Question

Calculate the value for the equilibrium constant for each of the following aqueous reactions.

1. ${\mathbf{NH}}_{\mathbf{3}}\mathbf{+}{\mathbf{H}}_{\mathbf{3}}{\mathbf{O}}^{\mathbf{+}}\mathbf{\rightleftharpoons }{{\mathbf{NH}}_{\mathbf{4}}}^{\mathbf{+}}\mathbf{+}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}$
2. ${{\mathbf{NO}}_{\mathbf{2}}}^{\mathbf{-}}\mathbf{+}{\mathbf{H}}_{\mathbf{3}}{\mathbf{O}}^{\mathbf{+}}\mathbf{\rightleftharpoons }{\mathbf{HNO}}_{\mathbf{2}}\mathbf{+}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}$
3. ${{\mathbf{NH}}_{\mathbf{4}}}^{\mathbf{+}}\mathbf{+}{\mathbf{CH}}_{\mathbf{3}}{{\mathbf{CO}}_{\mathbf{2}}}^{\mathbf{-}}\mathbf{\rightleftharpoons }{\mathbf{NH}}_{\mathbf{3}}\mathbf{+}{\mathbf{CH}}_{\mathbf{3}}\mathbf{COOH}$
4. ${\mathbf{H}}_{\mathbf{3}}{\mathbf{O}}^{\mathbf{+}}\mathbf{+}{\mathbf{OH}}^{\mathbf{-}}\mathbf{\rightleftharpoons }\mathbf{2}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}$
5. ${{\mathbf{NH}}_{\mathbf{4}}}^{\mathbf{+}}\mathbf{+}{\mathbf{OH}}^{\mathbf{-}}\mathbf{\rightleftharpoons }{\mathbf{NH}}_{\mathbf{3}}\mathbf{+}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}$
6. ${\mathbf{HNO}}_{\mathbf{2}}\mathbf{+}{\mathbf{OH}}^{\mathbf{-}}\mathbf{\rightleftharpoons }{{\mathbf{NO}}_{\mathbf{2}}}^{\mathbf{-}}\mathbf{+}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}$
   

Short answer

Equilibrium Constant for the given reactions are

1. 1.8× 10⁹
2. 2.5× 10^-3
3. 3.09 × 10^-5
4. 1 × 10¹⁴
5. 4 × 10¹⁰

Step-by-step solution

Equilibrium Constant for reaction   

Reaction a is given as

${\text{NH}}_{\text{3}}{\text{+H}}_{\text{3}}{\text{O}}^{\text{+}}\rightleftharpoons {{\text{NH}}_{\text{4}}}^{\text{+}}{\text{+H}}_{\text{2}}\text{O}$

$\text{K=}\frac{{{\text{[NH}}_{\text{4}}}^{\text{+}}\text{]}}{{\text{[NH}}_{\text{3}}{\text{][H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}}$

NH₃ reacts with water as

$\begin{array}{l}{\text{NH}}_{\text{3}}{\text{+H}}_{\text{2}}\text{O}\rightleftharpoons {{\text{NH}}_{\text{4}}}^{\text{+}}{\text{+OH}}^{\text{-}}\\ \text{K=}\frac{{{\text{[NH}}_{\text{4}}}^{\text{+}}{\text{][OH}}^{\text{-}}\text{]}}{{\text{[NH}}_{\text{3}}\text{]}}\end{array}$

From the two equilibrium constants we can find out that

$\begin{array}{l}\frac{\text{K'}}{\text{K}}\text{=}\frac{{{\text{[NH}}_{\text{4}}}^{\text{+}}{\text{][OH}}^{\text{-}}\text{]}}{{\text{[NH}}_{\text{3}}\text{]}}\text{×}\frac{{\text{[NH}}_{\text{3}}{\text{][H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}}{{{\text{[NH}}_{\text{4}}}^{\text{+}}\text{]}}{\text{=[H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{][OH}}^{\text{-}}{\text{]=K}}_{\text{w}}\\ \text{K=}\frac{\text{K'}}{{\text{K}}_{\text{w}}}\text{=}\frac{{\text{1.8×10}}^{\text{-5}}}{{\text{1×10}}^{\text{-14}}}{\text{=1.8×10}}^{\text{9}}\end{array}$

Hence, equilibrium constant for the reaction is 1.8× 10⁹.

Equilibrium Constant for reaction b  

Reaction b is given as

_{${{\text{NO}}_{\text{2}}}^{\text{-}}{\text{+H}}_{\text{3}}{\text{O}}^{\text{+}}\rightleftharpoons {\text{HNO}}_{\text{2}}{\text{+H}}_{\text{2}}\text{O}$}

_{$\text{K=}\frac{{\text{[HNO}}_{\text{2}}\text{]}}{{\text{[NO}}_{\text{2}}{\text{][H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}}$}

HNO₂ reacts with water as

${\text{HNO}}_{\text{2}}{\text{+H}}_{\text{2}}\text{O\hspace{0.17em}}\rightleftharpoons {{\text{NO}}_{\text{2}}}^{\text{-}}{\text{+H}}_{\text{3}}{\text{O}}^{\text{+}}$

$\begin{array}{l}{\text{K'}}_{}\text{=}\frac{{{\text{[NO}}_{\text{2}}}^{\text{-}}{\text{][H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}}{{\text{[HNO}}_{\text{2}}\text{]}}\text{=}\frac{\text{1}}{\text{K}}\\ \text{Fromthetwoequilibriumconstantswecanfindoutthat}\\ \text{K=}\frac{\text{1}}{\text{K'}}\text{=}\frac{\text{1}}{{\text{4×10}}^{\text{-4}}}{\text{=2.5×10}}^{\text{-3}}\end{array}$

Hence, equilibrium constant for the reaction is 2.5× 10^-3.

Equilibrium Constant for reaction c

Reaction c is given as

${{\text{NH}}_{\text{4}}}^{\text{+}}{\text{+CH}}_{\text{3}}{{\text{CO}}_{\text{2}}}^{\text{-}}\rightleftharpoons {\text{NH}}_{\text{3}}{\text{+CH}}_{\text{3}}\text{COOH}$

$\text{K=}\frac{{\text{[NH}}_{\text{3}}{\text{][CH}}_{\text{3}}\text{COOH]}}{{{\text{[NH}}_{\text{4}}}^{\text{+}}{\text{][CH}}_{\text{3}}{\text{COO}}^{\text{-}}\text{]}}$

For the reaction,

${\text{NH}}_{\text{3}}{\text{+H}}_{\text{3}}{\text{O}}^{\text{+}}\rightleftharpoons {{\text{NH}}_{\text{4}}}^{\text{+}}{\text{+H}}_{\text{2}}\text{O}$

$\begin{array}{l}\text{K'=}\frac{{{\text{[NH}}_{\text{4}}}^{\text{+}}\text{]}}{{\text{[NH}}_{\text{3}}{\text{][H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}}\\ {\text{K=1.8×10}}^{\text{9}}\end{array}$

For the reaction of acetic acid in water,

${\text{CH}}_{\text{3}}{\text{CO}}_{\text{2}}{\text{H+H}}_{\text{2}}\text{O}\rightleftharpoons {\text{H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{+CH}}_{\text{3}}{\text{COO}}^{\text{-}}$

$\begin{array}{l}\text{K"=}\frac{{\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{][CH}}_{\text{3}}{\text{COO}}^{\text{-}}\text{]}}{{\text{[CH}}_{\text{3}}\text{COOH]}}{\text{=1.8×10}}^{\text{-5}}\\ \text{Fromthethreeequilibriumconstantswecanfindoutthat}\\ \frac{\text{1}}{\text{K'×K"}}\text{=}\frac{{\text{[NH}}_{\text{3}}{\text{][H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}}{{{\text{[NH}}_{\text{4}}}^{\text{+}}\text{]}}\text{×}\frac{{\text{[CH}}_{\text{3}}\text{COOH]}}{{\text{[CH}}_{\text{3}}{\text{COO}}^{\text{-}}{\text{][H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}}\\ \text{K=}\frac{\text{1}}{{\text{1.8×10}}^{\text{-5}}{\text{×1.8×10}}^{\text{9}}}{\text{=3.09×10}}^{\text{-5}}\end{array}$

Hence, equilibrium constant for the reaction is 3.09 × 10^-5.

Equilibrium Constant for reaction d

Reaction d is given as

${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{+OH}}^{\text{-}}\rightleftharpoons {\text{2H}}_{\text{2}}\text{O}$

And Ionization constant of water is related as,

$\begin{array}{l}\text{K=}\frac{\text{1}}{{\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{][OH}}^{\text{-}}\text{]}}\\ \frac{1}{{\text{K}}_{\text{w}}}=\frac{1}{{\text{1×10}}^{\text{-14}}}\\ \frac{1}{{\text{K}}_{\text{w}}}={\text{1×10}}^{\text{14}}\end{array}$

Hence, equilibrium constant for the reaction is 1 × 10¹⁴.

Equilibrium Constant for reaction e  

Reaction e is given as

${{\text{NH}}_{\text{4}}}^{\text{+}}{\text{+OH}}^{\text{-}}\rightleftharpoons {\text{NH}}_{\text{3}}{\text{+H}}_{\text{2}}\text{O}$

$\text{K=}\frac{{\text{[NH}}_{\text{3}}\text{]}}{{{\text{[NH}}_{\text{4}}}^{\text{+}}{\text{][OH}}^{\text{-}}\text{]}}$

NH₃ reacts with water as

$\begin{array}{l}{\text{NH}}_{\text{3}}{\text{+H}}_{\text{2}}\text{O}\rightleftharpoons {{\text{NH}}_{\text{4}}}^{\text{+}}{\text{+OH}}^{\text{-}}\\ {\text{K'}}_{}\text{=}\frac{{{\text{[NH}}_{\text{4}}}^{\text{+}}{\text{][OH}}^{\text{-}}\text{]}}{{\text{[NH}}_{\text{3}}\text{]}}\text{=}\frac{\text{1}}{\text{K}}\end{array}$

$\text{K'=}\frac{\text{1}}{\text{K'}}\text{=}\frac{\text{1}}{{\text{1.8×10}}^{\text{-5}}}{\text{=5.5×10}}^{\text{4}}$

Hence, equilibrium constant for the reaction is 5.5 × 10⁴

Equilibrium Constant for reaction f

Reaction f is given as

${\text{HNO}}_{\text{2}}{\text{+OH}}^{\text{-}}\rightleftharpoons {{\text{NO}}_{\text{2}}}^{\text{-}}{\text{+H}}_{\text{2}}\text{O}$

$\text{K=}\frac{{{\text{[NO}}_{\text{2}}}^{\text{-}}\text{]}}{{\text{[HNO}}_{\text{2}}{\text{][OH}}^{\text{-}}\text{]}}$

Dissociation of HNO₂ is given as

${\text{HNO}}_{\text{2}}{\text{+H}}_{\text{2}}\text{O\hspace{0.17em}}\rightleftharpoons {{\text{NO}}_{\text{2}}}^{\text{-}}{\text{+H}}_{\text{3}}{\text{O}}^{\text{+}}$

$\text{K'=}\frac{{{\text{[NO}}_{\text{2}}}^{\text{-}}{\text{][H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}}{{\text{[HNO}}_{\text{2}}\text{]}}$

Ionisation constant of water is represented as

${\text{K}}_{\text{w}}{\text{=[H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{][OH}}^{\text{-}}\text{]}$

From these 3 equations we can say that

$\begin{array}{l}\text{K=}\frac{\text{K'}}{{\text{K}}_{\text{w}}}\\ \text{K=}\frac{{\text{4×10}}^{\text{-4}}}{{\text{1×10}}^{\text{-14}}}\\ {\text{K=4×10}}^{\text{10}}\end{array}$

Hence, equilibrium constant for the reaction is 4 × 10¹⁰