$> $\text{1}\text{.77 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-9}}}$> $\text{1}\text{.3 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-14}}}$

Decreasing order of basic strength. 

Dimethylamine > Ammonia > Pyridine > Urea

Hence among the given bases, the strongest base is Dimethylamine. 

Ans: As the Conjugate acid of a weak base is always stronger and conjugate base of weak acid is always strong.

The weak acids are:

⇒ ${{\text{H}}_{\text{2}}}\text{O,ROH,C}{{\text{H}}_{\text{3}}}\text{COOH,HCl}$

⇒ $\text{HClC}{{\text{H}}_{\text{3}}}\text{COOHROH}{{\text{H}}_{\text{2}}}\text{O}$

⇒ Conjugate base order ⇒ $\text{O}{{\text{H}}^{\text{-}}}\text{R}{{\text{O}}^{\text{-}}}\text{C}{{\text{H}}_{\text{3}}}\text{CO}{{\text{O}}^{\text{-}}}\text{C}{{\text{l}}^{\text{-}}}$

Ans: Correct order of pH is -

$\text{N}{{\text{H}}_{\text{4}}}\text{Cl}{{\text{C}}_{\text{6}}}{{\text{H}}_{\text{5}}}\text{COON}{{\text{H}}_{\text{4}}}\text{KN}{{\text{O}}_{\text{3}}}\text{C}{{\text{H}}_{\text{3}}}\text{COONa}$

Salts of strong acid and strong base do not hydrolyse and form neutral solution thus, pH will be nearly 7 of $\text{KN}{{\text{O}}_{\text{3}}}$​. 

In sodium acetate, acetic acid remains unionised this results in increase in $\text{O}{{\text{H}}^{\text{-}}}$ concentration and pH will be more than 7. 

$\text{N}{{\text{H}}_{\text{4}}}\text{Cl}$formed from weak base, $\text{N}{{\text{H}}_{\text{4}}}\text{OH}$and strong acid, HCl, in water dissociates completely, aq. ammonium ions undergo hydrolysis with water to form $\text{N}{{\text{H}}_{\text{4}}}\text{OH}$and ${{\text{H}}^{\text{+}}}$ions resulting in less pH value.

Ans:  We can calculate the reaction quotient Q for the reaction and that will be given by the expression:

\[ Q = \frac{[H_{2}][I_{2}]}{[HI]} = \frac{(2 \times 10 - 5)21 \times 10^{-5} }{(2 \times 10^{-5})^2 1}  = \frac{1}{4}\]

=0.25=2.5×$\text{1}{{\text{0}}^{\text{-1}}}$

As the value of the reaction quotient is greater than the value of ${{\text{K}}_{\text{c}}}$​ that is $\text{1  }\!\!\times\!\!\text{  1}{{\text{0}}^{\text{--4}}}$ the reaction will proceed in the reverse direction.

Ans: Here concentration of water cannot be neglected since the solution is very dilute. pH will be less than 7.0. Hence, the total concentration is given as -

$\text{ }\!\![\!\!\text{ }{{\text{H}}_{\text{3}}}{{\text{O}}^{\text{+}}}\text{ }\!\!]\!\!\text{ =1}{{\text{0}}^{\text{-8}}}\text{+1}{{\text{0}}^{\text{-7}}}\text{M}$.

Ans: Given with the pH = 5 as, $\text{pH=-- log  }\!\![\!\!\text{ }{{\text{H}}^{\text{+}}}\text{ }\!\!]\!\!\text{ }$

$\text{ }\!\![\!\!\text{ }{{\text{H}}^{\text{+}}}\text{ }\!\!]\!\!\text{ =1}{{\text{0}}^{\text{-5}}}\text{M}$

On 100 times dilution-

$\text{ }\!\![\!\!\text{ }{{\text{H}}^{\text{+}}}\text{ }\!\!]\!\!\text{ =1}{{\text{0}}^{\text{-7}}}\text{M}$.

Ans: The chemical equation is given below-  

$\text{BaS}{{\text{O}}_{\text{4}}}\left( \text{s} \right)\to \text{B}{{\text{a}}^{\text{2+}}}\left( \text{aq} \right)\text{ + S}{{\text{O}}_{\text{4}}}^{\text{2-}}\left( \text{aq} \right)$

At t = 0         1                 0                 0 

At equilibrium

 in water         1-S             S                 S 

At equilibrium in 

the presence     1-S           S            (S+0.01) 

of sulphuric acid 

${{\text{K}}_{\text{sp}}}\text{ for BaS}{{\text{O}}_{\text{4}}}\text{ in water = }\!\![\!\!\text{ B}{{\text{a}}^{\text{2+}}}\text{ }\!\!]\!\!\text{   }\!\![\!\!\text{ S}{{\text{O}}_{\text{2}}}^{\text{4-}}\text{ }\!\!]\!\!\text{  = }\left( \text{S} \right)\text{ }\left( \text{S} \right)\text{ = }{{\text{S}}^{\text{2}}}$

But S = 8 x $\text{1}{{\text{0}}^{\text{-4}}}\text{ mol d}{{\text{m}}^{\text{-3}}}$

${{\text{K}}_{\text{sp}}}\text{ = }{{\left( \text{8 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-4}}} \right)}^{\text{2}}}\text{ = 64 }\!\!\times\!\!\text{  1}{{\text{0}}^{\text{-8}}}\text{ }$... (1) 

The expression for ${{\text{K}}_{\text{sp}}}$ in the presence of sulphuric acid will be as follows: 

${{\text{K}}_{\text{sp}}}$ = (S) (5 + 0.01) ... (2) 

Since value of ${{\text{K}}_{\text{sp}}}$will not change in the presence of sulphuric acid, therefore from (1) and (2) 

(S) (S + 0.01) = $\text{64 }\!\!\times\!\!\text{  1}{{\text{0}}^{\text{-8}}}$

 ${{\text{S}}^{\text{2}}}$ + 0.01 S = $\text{64 }\!\!\times\!\!\text{  1}{{\text{0}}^{\text{-8}}}$

 ${{\text{S}}^{\text{2}}}$ + 0.01 S - $\text{64 }\!\!\times\!\!\text{  1}{{\text{0}}^{\text{-8}}}$= 0

$\text{S=}\frac{\text{-0}\text{.01 }\!\!\pm\!\!\text{ }\sqrt{{{\text{(0}\text{.01)}}^{\text{2}}}\text{+(}}\text{4 }\!\!\times\!\!\text{ 64 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-8}}}\text{)}}{\text{2}}$

When we solve the above equation, we will get the value of

 S is= $\text{6 }\!\!\times\!\!\text{  1}{{\text{0}}^{\text{-4}}}\text{ mo}{{\text{l}}^{\text{-1}}}\text{ d}{{\text{m}}^{\text{-3}}}$