Chapter 16, Problem 16.94QP

Interpretation Introduction

Interpretation:

The equilibrium constant for the given reaction has to be calculated.

Concept Information:

Acid ionization constant ${\text{K}}_{\text{a}}$:

Acids ionize in water.  Strong acids ionize completely whereas weak acids ionize to some limited extent.

The degree to which a weak acid ionizes depends on the concentration of the acid and the equilibrium constant for the ionization.

The ionization of a weak acid $\text{HA}$ can be given as follows,

  ${\text{HA}}_{\text{(aq)}}\to {\text{H}}^{\text{+}}{}_{\text{(aq)}}\text{+}{\text{A}}^{\text{-}}{}_{\text{(aq)}}$

The equilibrium expression for the above reaction is given below.

  $K_a=\frac{[{\text{H}}^{\text{+}}{\text{][A}}^{\text{-}}\text{]}}{[\text{HA]}}$

Where,

  $K_a$ is acid ionization constant,

  $[{\text{H}}^{\text{+}}\text{]}$  is concentration of hydrogen ion

  $[{\text{A}}^{\text{-}}\text{]}$  is concentration of acid anion

  $[\text{HA]}$ is concentration of the acid

Autoionization of water:

The equation of equilibrium for autoionization of water is,

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

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

The equilibrium expression for water at $\text{25}{}^{\text{o}}\text{C}$ is,

  ${\text{[H}}^{\text{+}}{\text{][OH}}^{\text{-}}\text{] = 1}\times {\text{10}}^{\text{-14}}$

Taking negative logarithm on both sides, we get

  $\begin{array}{l}-\log ({\text{[H}}^{\text{+}}{\text{][OH}}^{\text{-}}\text{])= -log(1}\times {\text{10}}^{\text{-14}})\\ (-\log {\text{[H}}^{\text{+}}{\text{])+(-log[OH}}^{\text{-}}\text{])= 14})\end{array}$

The relationship between the hydronium ion concentration and the hydroxide ion concentration is given by the equation,

  $\text{pH}\text{+}\text{pOH}\text{=}\text{14,}\text{at}\text{25}{}^{\text{o}}\text{C}$

As $\text{pOH}$ and $\text{pH}$ are opposite scale, the total of both has to be equal to 14.

Therefore,

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

To Calculate: The equilibrium constant for the given reaction

Answer to Problem 16.94QP

• The equilibrium constant for the given reaction is $1.7\times {10}^{10}$

Explanation of Solution

Record the given datas

  ${\text{CH}}_{\text{3}}{\text{COOH}}_{(aq)}+{\text{NO}}_{\text{2}}{}^{\text{-}}\to {\text{ CH}}_{\text{3}}{\text{COO}}^{\text{-}}{}_{(aq)}+{\text{HNO}}_{\text{2}}{}_{(aq)}$

From the given data, we know that the equilibrium reaction between acetic acid and ${\text{NO}}_{\text{2}}{}^{\text{-}}$ ion in aqueous medium gives ${\text{CH}}_{\text{3}}{\text{COO}}^{\text{-}}$ ions and nitrous acid as products.

Splitting of the given equilibrium reaction

Write the given equilibrium reaction into two equilibria as follows,

  $\begin{array}{l}{\text{CH}}_{\text{3}}{\text{COOH}}_{\text{(aq)}}\to {\text{ CH}}_{\text{3}}{\text{COO}}^{\text{-}}{}_{\text{(aq)}}{\text{+H}}^{\text{+}}{}_{\text{(aq)}}\\ {\text{H}}^{\text{+}}{}_{\text{(aq)}}{\text{+ NO}}_{\text{2}}{{}^{\text{-}}}_{\text{(aq)}}\to {\text{ HNO}}_{\text{2}}{}_{\text{(aq)}}\end{array}$

Calculation of $K_a$ for ${\text{CH}}_{\text{3}}{\text{COOH}}_{(aq)}\to {\text{ CH}}_{\text{3}}{\text{COO}}^{\text{-}}{}_{(aq)}{\text{+H}}^{\text{+}}{}_{(aq)}$

The acid ionization constant $K_a$ for ${\text{CH}}_{\text{3}}{\text{COOH}}_{(aq)}\to {\text{ CH}}_{\text{3}}{\text{COO}}^{\text{-}}{}_{(aq)}{\text{+H}}^{\text{+}}{}_{(aq)}$ can be calculated as follows,

  $K_a=\frac{[{\text{H}}^{\text{+}}{\text{][CH}}_{\text{3}}{\text{COO}}^{\text{-}}\text{]}}{[{\text{CH}}_{\text{3}}\text{COOH]}}$

The acid ionization constant for formic acid is $1.8\times {10}^{-5}$

Therefore,

  $\begin{array}{l}{K}_a=\frac{[{\text{H}}^{\text{+}}{\text{][CH}}_{\text{3}}{\text{COO}}^{\text{-}}\text{]}}{[{\text{CH}}_{\text{3}}\text{COOH]}}\\ 1.8\times {10}^{-5}=\frac{[{\text{H}}^{\text{+}}{\text{][CH}}_{\text{3}}{\text{COO}}^{\text{-}}\text{]}}{[{\text{CH}}_{\text{3}}\text{COOH]}}\end{array}$

Calculation of $K{\text{'}}_a$ for ${\text{H}}^{\text{+}}{}_{(aq)}{\text{+ NO}}_{\text{2}}{{}^{\text{-}}}_{(aq)}\to {\text{ HNO}}_{\text{2}}{}_{(aq)}$

The acid ionization constant $K{\text{'}}_a$ for ${\text{H}}^{\text{+}}{}_{(aq)}{\text{+ NO}}_{\text{2}}{{}^{\text{-}}}_{(aq)}\to {\text{ HNO}}_{\text{2}}{}_{(aq)}$ can be calculated as follows,

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

The acid ionization constant for nitrous acid is $4.5\times {10}^{-4}$

Therefore,

  $\begin{array}{l}K{\text{'}}_a=\frac{{\text{[HNO}}_{\text{2}}\text{]}}{[{\text{H}}^{\text{+}}][{\text{NO}}_{\text{2}}{}^{\text{-}}\text{]}}\\ K{\text{'}}_a=\frac{1}{K_a({\text{HNO}}_{\text{2}})}\\ =\frac{1}{4.5\times {10}^{-4}}\\ =2.2\times {10}^3\end{array}$

Equilibrium constant for the given reaction:

Adding up of two equilibria reaction in step 2, will result in the given equilibrium reaction.

  $\begin{array}{l}{\text{CH}}_{\text{3}}{\text{COOH}}_{(aq)}\to {\text{ CH}}_{\text{3}}{\text{COO}}^{\text{-}}{}_{(aq)}{\text{+H}}^{\text{+}}{}_{(aq)}\\ {\text{H}}^{\text{+}}{}_{(aq)}{\text{+NO}}_{\text{2}}{{}^{\text{-}}}_{(aq)}\to {\text{ HNO}}_{\text{2}}{}_{(aq)}\\ \text{On adding up the above two reactions, we get}\\ {\text{CH}}_{\text{3}}{\text{COOH}}_{(aq)}+{\text{NO}}_{\text{2}}{}^{\text{-}}\to {\text{ CH}}_{\text{3}}{\text{COO}}^{\text{-}}{}_{(aq)}+{\text{HNO}}_{\text{2}}{}_{(aq)}\end{array}$

The equilibrium constant for this sum is the product of the equilibrium constants of the component reactions,

  $\begin{array}{l}K=\frac{{\text{[CH}}_{\text{3}}{\text{COO}}^{\text{-}}\text{]}[{\text{HNO}}_{\text{2}}\text{]}}{[{\text{CH}}_{\text{3}}{\text{COOH][NO}}_{\text{2}}{}^{\text{-}}\text{]}}\\ =K_a\times K{\text{'}}_a\\ =(1.8\times {10}^{-5})(2.2\times {10}^3)\\ =4.0\times {10}^{-2}\end{array}$

Therefore, the equilibrium constant for the given reaction is $4.0\times {10}^{-2}$

Conclusion

The equilibrium constant for the given reaction was calculated.