Chapter 3, Problem 3.67AP

(a)

Interpretation Introduction

Interpretation:

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

Explanation of Solution

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

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

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

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

(b)

Interpretation Introduction

Interpretation:

The formula of the ionic compound formed when ${\text{Ba}}^{\text{2+}}$ reacts with ${\text{PO}}_{\text{4}}{}^{\text{3-}}$ has to be given.

Explanation of Solution

When barium cation $\left({\text{Ba}}^{\text{2+}}\right)$ reacts with phosphate ion $\left({\text{PO}}_{\text{4}}{}^{\text{3-}}\right)\text{,}$ the compound formed is barium phosphate $\left({\text{Ba}}_{\text{3}}{\left({\text{PO}}_{\text{4}}\right)}_{\text{2}}\right)\text{.}$

  ${\text{Ba}}^{\text{2+}}{\text{+PO}}_{\text{4}}{}^{\text{3-}}\stackrel{}{\to }{\text{Ba}}_{\text{3}}{\left({\text{PO}}_{\text{4}}\right)}_{\text{2}}$

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

The formula of the ionic compound formed when ${\text{Ba}}^{\text{2+}}$ reacts with ${\text{PO}}_{\text{4}}{}^{\text{3-}}$ is ${\text{Ba}}_{\text{3}}{\left({\text{PO}}_{\text{4}}\right)}_{\text{2}}\text{.}$

(c)

Interpretation Introduction

Interpretation:

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

Explanation of Solution

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

  ${\text{Ba}}^{\text{2+}}{\text{+HPO}}_{\text{4}}{}^{\text{2-}}\stackrel{}{\to }{\text{BaHPO}}_{\text{4}}$

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 ${\text{Ba}}^{\text{2+}}$ reacts with ${\text{HPO}}_{\text{4}}{}^{\text{2-}}$ is ${\text{BaHPO}}_{\text{4}}\text{.}$

(d)

Interpretation Introduction

Interpretation:

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

Explanation of Solution

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

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

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

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