Chapter 14, Problem 17QRT

(a)

Interpretation Introduction

Interpretation:

In the reactants, acid and base has to be identified and also in the product, conjugate acid and conjugate base has to be identified in the given reaction.

    ${\text{HS}}^{-}(aq)+{\text{H}}_{\text{2}}\text{O}(l)\to {\text{H}}_{\text{2}}\text{S}(aq)+{\text{OH}}^{-}(aq)$

Concept Introduction:

Brønsted-Lowry acid is the one which can donate ${\text{H}}^{\text{+}}$ ion to other substances.  It can also be said as proton donors.  Brønsted-Lowry base is the one which can accept ${\text{H}}^{\text{+}}$ ion from other substances.  It can also be said as proton acceptors.  In order to be a Brønsted-Lowry base, an unshared pair of electrons must be present on molecule or the ion.

When an acid donates a proton, the species formed from acid is known as conjugate base.  When a base accepts a proton from acid, the species formed from the base is known as conjugate acid.

Explanation of Solution

Given reaction is,

    ${\text{HS}}^{-}(aq)+{\text{H}}_{\text{2}}\text{O}(l)\to {\text{H}}_{\text{2}}\text{S}(aq)+{\text{OH}}^{-}(aq)$

In the above equation, water present in the reactant side acts as an acid and donates a proton to ${\text{HS}}^{-}$.  Therefore, water is an acid and ${\text{HS}}^{-}$ is a base.  The species formed from water by donation of proton is hydroxide ion.  The species formed from ${\text{HS}}^{-}$ by accepting a proton is dihydrogen sulfide.  Therefore, hydroxide ion is a conjugate base and dihydrogen sulfide is a conjugate acid.  This can be represented as,

    $\begin{array}{c}{\text{HS}}^{-}:\text{base}\\ {\text{H}}_{\text{2}}\text{O: acid}\\ {\text{H}}_{\text{2}}\text{S: conjugate acid}\\ {\text{OH}}^{-}:\text{conjugate base}\end{array}$

(b)

Interpretation Introduction

Interpretation:

In the reactants, acid and base has to be identified and also in the product, conjugate acid and conjugate base has to be identified in the given reaction.

    ${\text{S}}^{2-}(aq)+{\text{NH}}_{\text{4}}{}^{\text{+}}(aq)\to {\text{NH}}_{\text{3}}(g)+{\text{SH}}^{-}(aq)$

Concept Introduction:

Refer part (a).

Explanation of Solution

Given reaction is,

    ${\text{S}}^{2-}(aq)+{\text{NH}}_{\text{4}}{}^{\text{+}}(aq)\to {\text{NH}}_{\text{3}}(g)+{\text{HS}}^{-}(aq)$

In the above equation, ammonium ion present in the reactant side acts as an acid and donates a proton to ${\text{S}}^{2-}$.  Therefore, ammonium ion acts as an acid and ${\text{S}}^{2-}$ acts as a base.  The species formed from ammonium ion by donation of proton is ammonia gas.  The species formed from ${\text{S}}^{2-}$ by accepting a proton is ${\text{HS}}^{-}$.  Therefore, ${\text{HS}}^{-}$ is a conjugate acid and ammonia is a conjugate base.  This can be represented as,

    $\begin{array}{c}{\text{S}}^{2-}:\text{base}\\ {\text{NH}}_{\text{4}}{}^{\text{+}}\text{: acid}\\ {\text{HS}}^{-}\text{: conjugate acid}\\ {\text{NH}}_{\text{3}}:\text{conjugate base}\end{array}$

(c)

Interpretation Introduction

Interpretation:

In the reactants, acid and base has to be identified and also in the product, conjugate acid and conjugate base has to be identified in the given reaction.

    ${\text{HCO}}_{\text{3}}{}^{-}(aq)+{\text{HSO}}_{\text{4}}{}^{-}(aq)\to {\text{H}}_{\text{2}}{\text{CO}}_{\text{3}}(aq)+{\text{SO}}_{\text{4}}{}^{2-}(aq)$

Concept Introduction:

Refer part (a).

Explanation of Solution

Given reaction is,

    ${\text{HCO}}_{\text{3}}{}^{-}(aq)+{\text{HSO}}_{\text{4}}{}^{-}(aq)\to {\text{H}}_{\text{2}}{\text{CO}}_{\text{3}}(aq)+{\text{SO}}_{\text{4}}{}^{2-}(aq)$

In the above equation, ${\text{HSO}}_{\text{4}}{}^{-}$ ion present in the reactant side acts as an acid and donates a proton to ${\text{HCO}}_{\text{3}}{}^{-}$.  Therefore, ${\text{HSO}}_{\text{4}}{}^{-}$ ion acts as an acid and ${\text{HCO}}_{\text{3}}{}^{-}$ acts as a base.  The species formed from ${\text{HSO}}_{\text{4}}{}^{-}$ ion by donation of proton is ${\text{SO}}_{\text{4}}{}^{2-}$ ion.  The species formed from ${\text{HCO}}_{\text{3}}{}^{-}$ by accepting a proton is ${\text{H}}_{\text{2}}{\text{CO}}_{\text{3}}$.  Therefore, ${\text{H}}_{\text{2}}{\text{CO}}_{\text{3}}$ is a conjugate acid and ${\text{SO}}_{\text{4}}{}^{2-}$ is a conjugate base.  This can be represented as,

    $\begin{array}{c}{\text{HCO}}_{\text{3}}{}^{-}:\text{base}\\ {\text{HSO}}_{\text{4}}{}^{-}\text{: acid}\\ {\text{H}}_{\text{2}}{\text{CO}}_{\text{3}}\text{: conjugate acid}\\ {\text{SO}}_{\text{4}}{}^{2-}:\text{conjugate base}\end{array}$

(d)

Interpretation Introduction

Interpretation:

In the reactants, acid and base has to be identified and also in the product, conjugate acid and conjugate base has to be identified in the given reaction.

    ${\text{NH}}_{\text{3}}(aq)+{\text{NH}}_2{}^{-}(aq)\to {\text{NH}}_2{}^{-}(aq)+{\text{NH}}_{\text{3}}(aq)$

Concept Introduction:

Refer part (a).

Explanation of Solution

Given reaction is,

    ${\text{NH}}_{\text{3}}(aq)+{\text{NH}}_2{}^{-}(aq)\to {\text{NH}}_2{}^{-}(aq)+{\text{NH}}_{\text{3}}(aq)$

In the above equation, ${\text{NH}}_{\text{3}}$ present in the reactant side acts as an acid and donates a proton to ${\text{NH}}_2{}^{-}$.  Therefore, ${\text{NH}}_{\text{3}}$ ion acts as an acid and ${\text{NH}}_2{}^{-}$ acts as a base.  The species formed from ${\text{NH}}_{\text{3}}$ ion by donation of proton is ${\text{NH}}_2{}^{-}$ ion.  The species formed from ${\text{NH}}_2{}^{-}$ by accepting a proton is ${\text{NH}}_{\text{3}}$.  Therefore, ${\text{NH}}_{\text{3}}$ is a conjugate acid and ${\text{NH}}_2{}^{-}$ is a conjugate base.  This can be represented as,

    $\begin{array}{c}{\text{NH}}_2{}^{-}:\text{base}\\ {\text{NH}}_{\text{3}}\text{: acid}\\ {\text{NH}}_{\text{3}}\text{: conjugate acid}\\ {\text{NH}}_2{}^{-}:\text{conjugate base}\end{array}$