Chapter 4, Problem 10QAP

Follow the directions for Question 9 for the following pairs of solutions.

(a) sodium phosphate and barium chloride

(b) zinc sulfate and potassium hydroxide

(c) ammonium sulfate and sodium chloride

(d) cobalt(III) nitrate and sodium phosphate

Interpretation Introduction

(a)

Interpretation:

The net ionic equation should be written when sodium phosphate and barium chloride are mixed.

Concept introduction:

Solubility of any compound is predicted by below solubility chart:

	${\text{NO}}_3^{-}$	${\mathrm{Cl}}^{-}$	${\text{SO}}_4^{2-}$	${\text{OH}}^{-}$	${\text{CO}}_3^{2-}$	${\text{PO}}_4^{3-}$
Group-1 cations and ${\text{NH}}_4^{+}$
Group-2 cations			${\text{BaSO}}_{\text{4}}$	$\text{Mg}{\left(\text{OH}\right)}_{\text{2}}$	-	-
Transitions metal cations ${\text{Pb}}^{2+}$, ${\text{Hg}}_2^{2+}$		$\text{AgCl}$ ${\text{PbCl}}_2$ ${\text{Hg}}_2{\text{Cl}}_2$	${\text{PbSO}}_{\text{4}}$ ${\text{Ag}}_2{\text{SO}}_{\text{4}}$	-	-	-

Blank boxes indicate no precipitate formation occurs which means soluble in dilute solution.

Boxes with hyphen will form precipitate from dilute solutions and boxes where formula is written this is a cation-anion combination that will form precipitate.

Precipitation reactions: It is a type of chemical reactions where two soluble salts react with each other and formed different products, out of which one product must be insoluble in solution which is known as precipitate.

A chemical equation which shows only the species that are participated in the reaction is said to be net ionic equation.

Answer to Problem 10QAP

Precipitation occurs

The net ionic equation is:

${\text{3Ba}}^{2+}{\text{(aq)+ 2PO}}_4{}^{3-}(\text{aq})\to {\text{Ba}}_3{{\text{(PO}}_{\text{4}})}_2(\text{s})$

Explanation of Solution

Sodium phosphate: ${\text{Na}}_3{\text{PO}}_4$

Barium chloride: ${\text{BaCl}}_2$

Reaction for the solution of sodium phosphate and barium chloride is written as:

${\text{Na}}_3{\text{PO}}_4{\text{+BaCl}}_2\to {\text{NaCl+Ba}}_3{{\text{(PO}}_{\text{4}})}_2$

Reactants:

Ions in solution: ${\text{Na}}_3{\text{PO}}_4$ : ${\text{Na}}^{\text{+}}$ and ${\text{PO}}_4{}^{3-}$

Ions in solution: ${\text{BaCl}}_2$ : ${\text{Ba}}^{2+}$ and ${\text{Cl}}^{-}$

Products:

Ions in solution: $\text{NaCl}$ : ${\text{Na}}^{+}$ and ${\text{Cl}}^{-}$

Ions in solution: ${\text{Ba}}_3{{\text{(PO}}_{\text{4}})}_2$ : ${\text{Ba}}^{\text{2+}}$ and ${\text{PO}}_4{}^{3-}$

Now, ${\text{Ba}}_3{{\text{(PO}}_{\text{4}})}_2$ is solid, not soluble in water whereas $\text{NaCl}$ is soluble in water.

So, the equation will be:

${\text{Na}}^{\text{+}}{\text{(aq)+ PO}}_4{}^{3-}(\text{aq})+{\text{Ba}}^{2+}{\text{(aq)+Cl}}^{-}(\text{aq})\to {\text{Ba}}_3{{\text{(PO}}_{\text{4}})}_2(\text{s}){\text{+Na}}^{+}{\text{(aq)+Cl}}^{-}(\text{aq})$

Now, cancelling out the ions which appear on both sides of the equation ( ${\text{Cl}}^{\text{-}}$ , ${\text{Na}}^{+}$ ), the final or net ionic equation is:

${\text{3Ba}}^{2+}{\text{(aq)+ 2PO}}_4{}^{3-}(\text{aq})\to {\text{Ba}}_3{{\text{(PO}}_{\text{4}})}_2(\text{s})$

Interpretation Introduction

(b)

Interpretation:

The net ionic equation should be written when zinc sulfate and potassium hydroxide are mixed.

Concept introduction:

Solubility of any compound is predicted by above solubility chart.

	${\text{NO}}_3^{-}$	${\mathrm{Cl}}^{-}$	${\text{SO}}_4^{2-}$	${\text{OH}}^{-}$	${\text{CO}}_3^{2-}$	${\text{PO}}_4^{3-}$
Group-1 cations and ${\text{NH}}_4^{+}$
Group-2 cations			${\text{BaSO}}_{\text{4}}$	$\text{Mg}{\left(\text{OH}\right)}_{\text{2}}$	-	-
Transitions metal cations ${\text{Pb}}^{2+}$, ${\text{Hg}}_2^{2+}$		$\text{AgCl}$ ${\text{PbCl}}_2$ ${\text{Hg}}_2{\text{Cl}}_2$	${\text{PbSO}}_{\text{4}}$ ${\text{Ag}}_2{\text{SO}}_{\text{4}}$	-	-	-

Blank boxes indicate no precipitate formation occurs which means soluble in dilute solution.

Boxes with hyphen will form precipitate from dilute solutions and boxes where formula is written this is a cation-anion combination that will form precipitate.

Precipitation reactions: It is a type of chemical reactions where two soluble salts react with each other and formed different products, out of which one product must be insoluble in solution which is known as precipitate.

A chemical equation which shows only the species that are participated in the reaction is said to be net ionic equation.

Answer to Problem 10QAP

Precipitation occurs

The net ionic equation is:

${\text{Zn}}^{\text{+2}}\text{(aq)}+{\text{2OH}}^{-}(\text{aq})\to {\text{Zn(OH)}}_2(\text{s})$

Explanation of Solution

Zinc sulfate: ${\text{ZnSO}}_4$

Potassium hydroxide: $\text{KOH}$

Reaction for the solution of zinc sulfate and potassium hydroxide is written as:

${\text{ZnSO}}_4+\text{KOH}\to {\text{K}}_{\text{2}}{\text{SO}}_4\text{+Zn(OH}{)}_2$

Reactants:

Ions in solution: ${\text{ZnSO}}_4$ : ${\text{Zn}}^{\text{+2}}$ and ${\text{SO}}_4^{2-}$

Ions in solution: $\text{KOH}$ : ${\text{K}}^{\text{+}}$ and ${\text{OH}}^{-}$

Products:

Ions in solution: ${\text{K}}_{\text{2}}{\text{SO}}_4$ : ${\text{K}}^{\text{+}}$ and ${\text{SO}}_4^{2-}$

Ions in solution: ${\text{Zn(OH)}}_{\text{2}}$ : ${\text{Zn}}^{\text{+2}}$ and ${\text{OH}}^{-}$

Now, ${\text{Zn(OH)}}_{\text{2}}$ is solid, not soluble in water whereas ${\text{K}}_{\text{2}}{\text{SO}}_4$ is soluble in water.

So, the equation will be:

${\text{Zn}}^{\text{+2}}{\text{(aq)+ SO}}_4^{2-}(\text{aq})+{\text{K}}^{\text{+}}{\text{(aq)+OH}}^{-}(\text{aq})\to {\text{Zn(OH)}}_{\text{2}}(\text{s}){\text{+K}}^{\text{+}}{\text{(aq)+SO}}_4^{2-}(\text{aq})$

Now, cancelling out the ions which appear on both sides of the equation ( ${\text{SO}}_4{}^{\text{2-}}$ , ${\text{K}}^{\text{+}}$ ), the final or net ionic equation is:

${\text{Zn}}^{\text{+2}}\text{(aq)}+{\text{2OH}}^{-}(\text{aq})\to {\text{Zn(OH)}}_2(\text{s})$

Interpretation Introduction

(c)

Interpretation:

The net ionic equation should be written when ammonium sulfate and sodium chloride are mixed.

Concept introduction:

Solubility of any compound is predicted by below solubility chart:

	${\text{NO}}_3^{-}$	${\mathrm{Cl}}^{-}$	${\text{SO}}_4^{2-}$	${\text{OH}}^{-}$	${\text{CO}}_3^{2-}$	${\text{PO}}_4^{3-}$
Group-1 cations and ${\text{NH}}_4^{+}$
Group-2 cations			${\text{BaSO}}_{\text{4}}$	$\text{Mg}{\left(\text{OH}\right)}_{\text{2}}$	-	-
Transitions metal cations ${\text{Pb}}^{2+}$, ${\text{Hg}}_2^{2+}$		$\text{AgCl}$ ${\text{PbCl}}_2$ ${\text{Hg}}_2{\text{Cl}}_2$	${\text{PbSO}}_{\text{4}}$ ${\text{Ag}}_2{\text{SO}}_{\text{4}}$	-	-	-

Blank boxes indicate no precipitate formation occurs which means soluble in dilute solution.

Boxes with hyphen will form precipitate from dilute solutions and boxes where formula is written this is a cation-anion combination that will form precipitate.

Precipitation reactions: It is a type of chemical reactions where two soluble salts react with each other and formed different products, out of which one product must be insoluble in solution which is known as precipitate.

A chemical equation which shows only the species that are participated in the reaction is said to be net ionic equation.

Answer to Problem 10QAP

No precipitation occurs.

Explanation of Solution

Ammonium sulfate: ${{\text{(NH}}_4)}_{\text{2}}{\text{SO}}_{\text{4}}$

Sodium chloride: $\text{NaCl}$

Reaction for the solution of silver nitrate and hydrochloric acid is written as:

${{\text{(NH}}_4)}_{\text{2}}{\text{SO}}_{\text{4}}+\text{NaCl}\to {\text{Na}}_2{\text{SO}}_4{\text{+NH}}_4\text{Cl}$

Reactants:

Ions in solution: ${{\text{(NH}}_4)}_{\text{2}}{\text{SO}}_{\text{4}}$ : ${\text{NH}}_{\text{4}}^{\text{+}}$ and ${\text{SO}}_4{}^{2-}$

Ions in solution: $\text{NaCl}$ : ${\text{Na}}^{\text{+}}$ and ${\text{Cl}}^{-}$

Products:

Ions in solution: ${\text{Na}}_2{\text{SO}}_4$ : ${\text{Na}}^{\text{+}}$ and ${\text{SO}}_4{}^{2-}$

Ions in solution: ${\text{NH}}_4\text{Cl}$ : ${\text{NH}}_{\text{4}}^{\text{+}}$ and ${\text{Cl}}^{-}$

Now, ${\text{Na}}_2{\text{SO}}_4$ and ${\text{NH}}_4\text{Cl}$ both are soluble in water or in aqueous solution. Thus, no precipitation occurs.

Interpretation Introduction

(d)

Interpretation:

The net ionic equation should be written when cobalt(II) nitrate and sodium phosphate are mixed.

Concept introduction:

Solubility of any compound is predicted by below solubility chart:

	${\text{NO}}_3^{-}$	${\mathrm{Cl}}^{-}$	${\text{SO}}_4^{2-}$	${\text{OH}}^{-}$	${\text{CO}}_3^{2-}$	${\text{PO}}_4^{3-}$
Group-1 cations and ${\text{NH}}_4^{+}$
Group-2 cations			${\text{BaSO}}_{\text{4}}$	$\text{Mg}{\left(\text{OH}\right)}_{\text{2}}$	-	-
Transitions metal cations ${\text{Pb}}^{2+}$, ${\text{Hg}}_2^{2+}$		$\text{AgCl}$ ${\text{PbCl}}_2$ ${\text{Hg}}_2{\text{Cl}}_2$	${\text{PbSO}}_{\text{4}}$ ${\text{Ag}}_2{\text{SO}}_{\text{4}}$	-	-	-

Blank boxes indicate no precipitate formation occurs which means soluble in dilute solution.

Boxes with hyphen will form precipitate from dilute solutions and boxes where formula is written this is a cation-anion combination that will form precipitate.

Precipitation reactions: It is a type of chemical reactions where two soluble salts react with each other and formed different products, out of which one product must be insoluble in solution which is known as precipitate.

A chemical equation which shows only the species that are participated in the reaction is said to be net ionic equation.

Answer to Problem 10QAP

Precipitation occurs

The net ionic equation is:

${\text{3Co}}^{\text{2+}}{\text{(aq)+2PO}}_4^{3-}(\text{aq})\to {\text{Co}}_3{{\text{(PO}}_{\text{4}}\text{)}}_2(\text{s})$

Explanation of Solution

Cobalt(II) nitrate: ${\text{Co(NO}}_3{)}_2$

Sodium phosphate: ${\text{Na}}_{\text{3}}{\text{PO}}_{\text{4}}$

Reaction for the solution of silver nitrate and hydrochloric acid is written as:

${\text{Co(NO}}_3{)}_2+{\text{Na}}_{\text{3}}{\text{PO}}_{\text{4}}\to {\text{Co}}_3{{\text{(PO}}_{\text{4}}\text{)}}_2{\text{+NaNO}}_3$

Reactants:

Ions in solution: ${\text{Co(NO}}_3{)}_2$ : ${\text{Co}}^{\text{2+}}$ and ${\text{NO}}_3{}^{-}$

Ions in solution: ${\text{Na}}_{\text{3}}{\text{PO}}_{\text{4}}$ : ${\text{Na}}^{\text{+}}$ and ${\text{PO}}_4^{3-}$

Products:

Ions in solution: ${\text{Co}}_3{{\text{(PO}}_{\text{4}}\text{)}}_2$ : ${\text{Co}}^{\text{2+}}$ and ${\text{PO}}_4^{3-}$

Ions in solution: ${\text{NaNO}}_3$ : ${\text{Na}}^{\text{+}}$ and ${\text{NO}}_3{}^{-}$

Now, ${\text{Co}}_3{{\text{(PO}}_{\text{4}}\text{)}}_2$ is solid, not soluble in water whereas ${\text{NaNO}}_3$ is soluble in water.

So, the equation will be:

${\text{Co}}^{\text{2+}}{\text{(aq)+ NO}}_3{}^{-}(\text{aq})+{\text{Na}}^{\text{+}}{\text{(aq)+PO}}_4^{3-}(\text{aq})\to {\text{Co}}_3{{\text{(PO}}_{\text{4}}\text{)}}_2(\text{s}){\text{+Na}}^{+}{\text{(aq)+NO}}_3{}^{-}(\text{aq})$

Now, cancelling out the ions which appear on both sides of the equation ( ${\text{NO}}_3{}^{\text{-}}$ , ${\text{Na}}^{\text{+}}$ ), the final or net ionic equation is:

${\text{3Co}}^{\text{2+}}{\text{(aq)+2PO}}_4^{3-}(\text{aq})\to {\text{Co}}_3{{\text{(PO}}_{\text{4}}\text{)}}_2(\text{s})$