Chapter 18.3, Problem 18.4BFP

(a)

Interpretation Introduction

Interpretation:

The conjugate acid formula of ${\text{HSO}}_{\text{3}}^{\text{-}}$ has to be identified.

Concept introduction:

Conjugate acid-Base pairs:

When an acid is dissolved in water, the acid $\text{(HA)}$ donates a proton to water to form a new acid (conjugate acid) and a new base (conjugate base).

$\text{RCOOH}$ has one more ${\text{H}}^{\text{+}}$ than ${\text{RCOO}}^{\text{– }}$, similarly ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ has one more ${\text{H}}^{\text{+}}$ than ${\text{H}}_{\text{2}}\text{O}$ therefore, $\text{RCOOH}$ and ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ are acids, ${\text{RCOO}}^{\text{– }}$ and ${\text{H}}_{\text{2}}\text{O}$ bases.

The conjugate acid/base pairs are ${\text{RCOOH / RCOO}}^{\text{–}}{\text{ and H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{/ H}}_{\text{2}}\text{O}$.

(b)

Interpretation Introduction

Interpretation:

The conjugate base formula of ${\text{C}}_{\text{5}}{\text{H}}_{\text{5}}{\text{NH}}^{\text{+}}$ has to be identified.

Concept introduction:

Conjugate acid-Base pairs:

When an acid is dissolved in water, the acid $\text{(HA)}$ donates a proton to water to form a new acid (conjugate acid) and a new base (conjugate base).

$\text{RCOOH}$ has one more ${\text{H}}^{\text{+}}$ than ${\text{RCOO}}^{\text{– }}$, similarly ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ has one more ${\text{H}}^{\text{+}}$ than ${\text{H}}_{\text{2}}\text{O}$ therefore, $\text{RCOOH}$ and ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ are acids, ${\text{RCOO}}^{\text{– }}$ and ${\text{H}}_{\text{2}}\text{O}$ bases.

The conjugate acid/base pairs are ${\text{RCOOH / RCOO}}^{\text{–}}{\text{ and H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{/ H}}_{\text{2}}\text{O}$.

(c)

Interpretation Introduction

Interpretation:

The conjugate acid formula of ${\text{CO}}_{\text{3}}^{\text{2-}}$ has to be identified.

Concept introduction:

Conjugate acid-Base pairs:

When an acid is dissolved in water, the acid $\text{(HA)}$ donates a proton to water to form a new acid (conjugate acid) and a new base (conjugate base).

$\text{RCOOH}$ has one more ${\text{H}}^{\text{+}}$ than ${\text{RCOO}}^{\text{– }}$, similarly ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ has one more ${\text{H}}^{\text{+}}$ than ${\text{H}}_{\text{2}}\text{O}$ therefore, $\text{RCOOH}$ and ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ are acids, ${\text{RCOO}}^{\text{– }}$ and ${\text{H}}_{\text{2}}\text{O}$ bases.

The conjugate acid/base pairs are ${\text{RCOOH / RCOO}}^{\text{–}}{\text{ and H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{/ H}}_{\text{2}}\text{O}$.

(d)

Interpretation Introduction

Interpretation:

The conjugate base formula of $\text{HCN}$ has to be identified.

Concept introduction:

Conjugate acid-Base pairs:

When an acid is dissolved in water, the acid $\text{(HA)}$ donates a proton to water to form a new acid (conjugate acid) and a new base (conjugate base).

$\text{RCOOH}$ has one more ${\text{H}}^{\text{+}}$ than ${\text{RCOO}}^{\text{– }}$, similarly ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ has one more ${\text{H}}^{\text{+}}$ than ${\text{H}}_{\text{2}}\text{O}$ therefore, $\text{RCOOH}$ and ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ are acids, ${\text{RCOO}}^{\text{– }}$ and ${\text{H}}_{\text{2}}\text{O}$ bases.

The conjugate acid/base pairs are ${\text{RCOOH / RCOO}}^{\text{–}}{\text{ and H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{/ H}}_{\text{2}}\text{O}$.