Chapter 17, Problem 16CR

Interpretation Introduction

(a)

Interpretation:

The equations demonstrating the behavior of chosen species as an acid or base in water are to be stated.

Concept Introduction:

Bronsted-Lowry base are those species which accept a proton. They are also called proton acceptor. Base accepts a proton and forms conjugate acid. Bronsted-Lowry acids are those species which donate a proton. They are also called proton donor. Acid loses a proton and forms conjugate base.

Answer to Problem 16CR

The equation showing the basic behavior of ${\text{NO}}_3^{-}$ is represented as,

${\text{NO}}_3^{-}\left(\text{a}\text{q}\right)+{\text{H}}_2\text{O}\left(\text{l}\right)\to {\text{HNO}}_3\left(\text{a}\text{q}\right)+{\text{OH}}^{-}\left(\text{a}\text{q}\right)$

The equation showing the basic behavior of ${\text{HSO}}_4^{-}$ is represented as,

${\text{HSO}}_4^{-}\left(\text{a}\text{q}\right)+{\text{H}}_2\text{O}\left(\text{l}\right)\to {\text{H}}_2{\text{SO}}_4\left(\text{a}\text{q}\right)+{\text{OH}}^{-}\left(\text{a}\text{q}\right)$

The equation showing the basic behavior of ${\text{ClO}}_4^{-}$ is represented as,

${\text{ClO}}_4^{-}\left(\text{a}\text{q}\right)+{\text{H}}_2\text{O}\left(\text{l}\right)\to {\text{HClO}}_4\left(\text{a}\text{q}\right)+{\text{OH}}^{-}\left(\text{a}\text{q}\right)$

The equation showing the basic behavior of ${\text{NH}}_3$ is represented as,

${\text{NH}}_3\left(\text{a}\text{q}\right)+{\text{H}}_2\text{O}\left(\text{l}\right)\to {\text{NH}}_4^{+}\left(\text{a}\text{q}\right)+{\text{OH}}^{-}\left(\text{a}\text{q}\right)$

The equation showing the basic behavior of ${\text{HCO}}_3^{-}$ is represented as,

${\text{HCO}}_3^{-}\left(\text{a}\text{q}\right)+{\text{H}}_2\text{O}\left(\text{l}\right)\to {\text{H}}_2{\text{CO}}_3\left(\text{a}\text{q}\right)+{\text{OH}}^{-}\left(\text{a}\text{q}\right)$

The equation showing the acidic behavior of $\text{HCl}$ is represented as,

$\text{HCl}\left(\text{a}\text{q}\right)+{\text{H}}_2\text{O}\left(\text{l}\right)\to {\text{Cl}}^{-}\left(\text{a}\text{q}\right)+{\text{H}}_3{\text{O}}^{+}\left(\text{a}\text{q}\right)$

The equation showing the acidic behavior of ${\text{H}}_2{\text{SO}}_4$ is represented as,

${\text{H}}_2{\text{SO}}_4\left(\text{a}\text{q}\right)+{\text{H}}_2\text{O}\left(\text{l}\right)\to {\text{HSO}}_4^{-}\left(\text{a}\text{q}\right)+{\text{H}}_3{\text{O}}^{+}\left(\text{a}\text{q}\right)$

The equation showing the acidic behavior of ${\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}$ is represented as,

${\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}\left(\text{l}\right)+{\text{H}}_2\text{O}\left(\text{l}\right)\to {\text{H}}_2{\text{PO}}_{\text{4}}{}^{-}\left(\text{a}\text{q}\right)+{\text{H}}_3{\text{O}}^{+}\left(\text{a}\text{q}\right)$

The equation showing the acidic behavior of ${\text{NH}}_4^{+}$ is represented as,

${\text{NH}}_4^{+}\left(\text{a}\text{q}\right)+{\text{H}}_2\text{O}\left(\text{l}\right)\to {\text{NH}}_3\left(\text{a}\text{q}\right)+{\text{H}}_3{\text{O}}^{+}\left(\text{a}\text{q}\right)$

The equation showing the acidic behavior of ${\text{H}}_2{\text{CO}}_3$ is represented as,

${\text{H}}_2{\text{CO}}_3\left(\text{a}\text{q}\right)+{\text{H}}_2\text{O}\left(\text{l}\right)\to {\text{HCO}}_3^{-}\left(\text{a}\text{q}\right)+{\text{H}}_3{\text{O}}^{+}\left(\text{a}\text{q}\right)$.

Explanation of Solution

The base ${\text{NO}}_3^{-}$ will accept ${\text{H}}^{+}$ ion when dissolve in water. The chemical equation is represented as,

${\text{NO}}_3^{-}\left(\text{a}\text{q}\right)+{\text{H}}_2\text{O}\left(\text{l}\right)\to {\text{HNO}}_3\left(\text{a}\text{q}\right)+{\text{OH}}^{-}\left(\text{a}\text{q}\right)$

The base ${\text{HSO}}_4^{-}$ will accept ${\text{H}}^{+}$ ion when dissolve in water. The chemical equation is represented as,

${\text{HSO}}_4^{-}\left(\text{a}\text{q}\right)+{\text{H}}_2\text{O}\left(\text{l}\right)\to {\text{H}}_2{\text{SO}}_4\left(\text{a}\text{q}\right)+{\text{OH}}^{-}\left(\text{a}\text{q}\right)$

The base ${\text{ClO}}_4^{-}$ will accept ${\text{H}}^{+}$ ion when dissolve in water. The chemical equation is represented as,

${\text{ClO}}_4^{-}\left(\text{a}\text{q}\right)+{\text{H}}_2\text{O}\left(\text{l}\right)\to {\text{HClO}}_4\left(\text{a}\text{q}\right)+{\text{OH}}^{-}\left(\text{a}\text{q}\right)$

The base ${\text{NH}}_3$ will accept ${\text{H}}^{+}$ ion when dissolve in water. The chemical equation is represented as,

${\text{NH}}_3\left(\text{a}\text{q}\right)+{\text{H}}_2\text{O}\left(\text{l}\right)\to {\text{NH}}_4^{+}\left(\text{a}\text{q}\right)+{\text{OH}}^{-}\left(\text{a}\text{q}\right)$

The base ${\text{HCO}}_3^{-}$ will accept ${\text{H}}^{+}$ ion when dissolve in water. The chemical equation is represented as,

${\text{HCO}}_3^{-}\left(\text{a}\text{q}\right)+{\text{H}}_2\text{O}\left(\text{l}\right)\to {\text{H}}_2{\text{CO}}_3\left(\text{a}\text{q}\right)+{\text{OH}}^{-}\left(\text{a}\text{q}\right)$

The acid $\text{HCl}$ will loses ${\text{H}}^{+}$ ion when dissolve in water. The chemical equation is represented as,

$\text{HCl}\left(\text{a}\text{q}\right)+{\text{H}}_3\text{O}\left(\text{l}\right)\to {\text{Cl}}^{-}\left(\text{a}\text{q}\right)+{\text{H}}_3{\text{O}}^{+}\left(\text{a}\text{q}\right)$

The acid ${\text{H}}_2{\text{SO}}_4$ will loses ${\text{H}}^{+}$ ion when dissolve in water. The chemical equation is represented as,

${\text{H}}_2{\text{SO}}_4\left(\text{a}\text{q}\right)+{\text{H}}_2\text{O}\left(\text{l}\right)\to {\text{HSO}}_4^{-}\left(\text{a}\text{q}\right)+{\text{H}}_3{\text{O}}^{+}\left(\text{a}\text{q}\right)$

The acid ${\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}$ will loses ${\text{H}}^{+}$ ion when dissolve in water. The chemical equation is represented as,

${\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}\left(\text{l}\right)+{\text{H}}_2\text{O}\left(\text{l}\right)\to {\text{H}}_2{\text{PO}}_{\text{4}}{}^{-}\left(\text{a}\text{q}\right)+{\text{H}}_3{\text{O}}^{+}\left(\text{a}\text{q}\right)$

The acid ${\text{NH}}_4^{+}$ will loses ${\text{H}}^{+}$ ion when dissolve in water. The chemical equation is represented as,

${\text{NH}}_4^{+}\left(\text{a}\text{q}\right)+{\text{H}}_2\text{O}\left(\text{l}\right)\to {\text{NH}}_3\left(\text{a}\text{q}\right)+{\text{H}}_3{\text{O}}^{+}\left(\text{a}\text{q}\right)$

The acid ${\text{H}}_2{\text{CO}}_3$ will loses ${\text{H}}^{+}$ ion when dissolve in water. The chemical equation is represented as,

${\text{H}}_2{\text{CO}}_3\left(\text{a}\text{q}\right)+{\text{H}}_2\text{O}\left(\text{l}\right)\to {\text{HCO}}_3^{-}\left(\text{a}\text{q}\right)+{\text{H}}_3{\text{O}}^{+}\left(\text{a}\text{q}\right)$.

Interpretation Introduction

(b)

Interpretation:

The formula of conjugate base or acid for the chosen species is to be stated.

Concept Introduction:

Bronsted-Lowry base are those species which accept a proton. They are also called proton acceptor. Base accepts a proton and forms conjugate acid. Bronsted-Lowry acids are those species which donate a proton. They are also called proton donor. Acid loses a proton and forms conjugate base.

Answer to Problem 16CR

The conjugate acid of ${\text{NO}}_3^{-}$ is ${\text{HNO}}_3$. The conjugate acid of ${\text{HSO}}_4^{-}$ is ${\text{H}}_2{\text{SO}}_4$. The conjugate acid of ${\text{ClO}}_4^{-}$ is ${\text{HClO}}_4$. The conjugate acid of ${\text{NH}}_3$ is ${\text{NH}}_4^{+}$. The conjugate acid of ${\text{HCO}}_3^{-}$ is ${\text{H}}_2{\text{CO}}_3$. The conjugate base of $\text{HCl}$ is ${\text{Cl}}^{-}$. The conjugate base of ${\text{H}}_2{\text{SO}}_4$ is ${\text{HSO}}_4^{-}$. The conjugate base of ${\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}$ is ${\text{H}}_2{\text{PO}}_{\text{4}}{}^{-}$. The conjugate base of ${\text{NH}}_4^{+}$ is ${\text{NH}}_3$. The conjugate base of ${\text{H}}_2{\text{CO}}_3$ is ${\text{HCO}}_3^{-}$.

Explanation of Solution

The conjugate acid of ${\text{NO}}_3^{-}$ is ${\text{HNO}}_3$. ${\text{NO}}_3^{-}$ accepts ${\text{H}}^{+}$ ion to form ${\text{HNO}}_3$. The chemical equation is represented as,

${\text{NO}}_3^{-}\left(\text{a}\text{q}\right)+{\text{H}}^{+}\left(\text{a}\text{q}\right)\to {\text{HNO}}_3\left(\text{a}\text{q}\right)$

The conjugate acid of ${\text{HSO}}_4^{-}$ is ${\text{H}}_2{\text{SO}}_4$. ${\text{HSO}}_4^{-}$ accepts ${\text{H}}^{+}$ ion to form ${\text{H}}_2{\text{SO}}_4$ The chemical equation is represented as,

${\text{HSO}}_4^{-}\left(\text{a}\text{q}\right)+{\text{H}}^{+}\left(\text{a}\text{q}\right)\to {\text{H}}_2{\text{SO}}_4\left(\text{a}\text{q}\right)$

The conjugate acid of ${\text{ClO}}_4^{-}$ is ${\text{HClO}}_4$. ${\text{ClO}}_4^{-}$ accepts ${\text{H}}^{+}$ ion to form ${\text{HClO}}_4$. The chemical equation is represented as,

${\text{ClO}}_4^{-}\left(\text{a}\text{q}\right)+{\text{H}}^{+}\left(\text{a}\text{q}\right)\to {\text{HClO}}_4\left(\text{a}\text{q}\right)$

The conjugate acid of ${\text{NH}}_3$ is ${\text{NH}}_4^{+}$. ${\text{NH}}_3$ accepts ${\text{H}}^{+}$ ion to form ${\text{NH}}_4^{+}$. The chemical equation is represented as,

${\text{NH}}_3\left(\text{a}\text{q}\right)+{\text{H}}^{+}\left(\text{a}\text{q}\right)\to {\text{NH}}_4^{+}\left(\text{a}\text{q}\right)$

The conjugate acid of ${\text{HCO}}_3^{-}$ is ${\text{H}}_2{\text{CO}}_3$. ${\text{HCO}}_3^{-}$ accepts ${\text{H}}^{+}$ ion to form ${\text{H}}_2{\text{CO}}_3$ The chemical equation is represented as,

${\text{HCO}}_3^{-}\left(\text{a}\text{q}\right)+{\text{H}}^{+}\left(\text{a}\text{q}\right)\to {\text{H}}_2{\text{CO}}_3\left(\text{a}\text{q}\right)$

The conjugate base of $\text{HCl}$ is ${\text{Cl}}^{-}$. $\text{HCl}$ loses ${\text{H}}^{+}$ ion to form ${\text{Cl}}^{-}$. The chemical equation is represented as,

$\text{HCl}\left(\text{a}\text{q}\right)\to {\text{Cl}}^{-}\left(\text{a}\text{q}\right)+{\text{H}}^{+}\left(\text{a}\text{q}\right)$

The conjugate base of ${\text{H}}_2{\text{SO}}_4$ is ${\text{HSO}}_4^{-}$. ${\text{H}}_2{\text{SO}}_4$ loses ${\text{H}}^{+}$ ion to form ${\text{HSO}}_4^{-}$. The chemical equation is represented as,

${\text{H}}_2{\text{SO}}_4\left(\text{a}\text{q}\right)\to {\text{HSO}}_4^{-}\left(\text{a}\text{q}\right)+{\text{H}}^{+}\left(\text{a}\text{q}\right)$

The conjugate base of ${\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}$ is ${\text{H}}_2{\text{PO}}_{\text{4}}{}^{-}$. ${\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}$ loses ${\text{H}}^{+}$ ion to form ${\text{H}}_2{\text{PO}}_{\text{4}}{}^{-}$. The chemical equation is represented as,

${\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}\left(\text{a}\text{q}\right)\to {\text{H}}_2{\text{PO}}_{\text{4}}{}^{-}\left(\text{a}\text{q}\right)+{\text{H}}^{+}\left(\text{a}\text{q}\right)$

The conjugate base of ${\text{NH}}_4^{+}$ is ${\text{NH}}_3$. ${\text{NH}}_4^{+}$ loses ${\text{H}}^{+}$ ion to form ${\text{NH}}_3$. The chemical equation is represented as,

${\text{NH}}_4^{+}\left(\text{a}\text{q}\right)\to {\text{NH}}_3\left(\text{a}\text{q}\right)+{\text{H}}^{+}\left(\text{a}\text{q}\right)$

The conjugate base of ${\text{H}}_2{\text{CO}}_3$ is ${\text{HCO}}_3^{-}$. ${\text{H}}_2{\text{CO}}_3$ loses ${\text{H}}^{+}$ ion to form ${\text{HCO}}_3^{-}$. The chemical equation is represented as,

${\text{H}}_2{\text{CO}}_3\left(\text{a}\text{q}\right)\to {\text{HCO}}_3^{-}\left(\text{a}\text{q}\right)+{\text{H}}^{+}\left(\text{a}\text{q}\right)$.