Chapter 15, Problem 15.85QP

Write a reaction for each of the following in which the species acts as a Brønsted acid. The equilibrium should favor the product side.

1. a H₂O₂
2. b HCO₃⁻
3. c NH₄⁻
4. d H₂PO₄⁻

Interpretation Introduction

Interpretation:

A reaction for each of the given species acts as a Bronsted acid with equilibrium should favor the product side has to be written.

Concept introduction:

Acid:

According to Bronsted-Lowry theory, a species which donates a proton in a proton transfer-reaction is said to be an acid.

Base:

According to Bronsted-Lowry theory, a species which accepts a proton in a proton transfer-reaction is said to be base.

Answer to Problem 15.85QP

(a)

The reaction for each of the given species acts as Bronsted acid is

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

(b)

The reaction for each of the given species acts as Bronsted acid is

${\text{HCO}}_{\text{3}}{{}^{\text{-}}}_{\text{(aq)}}{\text{ + OH}}^{\text{-}}{}_{\text{(aq)}}\stackrel{}{\to }{\text{ CO}}_{\text{3}}{{}^{\text{2-}}}_{\text{(aq)}}{\text{ + H}}_{\text{2}}{\text{O}}_{\text{(l)}}$

(c)

The reaction for each of the given species acts as Bronsted acid is

${\text{NH}}_{\text{4}}{{}^{\text{+}}}_{\text{(aq)}}{\text{ + CN}}^{\text{-}}{}_{\text{(aq)}}\stackrel{}{\to }{\text{ NH}}_{\text{3(aq) }}{\text{+ HCN}}_{\text{(aq)}}$

(d)

The reaction for each of the given species acts as Bronsted acid is

${\text{H}}_{\text{2}}{\text{PO}}_{\text{4}}{{}^{\text{-}}}_{\text{(aq) }}{\text{+ OH}}^{\text{-}}{}_{\text{(aq)}}\stackrel{}{\to }{\text{ HPO}}_{\text{4}}{{}^{\text{2-}}}_{\text{(aq) }}{\text{+ H}}_{\text{2}}{\text{O}}_{\text{(l)}}$

Explanation of Solution

(a)

According to Bronsted-Lowry theory, the equilibrium reaction for the reaction between water and ${\text{ S}}^{\text{2-}}$ ion is represented as follows.

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

(b)

According to Bronsted-Lowry theory, the equilibrium reaction for the reaction between  ${\text{HCO}}_{\text{3}}{}^{\text{-}}$ and ${\text{OH}}^{\text{-}}$ ion is represented as follows.

${\text{HCO}}_{\text{3}}{{}^{\text{-}}}_{\text{(aq)}}{\text{ + OH}}^{\text{-}}{}_{\text{(aq)}}\stackrel{}{\to }{\text{ CO}}_{\text{3}}{{}^{\text{2-}}}_{\text{(aq)}}{\text{ + H}}_{\text{2}}{\text{O}}_{\text{(l)}}$

(c)

According to Bronsted-Lowry theory, the equilibrium reaction for the reaction between ${\text{NH}}_{\text{4}}{}^{\text{+}}$ and ${\text{CN}}^{\text{-}}$ ion is represented as follows.

${\text{NH}}_{\text{4}}{{}^{\text{+}}}_{\text{(aq)}}{\text{ + CN}}^{\text{-}}{}_{\text{(aq)}}\stackrel{}{\to }{\text{ NH}}_{\text{3(aq) }}{\text{+ HCN}}_{\text{(aq)}}$

(d)

According to Bronsted-Lowry theory, the equilibrium reaction for the reaction between ${\text{H}}_{\text{2}}{\text{PO}}_{\text{4}}{}^{\text{-}}$ and ${\text{OH}}^{\text{-}}$ ion is represented as follows.

${\text{H}}_{\text{2}}{\text{PO}}_{\text{4}}{{}^{\text{-}}}_{\text{(aq) }}{\text{+ OH}}^{\text{-}}{}_{\text{(aq)}}\stackrel{}{\to }{\text{ HPO}}_{\text{4}}{{}^{\text{2-}}}_{\text{(aq) }}{\text{+ H}}_{\text{2}}{\text{O}}_{\text{(l)}}$

Conclusion

A reaction for each of the given species acts as a Bronsted acid with equilibrium should favor the product side was written.