Chapter 9, Problem 57P

Interpretation Introduction

(a)

Interpretation:

The acid and conjugate acid in below reaction equation needs to be identified. The direction of reaction at equilibrium condition needs to be determined.

  ${\text{H}}_{\text{3}}{\text{PO}}_{\text{4(aq)}}{\text{ + CN}}^{\text{-}}{}_{\text{(aq)}}\underset{}{\rightleftarrows }{\text{H}}_{\text{2}}{\text{PO}}_{\text{4}}{{}^{\text{-}}}_{\text{(aq)}}{\text{ + HCN}}_{\text{(aq)}}$

Concept Introduction:

The Bronsted-Lowry acid-base theory was purposed by Bronsted and Lowery is called Bronsted-Lowry acid-base theory. It states that acid can give ${\text{H}}^{\text{+}}$ ions whereas a base can accept the ${\text{H}}^{\text{+}}$ ion in its solution. Hence this theory is entirely based on the presence of ${\text{H}}^{\text{+}}$ ion in the given substance. It purposed the concept of conjugated acid-base pair. A Bronsted acid gives ${\text{H}}^{\text{+}}$ ion to form conjugated base whereas a Bronsted base accepts ${\text{H}}^{\text{+}}$ ion to form its conjugated acid.

  $\begin{array}{l}{\text{ HA + H}}_{\text{2}}\text{O }\underset{}{\to }{\text{ A}}^{\text{-}}{\text{ + H}}_{\text{3}}{\text{O}}^{\text{+}}\\ \text{Bronsted + Bronsted Conjugated Conjugated}\\ \text{ Acid base base acid}\end{array}$

A strong acid shows complete dissociation to respective anion and ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ whereas a weak acid can only partially ionize to its respective ions. A strong acid forms a weak conjugated base whereas a weak acid forms a strong conjugated base.

Answer to Problem 57P

Acid = ${\text{H}}_{\text{3}}{\text{PO}}_{\text{4(aq)}}$

Conjugate acid= $\text{HCN}$

Reaction will favor right side towards product.

Explanation of Solution

According to Conjugated acid-base pair concept. A strong acid forms a weak conjugated base whereas a weak acid forms a strong conjugated base.

  $\begin{array}{l}{\text{ HA + H}}_{\text{2}}\text{O }\underset{}{\to }{\text{ A}}^{\text{-}}{\text{ + H}}_{\text{3}}{\text{O}}^{\text{+}}\\ \text{Bronsted + Bronsted Conjugated Conjugated}\\ \text{ Acid base base acid}\\ \text{(weak) (strong) (strong) (weak)}\end{array}$

  ${\text{H}}_{\text{3}}{\text{PO}}_{\text{4(aq)}}{\text{ + CN}}^{\text{-}}{}_{\text{(aq)}}\underset{}{\rightleftarrows }{\text{H}}_{\text{2}}{\text{PO}}_{\text{4}}{{}^{\text{-}}}_{\text{(aq)}}{\text{ + HCN}}_{\text{(aq)}}$

In the given reaction ${\text{H}}_{\text{3}}{\text{PO}}_{\text{4(aq)}}$ is an acid as it gives ${\text{H}}^{\text{+}}$ ion to form its conjugate base ${\text{H}}_{\text{2}}{\text{PO}}_{\text{4}}{{}^{\text{-}}}_{\text{(aq)}}$ whereas ${\text{CN}}^{\text{-}}{}_{\text{(aq)}}$ accept ${\text{H}}^{\text{+}}$ ion to form its conjugateacid as ${\text{HCN}}_{\text{(aq)}}$.

The equilibrium constant for acid dissociation is denoted as ${\text{K}}_{\text{a}}$ . It represents the ratio of the equilibrium concentration of product and reactant molecule. For the given acid dissociation, the ${\text{K}}_{\text{a}}$ expression can be written as:

  ${\text{HA + H}}_{\text{2}}\text{O }\underset{}{\rightleftarrows }{\text{ A}}^{-}+{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$

  ${\text{K}}_{\text{a}}\text{ = }\frac{{\text{[A}}^{-}][{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}}{\text{[HA]}}$

The ${\text{K}}_{\text{a}}$ value of ${\text{H}}_{\text{3}}{\text{PO}}_{\text{4(aq)}}$ ( $7.5\times {10}^{-3}$ )is larger than ${\text{HCN}}_{\text{(aq)}}$ ( $4.9\times {10}^{-10}$ ), therefore the reaction will favor the weak acid and will shift the equilibrium towards product.

Interpretation Introduction

(b)

Interpretation:

The acid and conjugate acid in below reaction equation needs to be identified. The direction of reaction at equilibrium condition needs to be determined.

  ${\text{Br}}^{\text{-}}{}_{\text{(aq)}}{\text{ + HSO}}_{\text{4}}{{}^{\text{-}}}_{\text{(aq)}}\underset{}{\rightleftarrows }{\text{SO}}_{\text{4}}{{}^{\text{2-}}}_{\text{(aq)}}{\text{ + HBr}}_{\text{(aq)}}$

Concept Introduction:

The Bronsted-Lowry acid-base theory was purposed by Bronsted and Lowery is called Bronsted-Lowry acid-base theory. It states that acid can give ${\text{H}}^{\text{+}}$ ions whereas a base can accept the ${\text{H}}^{\text{+}}$ ion in its solution. Hence, this theory is entirely based on the presence of ${\text{H}}^{\text{+}}$ ion in the given substance. It purposed the concept of conjugated acid-base pair. A Bronsted acid gives ${\text{H}}^{\text{+}}$ ion to form conjugated base whereas a Bronsted base accepts ${\text{H}}^{\text{+}}$ ion to form its conjugated acid.

  $\begin{array}{l}{\text{ HA + H}}_{\text{2}}\text{O }\underset{}{\to }{\text{ A}}^{\text{-}}{\text{ + H}}_{\text{3}}{\text{O}}^{\text{+}}\\ \text{Bronsted + Bronsted Conjugated Conjugated}\\ \text{ Acid base base acid}\end{array}$

A strong acid shows complete dissociation to respective anion and ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ whereas a weak acid can only partially ionize to its respective ions. A strong acid forms a weak conjugated base whereas a weak acid forms a strong conjugated base.

Answer to Problem 57P

Acid = ${\text{HSO}}_{\text{4}}{}^{\text{-}}$

Conjugate acid= $\text{HBr}$

Reaction will favor left side towards reactants.

Explanation of Solution

According to Conjugated acid-base pair concept, a strong acid forms a weak conjugated base whereas a weak acid forms a strong conjugated base.

  $\begin{array}{l}{\text{ HA + H}}_{\text{2}}\text{O }\underset{}{\to }{\text{ A}}^{\text{-}}{\text{ + H}}_{\text{3}}{\text{O}}^{\text{+}}\\ \text{Bronsted + Bronsted Conjugated Conjugated}\\ \text{ Acid base base acid}\\ \text{(weak) (strong) (strong) (weak)}\end{array}$

  ${\text{Br}}^{\text{-}}{}_{\text{(aq)}}{\text{ + HSO}}_{\text{4}}{{}^{\text{-}}}_{\text{(aq)}}\underset{}{\rightleftarrows }{\text{SO}}_{\text{4}}{{}^{\text{2-}}}_{\text{(aq)}}{\text{ + HBr}}_{\text{(aq)}}$

In the given reaction ${\text{ HSO}}_{\text{4}}{{}^{\text{-}}}_{\text{(aq)}}$ is an acid as it gives ${\text{H}}^{\text{+}}$ ion to form its conjugate base ${\text{SO}}_{\text{4}}{}^{\text{2-}}$ whereas ${\text{Br}}^{\text{-}}$ accept ${\text{H}}^{\text{+}}$ ion to form its conjugate acid as $\text{HBr}$.

The equilibrium constant for acid dissociation is denoted as ${\text{K}}_{\text{a}}$ . It represents the ratio of the equilibrium concentration of product and reactant molecule. For the given acid dissociation the ${\text{K}}_{\text{a}}$ expression can be written as:

  ${\text{HA + H}}_{\text{2}}\text{O }\underset{}{\rightleftarrows }{\text{ A}}^{-}+{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$

  ${\text{K}}_{\text{a}}\text{ = }\frac{{\text{[A}}^{-}][{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}}{\text{[HA]}}$

The ${\text{K}}_{\text{a}}$ value of ${\text{ HSO}}_{\text{4}}{{}^{\text{-}}}_{\text{(aq)}}$ ( $1.2\times {10}^{-2}$ )is lesser than $\text{HBr}$ ( $1.0\times {10}^9$ ), therefore the reaction will favor the weak acid and will shift the equilibrium towards reactants.

Interpretation Introduction

(c)

Interpretation:

The acid and conjugate acid in below reaction equation needs to be identified. The direction of reaction at equilibrium condition needs to be determined.

  ${\text{CH}}_{\text{3}}{\text{COO}}^{\text{-}}{}_{\text{(aq)}}{\text{ + H}}_{\text{2}}{\text{CO}}_{\text{3}}{}_{\text{(aq)}}\underset{}{\rightleftarrows }{\text{CH}}_{\text{3}}{\text{COOH}}_{\text{(aq)}}{\text{ + HCO}}_3{{}^{-}}_{\text{(aq)}}$

Concept Introduction:

The Bronsted-Lowry acid-base theory was purposed by Bronsted and Lowery is called Bronsted-Lowry acid-base theory. It states that acid can give ${\text{H}}^{\text{+}}$ ions whereas a base can accept the ${\text{H}}^{\text{+}}$ ion in its solution. Hence, this theory is entirely based on the presence of ${\text{H}}^{\text{+}}$ ion in the given substance. It purposed the concept of conjugated acid-base pair. A Bronsted acid gives ${\text{H}}^{\text{+}}$ ion to form conjugated base whereas a Bronsted base accepts ${\text{H}}^{\text{+}}$ ion to form its conjugated acid.

  $\begin{array}{l}{\text{ HA + H}}_{\text{2}}\text{O }\underset{}{\to }{\text{ A}}^{\text{-}}{\text{ + H}}_{\text{3}}{\text{O}}^{\text{+}}\\ \text{Bronsted + Bronsted Conjugated Conjugated}\\ \text{ Acid base base acid}\end{array}$

A strong acid shows complete dissociation to respective anion and ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ whereas a weak acid can only partially ionize to its respective ions. A strong acid forms a weak conjugated base whereas a weak acid forms a strong conjugated base.

Answer to Problem 57P

Acid = ${\text{H}}_{\text{2}}{\text{CO}}_{\text{3}}$

Conjugate acid= ${\text{CH}}_{\text{3}}\text{COOH}$

Reaction will favor left side towards reactants.

Explanation of Solution

According to Conjugated acid-base pair concept; a strong acid forms a weak conjugated base whereas a weak acid forms a strong conjugated base.

  $\begin{array}{l}{\text{ HA + H}}_{\text{2}}\text{O }\underset{}{\to }{\text{ A}}^{\text{-}}{\text{ + H}}_{\text{3}}{\text{O}}^{\text{+}}\\ \text{Bronsted + Bronsted Conjugated Conjugated}\\ \text{ Acid base base acid}\\ \text{(weak) (strong) (strong) (weak)}\end{array}$

  ${\text{CH}}_{\text{3}}{\text{COO}}^{\text{-}}{}_{\text{(aq)}}{\text{ + H}}_{\text{2}}{\text{CO}}_{\text{3}}{}_{\text{(aq)}}\underset{}{\rightleftarrows }{\text{CH}}_{\text{3}}{\text{COOH}}_{\text{(aq)}}{\text{ + HCO}}_3{{}^{-}}_{\text{(aq)}}$

In the given reaction ${\text{ H}}_{\text{2}}{\text{CO}}_{\text{3}}$ is an acid as it gives ${\text{H}}^{\text{+}}$ ion to form its conjugate base ${\text{HCO}}_3{}^{-}$ whereas ${\text{CH}}_{\text{3}}{\text{COO}}^{\text{-}}$ accept ${\text{H}}^{\text{+}}$ ion to form its conjugate acid as ${\text{CH}}_{\text{3}}\text{COOH}$.

The equilibrium constant for acid dissociation is denoted as ${\text{K}}_{\text{a}}$ . It represents the ratio of the equilibrium concentration of product and reactant molecule. For the given acid dissociation the ${\text{K}}_{\text{a}}$ expression can be written as:

  ${\text{HA + H}}_{\text{2}}\text{O }\underset{}{\rightleftarrows }{\text{ A}}^{-}+{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$

  ${\text{K}}_{\text{a}}\text{ = }\frac{{\text{[A}}^{-}][{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}}{\text{[HA]}}$

The ${\text{K}}_{\text{a}}$ value of ${\text{CH}}_{\text{3}}\text{COOH}$ ( $1.8\times {10}^{-5}$ ) is larger than ${\text{ H}}_{\text{2}}{\text{CO}}_{\text{3}}$ ( $4.3\times {10}^{-7}$ ), therefore the reaction will favor the weak acid and will shift the equilibrium towards reactants.