Chapter 3, Problem 3.65AP

(a)

Interpretation Introduction

Interpretation:

The formula of the ionic compound formed when ${\text{K}}^{\text{+}}$ reacts with ${\text{HSO}}_{\text{4}}{}^{\text{-}}$ has to be given.

Answer to Problem 3.65AP

The formula of the ionic compound formed when ${\text{K}}^{\text{+}}$ reacts with ${\text{HSO}}_{\text{4}}{}^{\text{-}}$ is ${\text{KHSO}}_{\text{4}}\text{.}$

Explanation of Solution

When potassium cation $\left({\text{K}}^{\text{+}}\right)$ reacts with bisulfate ion $\left({\text{HSO}}_{\text{4}}{}^{\text{-}}\right)\text{,}$ the compound formed is potassium bisulfate $\left({\text{KHSO}}_{\text{4}}\right)\text{.}$

  ${\text{K}}^{\text{+}}{\text{+HSO}}_{\text{4}}{}^{\text{-}}\stackrel{}{\to }{\text{KHSO}}_{\text{4}}$

The charges are equal in magnitude, and one of each ion is needed to balance the charge.

The formula of the ionic compound formed when ${\text{K}}^{\text{+}}$ reacts with ${\text{HSO}}_{\text{4}}{}^{\text{-}}$ is ${\text{KHSO}}_{\text{4}}\text{.}$

(b)

Interpretation Introduction

Interpretation:

The formula of the ionic compound formed when ${\text{Ba}}^{\text{2+}}$ reacts with ${\text{HSO}}_{\text{4}}{}^{\text{-}}$ has to be given.

Answer to Problem 3.65AP

The formula of the ionic compound formed when ${\text{Ba}}^{\text{2+}}$ reacts with ${\text{HSO}}_{\text{4}}{}^{\text{-}}$ is $\text{Ba}{\left({\text{HSO}}_{\text{4}}\right)}_{\text{2}}\text{.}$

Explanation of Solution

When barium cation $\left({\text{Ba}}^{\text{2+}}\right)$ reacts with bisulfate ion $\left({\text{HSO}}_{\text{4}}{}^{\text{-}}\right)\text{,}$ the compound formed is barium bisulfate $\left(\text{Ba}{\left({\text{HSO}}_{\text{4}}\right)}_{\text{2}}\right)\text{.}$

  ${\text{Ba}}^{\text{2+}}{\text{+HSO}}_{\text{4}}{}^{\text{-}}\stackrel{}{\to }\text{Ba}{\left({\text{HSO}}_{\text{4}}\right)}_{\text{2}}$

The charges are not equal in magnitude, and two bisulfate ions are required to balance the charge.

The formula of the compound formed when ${\text{Ba}}^{\text{2+}}$ reacts with ${\text{HSO}}_{\text{4}}{}^{\text{-}}$ is $\text{Ba}{\left({\text{HSO}}_{\text{4}}\right)}_{\text{2}}\text{.}$

(c)

Interpretation Introduction

Interpretation:

The formula of the ionic compound formed when ${\text{Al}}^{\text{3+}}$ reacts with ${\text{HSO}}_{\text{4}}{}^{\text{-}}$ has to be given.

Answer to Problem 3.65AP

The formula of the ionic compound formed when ${\text{Al}}^{\text{3+}}$ reacts with ${\text{HSO}}_{\text{4}}{}^{\text{-}}$ is $\text{Al}{\left({\text{HSO}}_{\text{4}}\right)}_{\text{3}}\text{.}$

Explanation of Solution

When aluminum cation $\left({\text{Al}}^{\text{3+}}\right)$ reacts with bisulfate ion $\left({\text{HSO}}_{\text{4}}{}^{\text{-}}\right)\text{,}$ the compound formed is aluminum bisulfate $\left(\text{Al}{\left({\text{HSO}}_{\text{4}}\right)}_{\text{3}}\right)\text{.}$

  ${\text{Al}}^{\text{3+}}{\text{+HSO}}_{\text{4}}{}^{\text{-}}\stackrel{}{\to }\text{Al}{\left({\text{HSO}}_{\text{4}}\right)}_{\text{3}}$

The charges are not equal in magnitude, and three bisulfate ions are required to balance the charge.

The formula of the ionic compound formed when ${\text{Al}}^{\text{3+}}$ reacts with ${\text{HSO}}_{\text{4}}{}^{\text{-}}$ is $\text{Al}{\left({\text{HSO}}_{\text{4}}\right)}_{\text{3}}\text{.}$

(d)

Interpretation Introduction

Interpretation:

The formula of the ionic compound formed when ${\text{Zn}}^{\text{2+}}$ reacts with ${\text{HSO}}_{\text{4}}{}^{\text{-}}$ has to be given.

Answer to Problem 3.65AP

The formula of the ionic compound formed when ${\text{Zn}}^{\text{2+}}$ reacts with ${\text{HSO}}_{\text{4}}{}^{\text{-}}$ is $\text{Zn}{\left({\text{HSO}}_{\text{4}}\right)}_{\text{2}}.$

Explanation of Solution

When zinc cation $\left({\text{Zn}}^{\text{2+}}\right)$ reacts with bisulfate ion $\left({\text{HSO}}_{\text{4}}{}^{\text{-}}\right)\text{,}$ the compound formed is zinc bisulfate $\text{Zn}{\left({\text{HSO}}_{\text{4}}\right)}_{\text{2}}\text{.}$

  ${\text{Zn}}^{\text{2+}}{\text{+HSO}}_{\text{4}}{}^{\text{-}}\stackrel{}{\to }\text{Zn}{\left({\text{HSO}}_{\text{4}}\right)}_{\text{2}}$

The charges are not equal in magnitude, and two bisulfate ions are required to balance the charge.

The formula of the ionic compound formed when ${\text{Zn}}^{\text{2+}}$ reacts with ${\text{HSO}}_{\text{4}}{}^{\text{-}}$ is $\text{Zn}{\left({\text{HSO}}_{\text{4}}\right)}_{\text{2}}.$