Chapter 4, Problem 8QAP

Follow the directions of Question 7 for solutions of the following:

(a) silver nitrate and sodium chloride

(b) cobalt(II) nitrate and sodium hydroxide

(c) ammonium phosphate and potassium hydroxide

(d) copper(II) sulfate and sodium carbonate

(e) lithium sulfate and barium hydroxide

Interpretation Introduction

(a)

Interpretation:

Whether a precipitate will form when the given solutions are mixed should be determined along with the net ionic equation should be written, if precipitate will form.

Silver nitrate and sodium chloride

Concept introduction:

Solubility of any compound is predicted by above solubility chart.

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

Boxes with grey small box 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 8QAP

Precipitation occurs

The net ionic equation is:

${\text{Ag}}^{\text{+}}{\text{(aq)+Cl}}^{-}(\text{aq})\to \text{AgCl(s)}$

Explanation of Solution

Silver nitrate: ${\text{AgNO}}_{\text{3}}$

Sodium chloride: $\text{NaCl}$

Reaction for the solution of silver nitrate and sodium chloride is written as:

${\text{AgNO}}_{\text{3}}\text{+NaCl}\to \text{NaNO}\text{+}\text{AgCl}$

Reactants:

Ions in solution: ${\text{AgNO}}_{\text{3}}$ : ${\text{Ag}}^{\text{+}}$ and ${\text{NO}}_3{}^{-}$

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

Products:

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

Ions in solution: $\text{AgCl}$ : ${\text{Ag}}^{\text{+}}$ and ${\text{Cl}}^{-}$

Now, $\text{AgCl}$ is solid, not soluble in water whereas $\text{NaNO}$ is soluble in water.

So, the equation will be:

${\text{Ag}}^{\text{+}}{\text{(aq)+NO}}_3{}^{-}(\text{aq})+{\text{Na}}^{+}{\text{(aq)+Cl}}^{-}(\text{aq})\to \text{AgCl(s)}+{\text{Na}}^{\text{+}}{\text{(aq)+NO}}_{\text{3}}{}^{-}(\text{aq})$

Now, cancelling out the ions which appear on both sides of the equation ( ${\text{NO}}_{\text{3}}{}^{\text{-}}$, ${\text{Na}}^{\text{+}}$), the final or net ionic equation is:

${\text{Ag}}^{\text{+}}{\text{(aq)+Cl}}^{-}(\text{aq})\to \text{AgCl(s)}$

Interpretation Introduction

(b)

Interpretation:

Whether a precipitate will form when the given solutions are mixed should be determined along with the net ionic equation should be written, if precipitate will form.

Cobalt(II) nitrate and sodium hydroxide

Concept introduction:

Solubility of any compound is predicted by above solubility chart.

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

Boxes with grey small box 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 8QAP

Precipitation occurs

The net ionic equation is:

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

Explanation of Solution

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

Sodium hydroxide: $\text{NaOH}$

Reaction for the solution of cobalt(II) nitrate and sodium hydroxide is written as:

${\text{Co(NO}}_3{)}_2+\text{NaOH}\to {\text{NaNO}}_{\text{3}}\text{+Co(OH}{)}_2$

Reactants:

Ions in solution: ${\text{Co(NO}}_3{)}_2$ : ${\text{Co}}^{\text{+2}}$ and ${\text{NO}}_{\text{3}}{}^{-}$

Ions in solution: $\text{NaOH}$ : ${\text{Na}}^{\text{+}}$ and ${\text{OH}}^{-}$

Products:

Ions in solution: ${\text{NaNO}}_{\text{3}}$ : ${\text{Na}}^{\text{+}}$ and ${\text{NO}}_{\text{3}}{}^{-}$

Ions in solution: $\text{Co(OH}{)}_2$ : ${\text{Co}}^{\text{+2}}$ and ${\text{OH}}^{-}$

Now, $\text{Co(OH}{)}_2$ is solid, not soluble in water whereas ${\text{NaNO}}_{\text{3}}$ is soluble in water.

So, the equation will be:

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

Now, cancelling out the ions which appear on both sides of the equation ( ${\text{NO}}_{\text{3}}{}^{\text{-}}$, ${\text{Na}}^{\text{+}}$), the final or net ionic equation is:

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

Interpretation Introduction

(c)

Interpretation:

Whether a precipitate will form when the given solutions are mixed should be determined along with the net ionic equation should be written, if precipitate will form.

Ammonium phosphate and potassium hydroxide

Concept introduction:

Solubility of any compound is predicted by above solubility chart.

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

Boxes with grey small box 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 8QAP

No precipitation occurs.

Explanation of Solution

Ammonium phosphate: ${{\text{(NH}}_{\text{4}})}_{\text{3}}{\text{PO}}_{\text{4}}$

Potassium hydroxide: $\text{KOH}$

Reaction for the solution of ammonium phosphate and potassium hydroxide is written as:

${{\text{(NH}}_{\text{4}})}_{\text{3}}{\text{PO}}_{\text{4}}+\text{KOH}\to {\text{NH}}_4{\text{OH+K}}_3{\text{PO}}_4$

Reactants:

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

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

Products:

Ions in solution: ${\text{NH}}_4\text{OH}$ : ${\text{NH}}_4{}^{\text{+}}$ and ${\text{OH}}^{-}$

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

Now, ${\text{NH}}_4\text{OH}$ and ${\text{K}}_3{\text{PO}}_4$ both are soluble in water. No precipitation occurs.

Interpretation Introduction

(d)

Interpretation:

Whether a precipitate will form when the given solutions are mixed should be determined along with the net ionic equation should be written, if precipitate will form.

Copper(II) sulfate and sodium carbonate

Concept introduction:

Solubility of any compound is predicted by above solubility chart.

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

Boxes with grey small box 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 8QAP

Precipitation occurs

The net ionic equation is:

${\text{Cu}}^{\text{+2}}{\text{(aq)+CO}}_3{}^{-2}(\text{aq})\to {\text{CuCO}}_3(\text{s})$

Explanation of Solution

Copper(II) sulfate: ${\text{CuSO}}_{\text{4}}$

Sodium carbonate: ${\text{Na}}_2{\text{CO}}_3$

Reaction for the solution of copper(II) sulfate and sodium carbonate is written as:

${\text{CuSO}}_{\text{4}}+{\text{Na}}_2{\text{CO}}_3\to {\text{CuCO}}_3{\text{+Na}}_2{\text{SO}}_{\text{4}}$

Reactants:

Ions in solution: ${\text{CuSO}}_{\text{4}}$ : ${\text{Cu}}^{\text{2+}}$ and ${\text{SO}}_4{}^{2-}$

Ions in solution: ${\text{Na}}_2{\text{CO}}_3$ : ${\text{Na}}^{\text{+}}$ and ${\text{CO}}_3^{2-}$

Products:

Ions in solution: ${\text{CuCO}}_3$ : ${\text{Cu}}^{\text{+2}}$ and ${\text{CO}}_3^{2-}$

Ions in solution: ${\text{Na}}_2{\text{SO}}_{\text{4}}$ : ${\text{Na}}^{\text{+}}$ and ${\text{SO}}_4{}^{2-}$

Now, ${\text{CuCO}}_3$ is solid, not soluble in water whereas ${\text{Na}}_2{\text{SO}}_{\text{4}}$ is soluble in water.

So, the equation will be:

${\text{2Na}}^{\text{+}}{\text{(aq)+CO}}_3{}^{2-}(\text{aq})+{\text{Cu}}^{\text{2+}}{\text{(aq)+SO}}_4{}^{2-}(\text{aq})\to {\text{CuCO}}_3(\text{s})+2{\text{Na}}^{\text{+}}{\text{(aq)+SO}}_4{}^{2-}(\text{aq})$

Now, cancelling out the ions which appear on both sides of the equation ( ${\text{2Na}}^{+},$

${\text{SO}}_4^{2-}$), the final or net ionic equation is:

${\text{Cu}}^{\text{+2}}{\text{(aq)+CO}}_3{}^{-2}(\text{aq})\to {\text{CuCO}}_3(\text{s})$

Interpretation Introduction

(e)

Interpretation:

Whether a precipitate will form when the given solutions are mixed should be determined along with the net ionic equation should be written, if precipitate will form.

Lithium sulfate and barium hydroxide

Concept introduction:

Solubility of any compound is predicted by above solubility chart.

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

Boxes with grey small box 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 8QAP

Precipitation occurs

The net ionic equation is:

${\text{Ba}}^{\text{+2}}\text{(aq)}+{\text{SO}}_4{}^{2-}(\text{aq})+\to {\text{BaSO}}_4(\text{s})$

Explanation of Solution

Lithium sulfate: ${\text{Li}}_2{\text{SO}}_4$

Barium hydroxide: ${\text{Ba(OH)}}_2$

Reaction for the solution of lithium sulfate and barium hydroxide is written as:

${\text{Li}}_2{\text{SO}}_4+{\text{Ba(OH)}}_2\to {\text{LiOH+BaSO}}_4$

Reactants:

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

Ions in solution: ${\text{Ba(OH)}}_2$ : ${\text{Ba}}^{\text{+2}}$ and ${\text{OH}}^{-}$

Products:

Ions in solution: $\text{LiOH}$ : ${\text{Li}}^{\text{+}}$ and ${\text{OH}}^{-}$

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

Now, ${\text{BaSO}}_4$ is solid, not soluble in water whereas $\text{LiOH}$ is soluble in water.

So, the equation will be:

${\text{2Li}}^{\text{+}}{\text{(aq)+SO}}_4{}^{2-}(\text{aq})+{\text{Ba}}^{\text{+2}}{\text{(aq)+2OH}}^{-}(\text{aq})\to {\text{BaSO}}_4(\text{s})+2{\text{Li}}^{\text{+}}{\text{(aq)+2OH}}^{-}(\text{aq})$

Now, cancelling out the ions which appear on both sides of the equation ( ${\text{2Li}}^{+},$

$2{\text{OH}}^{-}$), the final or net ionic equation is:

${\text{Ba}}^{\text{+2}}\text{(aq)}+{\text{SO}}_4{}^{2-}(\text{aq})+\to {\text{BaSO}}_4(\text{s})$