Chapter 16, Problem 40PS

For each of the following reactions, predict whether the equilibrium lies predominantly to the left or to the right. Explain your predictions briefly.

1. (a) H₂S(aq) + CO₃²⁻(aq) ⇄ HS⁻(aq) + HCO₃⁻(aq)
2. (b) HCN(aq) + SO₄²⁻(aq) ⇄ CN(aq) + HSO₄(aq)
3. (c) SO₄²⁻(aq) + CH₃CO₂H(aq) ⇄ HSO₄⁻(aq) + CH₃CO₂⁻(aq)

Interpretation Introduction

Interpretation: The direction of the equilibrium for the given reaction is to be determined.

Concept introduction: An acid-base reaction reaction is represented as written below.

  $\begin{array}{l}\text{HA}\left(\text{aq}\right)+\text{B}\left(\text{aq}\right)\rightleftharpoons {\text{ BH}}^{\text{+}}\left(\text{aq}\right)+{\text{ A}}^{-}\left(\text{aq}\right)\\ \left(\text{acid}\right)\left(\text{base}\right)\left(\begin{array}{l}\text{conjugate }\\ \text{ acid}\end{array}\right)\left(\begin{array}{l}\text{conjugate }\\ \text{ base}\end{array}\right)\end{array}$

The base will take up the proton from acid and form its conjugate acid and simultaneously acid will form its conjugate base. The equilibrium will be forward or backward can be determined by using the dissociation constants $\left(K_{\text{a}}\text{and }{K}_{\text{b}}\right)$ for reactants as well as of the products. The more the value of $K_{\text{a}}$ for an acid, stronger will be the acid and it will undergo faster ionization. Similarly, higher the value of $K_{\text{b}}$ for a base, stronger will be the base and it will undergo faster ionization.

Answer to Problem 40PS

The equilibrium for the given reaction will lie in right direction predominantly.

  $\begin{array}{l}{\text{H}}_2\text{S}\left(\text{aq}\right)+{\text{ CO}}_3{}^{2-}\left(\text{aq}\right)\rightleftharpoons {\text{ HS}}^{-}\left(\text{aq}\right)+{\text{ HCO}}_3{}^{-}\left(\text{aq}\right)\\ \left(\text{acid}\right)\left(\text{base}\right)\left(\begin{array}{l}\text{conjugate }\\ \text{ base}\end{array}\right)\left(\begin{array}{l}\text{conjugate }\\ \text{ acid}\end{array}\right)\end{array}$

Explanation of Solution

The equilibrium for the given reaction will move in left side is explained below.

  $\begin{array}{l}{\text{H}}_2\text{S}\left(\text{aq}\right)+{\text{ CO}}_3{}^{2-}\left(\text{aq}\right)\rightleftharpoons {\text{ HS}}^{-}\left(\text{aq}\right)+{\text{ HCO}}_3{}^{-}\left(\text{aq}\right)\\ \left(\text{acid}\right)\left(\text{base}\right)\left(\begin{array}{l}\text{conjugate }\\ \text{ base}\end{array}\right)\left(\begin{array}{l}\text{conjugate }\\ \text{ acid}\end{array}\right)\end{array}$

Here, the acid ${\text{ H}}_2\text{S}$$\left(K_{\text{a}}=1.2\times {10}^{-7}\right)$ is stronger acid as compared to conjugate acid ${\text{HCO}}_3{}^{-}$$\left(K_{\text{a}}=4.8\times {10}^{-11}\right)$, hence ${\text{ H}}_2\text{S}$ will ionize faster as compared to ${\text{HCO}}_3{}^{-}$ due to which the equilibrium will shift in left direction. So it can be stated that the given reaction is reactant favoured.

Interpretation Introduction

Interpretation: The direction of the equilibrium for the given reaction is to be determined.

Concept introduction: An acid-base reaction reaction is represented as written below.

  $\begin{array}{l}\text{HA}\left(\text{aq}\right)+\text{B}\left(\text{aq}\right)\rightleftharpoons {\text{ BH}}^{\text{+}}\left(\text{aq}\right)+{\text{ A}}^{-}\left(\text{aq}\right)\\ \left(\text{acid}\right)\left(\text{base}\right)\left(\begin{array}{l}\text{conjugate }\\ \text{ acid}\end{array}\right)\left(\begin{array}{l}\text{conjugate }\\ \text{ base}\end{array}\right)\end{array}$

The base will take up the proton from acid and form its conjugate acid and simultaneously acid will form its conjugate base. The equilibrium will be forward or backward can be determined by using the dissociation constants $\left(K_{\text{a}}\text{and }{K}_{\text{b}}\right)$ for reactants as well as of the products. The more the value of $K_{\text{a}}$ for an acid, stronger will be the acid and it will undergo faster ionization. Similarly, higher the value of $K_{\text{b}}$ for a base, stronger will be the base and it will undergo faster ionization.

Answer to Problem 40PS

The equilibrium for the given reaction will lie in left direction predominantly.

  $\begin{array}{l}\text{HCN}\left(\text{aq}\right)+{\text{ SO}}_4{}^{2-}\left(\text{aq}\right)\rightleftharpoons {\text{ HSO}}_4{}^{-}\left(\text{aq}\right)+{\text{ CN}}^{-}\left(\text{aq}\right)\\ \left(\text{acid}\right)\left(\text{base}\right)\left(\begin{array}{l}\text{conjugate }\\ \text{ acid}\end{array}\right)\left(\begin{array}{l}\text{conjugate }\\ \text{ base}\end{array}\right)\end{array}$

Explanation of Solution

The equilibrium for the given reaction will move in left side is explained below.

  $\begin{array}{l}\text{HCN}\left(\text{aq}\right)+{\text{ SO}}_4{}^{2-}\left(\text{aq}\right)\rightleftharpoons {\text{ HSO}}_4{}^{-}\left(\text{aq}\right)+{\text{ CN}}^{-}\left(\text{aq}\right)\\ \left(\text{acid}\right)\left(\text{base}\right)\left(\begin{array}{l}\text{conjugate }\\ \text{ acid}\end{array}\right)\left(\begin{array}{l}\text{conjugate }\\ \text{ base}\end{array}\right)\end{array}$

Here, the conjugate acid ${\text{ HSO}}_4{}^{-}\left(K_{\text{a}}=1.2\times {10}^{-2}\right)$ is stronger acid as compared to $\text{HCN}$$\left(K_{\text{a}}=4.0\times {10}^{-10}\right)$, hence ${\text{ HSO}}_4{}^{-}$ will ionize faster as compared to $\text{HCN}$ due to which the equilibrium will shift in left direction. Also ${\text{SO}}_4{}^{2-}$ conjugate base is stronger base than ${\text{CN}}^{-}$ and it ${\text{SO}}_4{}^{2-}$ will also ionize faster as compared to ${\text{CN}}^{-}$. So it can be stated that the given reaction is reactant favoured.

Interpretation Introduction

Interpretation: The direction of the equilibrium for the given reaction is to be determined.

Concept introduction: An acid-base reaction reaction is represented as written below.

  $\begin{array}{l}\text{HA}\left(\text{aq}\right)+\text{B}\left(\text{aq}\right)\rightleftharpoons {\text{ BH}}^{\text{+}}\left(\text{aq}\right)+{\text{ A}}^{-}\left(\text{aq}\right)\\ \left(\text{acid}\right)\left(\text{base}\right)\left(\begin{array}{l}\text{conjugate }\\ \text{ acid}\end{array}\right)\left(\begin{array}{l}\text{conjugate }\\ \text{ base}\end{array}\right)\end{array}$

The base will take up the proton from acid and form its conjugate acid and simultaneously acid will form its conjugate base. The equilibrium will be forward or backward can be determined by using the dissociation constants $\left(K_{\text{a}}\text{and }{K}_{\text{b}}\right)$ for reactants as well as of the products. The more the value of $K_{\text{a}}$ for an acid, stronger will be the acid and it will undergo faster ionization. Similarly, higher the value of $K_{\text{b}}$ for a base, stronger will be the base and it will undergo faster ionization.

Answer to Problem 40PS

The equilibrium for the given reaction will lie in left side predominantly.

  $\begin{array}{l}{\text{CH}}_3\text{COOH}\left(\text{aq}\right)+{\text{ SO}}_4{}^{2-}\left(\text{aq}\right)\rightleftharpoons {\text{ HSO}}_4{}^{-}\left(\text{aq}\right)+{\text{ CH}}_3{\text{COO}}^{-}\left(\text{aq}\right)\\ \left(\text{acid}\right)\left(\text{base}\right)\left(\begin{array}{l}\text{conjugate }\\ \text{ acid}\end{array}\right)\left(\begin{array}{l}\text{conjugate }\\ \text{ base}\end{array}\right)\end{array}$

Explanation of Solution

The equilibrium for the given reaction will move in left side is explained below.

  $\begin{array}{l}{\text{CH}}_3\text{COOH}\left(\text{aq}\right)+{\text{ SO}}_4{}^{2-}\left(\text{aq}\right)\rightleftharpoons {\text{ HSO}}_4{}^{-}\left(\text{aq}\right)+{\text{ CH}}_3{\text{COO}}^{-}\left(\text{aq}\right)\\ \left(\text{acid}\right)\left(\text{base}\right)\left(\begin{array}{l}\text{conjugate }\\ \text{ acid}\end{array}\right)\left(\begin{array}{l}\text{conjugate }\\ \text{ base}\end{array}\right)\end{array}$

Here, the conjugate acid ${\text{ HSO}}_4{}^{-}\left(K_{\text{a}}=1.2\times {10}^{-2}\right)$ is stronger as compared to acid ${\text{CH}}_{\text{3}}\text{COOH}\left(K_{\text{a}}=1.8\times {10}^{-5}\right)$, hence ${\text{HSO}}_4{}^{-}$ will ionize faster as compared to ${\text{CH}}_3\text{COOH}$, due to which the equilibrium will shift in left direction. Also the base ${\text{ SO}}_4{}^{2-}$ ion is stronger base than conjugate base ${\text{CH}}_{\text{3}}{\text{COO}}^{-}$ and hence ${\text{SO}}_4{}^{2-}$ will also ionize faster as compared to ${\text{CH}}_{\text{3}}{\text{COO}}^{-}$. So it can be stated that the given reaction is reactant favoured.