Chapter 12, Problem 12.65PAE

(a)

Interpretation Introduction

To determine:

Write chemical equations and equilibrium expressions for the reactions for acetic acid with water.

Explanation of Solution

$C{H}_{3}COOH\left(aq\right)+H_{2}O\left(l\right)\to C{H}_{3}CO{O}^{-}\left(aq\right)+H_3{O}^{+}\left(aq\right)$

Rate of equilibrium expression $\left(K\right)=\frac{\left[C{H}_{3}CO{O}^{-}\right]\left[H_3{O}^{+}\right]}{\left[C{H}_{3}COOH\right]\left[H_{2}O\right]}$

In this equilibrium, water is pure liquid and its concentration is constant which is called acid dissociation constant $K_a$

$\left(K_a\right)=\frac{\left[C{H}_{3}CO{O}^{-}\right]\left[H_3{O}^{+}\right]}{\left[C{H}_{3}COOH\right]}$

So, final concentration of product $\left[C{H}_{3}CO{O}^{-}\right]\left[H_3{O}^{+}\right]=K_a\left[C{H}_{3}COOH\right]$.

(b)

Interpretation Introduction

To determine:

Write chemical equations and equilibrium expressions for the reactions for ethanoic acid and water.

Explanation of Solution

The given expression is expressed below.

$C_2{H}_{5}COOH\left(aq\right)+H_{2}O\left(l\right)\to C_2{H}_{5}CO{O}^{-}\left(aq\right)+H_3{O}^{+}\left(aq\right)$

Rate of constant $\left(K\right)=\frac{\left[C_2{H}_{5}CO{O}^{-}\right]\left[H_3{O}^{+}\right]}{\left[C_2{H}_{5}COOH\right]\left[H_{2}O\right]}$

In this equilibrium water is pure liquid and its concentration is constant so, $\left[H_{2}O\right]=1$ and can be dropped.

$\left(K_a\right)=\frac{\left[C_2{H}_{5}CO{O}^{-}\right]\left[H_3{O}^{+}\right]}{\left[C_2{H}_{5}COOH\right]}$

$K_a$ : Dissociation constant

$\left[C_2{H}_{5}CO{O}^{-}\right]\left[H_3{O}^{+}\right]=K_a\left[C_2{H}_{5}COOH\right]$

(c)

Interpretation Introduction

To determine:

Write chemical equation and equilibrium expressions for the reactions of hydrogen fluoride acid and water.

Explanation of Solution

$HF\left(aq\right)+H_{2}O\left(l\right)\to F^{-}\left(aq\right)+H_3{O}^{+}\left(aq\right)$

Equilibrium rate constant $\left(K\right)=\frac{\left[F^{-}\right]\left[H_3{O}^{+}\right]}{\left[HF\right]\left[H_{2}O\right]}$

Here, water is a pure is a pure liquid and its concentration is constant. $\left[H_{2}O\right]=1$ Then, it is called dissociation constant.

$\left(K_a\right)=\frac{\left[F^{-}\right]\left[H_3{O}^{+}\right]}{\left[HF\right]}$

Arranging the equation, you get

$\left[F^{-}\right]\left[H_3{O}^{+}\right]=K_a\left[HF\right]$

(d)

Interpretation Introduction

To determine:

Write chemical equations and equilibrium expressions for the reactions for hydrogen fluoride acid and water.

Explanation of Solution

$HClO\left(aq\right)+H_{2}O\left(l\right)\to Cl{O}^{-}\left(aq\right)+H_3{O}^{+}\left(aq\right)$

Rate constant of equilibrium reaction $\left(K\right)=\frac{\left[Cl{o}^{-}\right]\left[H_3{O}^{+}\right]}{\left[HClO\right]\left[H_{2}O\right]}$

Here water is pure liquid and its concentration is constant $\left[H_{2}O\right]=1$.

Then, it is called dissociation constant.

$\left(K_a\right)=\frac{\left[Cl{o}^{-}\right]\left[H_3{O}^{+}\right]}{\left[HClO\right]}$

Arranging the equation you get

$\left[Cl{o}^{-}\right]\left[H_3{O}^{+}\right]=K_a\left[HClO\right]$

(e)

Interpretation Introduction

To determine:

Write chemical equation and equilibrium expressions for the reactions of carbonic acid and water.

Explanation of Solution

$H_{2}C{O}_3\left(aq\right)+H_{2}O\left(l\right)\to H_3{O}^{+}\left(aq\right)+HC{O}_3^{-}\left(aq\right)$

Rate of equilibrium constant $\left(k\right)=\frac{\left[HC{O}_3^{-}\right]\left[H_3{O}^{+}\right]}{\left[H_{2}C{O}_3\right]\left[H_{2}O\right]}$

In this equilibrium water is pure liquid and its concentration is constant $\left[H_{2}O\right]=1$

So, dissociation constant $\left(k_a\right)=\frac{\left[HC{O}_3^{-}\right]\left[H_3{O}^{+}\right]}{\left[H_{2}C{O}_3\right]}$

Arranging the equation, you get.

$\left[HC{O}_3^{-}\right]\left[H_3{O}^{+}\right]=K_a\left[H_{2}C{O}_3\right]$.

Conclusion

Hence, all the parts have been described.