Chapter 7: Equilibrium

Ans: Ionisation constant can be calculated as given below-

pH of HOCl=2.85

But \[pH = log[H^{+}]\]

\[ \therefore -2.85 = log [H^{+}]\]

\[\Rightarrow 3^{-} .15 = log [H^+] \]

\[\Rightarrow [H^+] antilog 3^{-} .15 \]

\[\Rightarrow [H^+] = 1.413 \times 10^{-5} \]

For weak mono basic acid \[[H^+] = K_{a} \times C \]

\[ K_{a} = C[H^+]^2 = 0.08(.413 \times 10^{-3})^{2} \]

\[ = 24.957 \times 10^{-6}  = 2.4957 \times 10^{-5}\]

Ans: we can find the pH of the solution as-

pH of solution A=6

Hence, concentration of [[{{\text{H}}^{\text{+}}}]] ion in solution A=$\text{1}{{\text{0}}^{\text{-6}}}$$\text{mol }{{\text{L}}^{\text{-1}}}$

pH of solution B=4

Therefore, concentration of [{{\text[{H}]}^{\text{+}}}] ion in solution B=$\text{1}{{\text{0}}^{\text{-4}}}$$\text{mol }{{\text{L}}^{\text{-1}}}$

On mixing one litre of each solution, total volume = 1L+1L=2L.

Amount of ${{\text{H}}^{\text{+}}}$ ions in 1L of solution

 A= concentration× Volume(V)= $\text{1}{{\text{0}}^{\text{-6}}}$mol× IL 

Amount of ${{\text{H}}^{\text{+}}}$ ions in 1L of solution B=$\text{1}{{\text{0}}^{\text{-4}}}$mol×1L

∴ Total amount of ${{\text{H}}^{\text{+}}}$ ions in the solution formed by mixing solutions A and B is ($\text{1}{{\text{0}}^{\text{-5}}}$mol+$\text{1}{{\text{0}}^{\text{-4}}}$mol)

This amount is present in 2L solution,

∴ Total \[{{\text{H}}^{\text{+}}}\]=2$\text{1}{{\text{0}}^{\text{-4}}}$ (1+0.01)​=21.01×$\text{1}{{\text{0}}^{\text{-4}}}$​$\text{mol }{{\text{L}}^{\text{-1}}}$

=21.01×$\text{1}{{\text{0}}^{\text{-4}}}$​$\text{mol }{{\text{L}}^{\text{-1}}}$ =0.5×10−4mol L−1 =5×$\text{1}{{\text{0}}^{\text{-5}}}$$\text{mol }{{\text{L}}^{\text{-1}}}$.

pH=−log[{{\text[{H}]}^{\text{+}}}]=−log(5×$\text{1}{{\text{0}}^{\text{-5}}}$)=−[log5+(−5log10)]

=−log5+5=5−log5=5−0.6990=4.3010=4.3

Thus, the pH of the solution will be 4.3.

Ans: Let us assume S be the solubility of $\text{Al}{{\left( \text{OH} \right)}_{\text{3}}}$.

${{\text{K}}_{\text{s}}}_{\text{p}}\text{=}\left[ \text{A}{{\text{l}}^{\text{3+}}} \right]{{\left[ \text{O}{{\text{H}}^{\text{-}}} \right]}^{\text{3}}}\text{=}\left( \text{S} \right){{\left( \text{3S} \right)}^{\text{3}}}\text{=27}{{\text{S}}^{\text{4}}}$

${{\text{S}}^{\text{4}}}\text{=27}{{\text{K}}_{\text{sp}}}\text{=27 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{27}}}\text{ }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-11}}}\text{=1 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-12}}}$

S=1×${{10}^{-3}}$ $\text{mol }{{\text{L}}^{\text{-1}}}$.

(i) Solubility of $\text{Al}{{\left( \text{OH} \right)}_{\text{3}}}\text{ }$​: Molar mass of $\text{Al}{{\left( \text{OH} \right)}_{\text{3}}}\text{ }$ is 78g. Hence, solubility of $\text{Al}{{\left( \text{OH} \right)}_{\text{3}}}\text{ }$​in $\text{g}{{\text{L}}^{\text{-1}}}\text{=1 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{3}}}\text{ }\!\!\times\!\!\text{ 78g}{{\text{L}}^{\text{-1}}}\text{=78 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-3}}}\text{g}{{\text{L}}^{\text{-1}}}\text{=7}\text{.8 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-2}}}\text{g}{{\text{L}}^{\text{-1}}}$
(ii) pH of the solution: S=1× ${{10}^{-3}}$mol ${{\text{L}}^{\text{-1}}}$

[OH]=3S=3×1× ${{10}^{-3}}$=3× ${{10}^{-3}}$

p[OH]=3−log3

pH=14−pOH=11+log3=11.4771.

Ans:

${{\text{K}}_{\text{sp}}}\text{=3}\text{.2 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-8}}}$

$\text{PbC}{{\text{l}}_{\text{2}}}\rightleftharpoons \text{P}{{\text{b}}^{\text{2+}}}\text{+2C}{{\text{l}}^{\text{-}}}$

${{\text{K}}_{\text{sp}}}\text{=}\left( \text{S} \right){{\left( \text{2S} \right)}^{\text{2}}}$

${{\text{K}}_{\text{sp}}}\text{=4}{{\text{S}}^{\text{3}}}$

${{\text{s}}^{\text{3}}}\text{=43}\text{.2 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-8}}}$

${{\text{s}}^{\text{3}}}\text{=8 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-9}}}$

$\text{s=2 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-3}}}$

$\text{Molar }\!\!~\!\!\text{ mass }\!\!~\!\!\text{ of }\!\!~\!\!\text{ PbC}{{\text{l}}_{\text{2}}}\text{=278g/mol }\!\!~\!\!\text{ }$

$\text{By using Molarity formula, }\!\!~\!\!\text{  }\!\!~\!\!\text{ }$

$\text{2 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-3}}}\text{=278 }\!\!\times\!\!\text{ x0}\text{.1}$

$\text{x=}\frac{\text{0}\text{.1}}{\text{278 }\!\!\times\!\!\text{ 2 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-3}}}}$

$\text{ }\!\!~\!\!\text{ =}\frac{\text{100}}{\text{278 }\!\!\times\!\!\text{ 2}}$

$\text{x=0}\text{.18L }\!\!~\!\!\text{ }$

Ans: As, $\text{N}{{\text{H}}_{\text{3}}}$ is Lewis base while $\text{B}{{\text{F}}_{\text{3}}}\text{ }$ is Lewis acid. Lewis electronic theory of acids and bases explains it. Hybridisation state: N in $\text{N}{{\text{H}}_{\text{3}}}$ is $\text{s}{{\text{p}}^{\text{3}}}$hybridised and Boron in $\text{B}{{\text{F}}_{\text{3}}}$ is $\text{s}{{\text{p}}^{2}}$ hybridised.

Ans: As, enthalpy of formation is given by-

${{\text{ }\!\!\Delta\!\!\text{ }}_{\text{r}}}{{\text{H}}^{\odot }}\text{=}{{\text{ }\!\!\Delta\!\!\text{ }}_{\text{f}}}{{\text{H}}^{\odot }}\left[ \text{CaO}\left( \text{s} \right) \right]\text{+}{{\text{ }\!\!\Delta\!\!\text{ }}_{\text{f}}}{{\text{H}}^{\odot }}\left[ \text{C}{{\text{O}}_{\text{2}}}\left( \text{g} \right) \right]\text{-}{{\text{ }\!\!\Delta\!\!\text{ }}_{\text{f}}}{{\text{H}}^{\odot }}\left[ \text{CaC}{{\text{O}}_{\text{3}}}\left( \text{s} \right) \right]$

${{\text{ }\!\!\Delta\!\!\text{ }}_{\text{r}}}{{\text{H}}^{\odot }}\text{=178}\text{.3kJ }\!\!~\!\!\text{ mo}{{\text{l}}^{\text{-1}}}$.

The reaction is endothermic. Therefore, according to Le Chatelier’s principle, reaction will proceed in forward direction on increasing temperature.

Ans: (i) $\to $ (b); (ii) $\to $ (d); (iii) $\to $ (c); (iv) $\to $ (a).

Ans: (i) $\to $ (d); (ii) $\to $ (c); (iii) $\to $ (b).

Ans: (i) $\to $ (d); (ii) $\to $ (a); (iii) $\to $ (b).

Ans: (i) $\to $ (b); (ii) $\to $ (e); (iii) $\to $ (c); (iv) $\to $ (d)