Chapter 4, Problem 93AE

Using the general solubility rules given in Table 6-1. name three reagents that would form precipitates with each of the following ions in aqueous solution. Write the net ionic equation for each of your suggestions.

a. chloride ion

b. calcium ion

c. iron(III) ion

d. sulfate ion

e. mercury(I) ion, Hg₂²⁺

f. silver ion

Interpretation Introduction

Interpretation: The reagent that helps in the formation of precipitate with their net ionic equation has to be written.

Concept introduction: Salts are formed by the neutralization reaction of acid and base.

The rules for salts to be soluble in water are as follows

1. 1. Most of the nitrate salts are soluble
2. 2. Salts containing the cation as alkali metal such as Sodium, Magnesium, Rubidium etc and Ammonium are said to be water soluble.
3. 3. Salts of Bromide, Chloride and iodide are said to be water soluble except in case of cations such as silver, lead and mercury.
4. 4. Most of the Sulphate salts are soluble except Barium sulphate, Mercury sulphate, Lead sulphate and Calcium sulphate.
5. 5. Most of the Hydroxides are sparingly soluble except Sodium hydroxide and Potassium hydroxide.
6. 6. Most of the Sulphide, Carbonate, Chromates and Phosphates are less soluble except those include in the rule 2.

Answer to Problem 93AE

Chloride: ${\text{AgNO}}_{\text{3}}{\text{,Pb(NO}}_{\text{3}}{\text{)}}_{\text{2}}\text{and}{\text{Hg}}_{\text{2}}{{\text{(NO}}_{\text{3}}\text{)}}_{\text{2}}$ forms precipitate with ${\text{Cl}}^{\text{-}}$ . The net ionic equations are

$\begin{array}{l}{\text{Ag}}^{\text{+}}{}_{\left(\text{aq}\right)}{\text{+Cl}}^{\text{-}}{}_{\left(\text{aq}\right)}\to {\text{AgCl}}_{\left(\text{s}\right)}\\ {\text{Pb}}^{\text{2+}}{}_{\left(\text{aq}\right)}{\text{+2Cl}}^{\text{-}}{}_{\left(\text{aq}\right)}\to {\text{PbCl}}_{\text{2(s)}}\\ {\text{Hg}}_{\text{2}}{{}^{\text{2+}}}_{\left(\text{aq}\right)}{\text{+2Cl}}^{\text{-}}{}_{\left(\text{aq}\right)}\to {\text{Hg}}_{\text{2}}{\text{Cl}}_{\text{2(s)}}\end{array}$

Explanation of Solution

To name three reagents that precipitates with ${\text{Cl}}^{\text{-}}$ ion.

The three reagents that precipitates chlorine ion are ${\text{AgNO}}_{\text{3}}{\text{,Pb(NO}}_{\text{3}}{\text{)}}_{\text{2}}\text{and}{\text{Hg}}_{\text{2}}{{\text{(NO}}_{\text{3}}\text{)}}_{\text{2}}$.

${\text{AgNO}}_{\text{3}}$ on reaction with salt of Chloride gives a precipitate of Silver chloride, that’s insoluble salt by rule 3. The net ionic equation can be given as,

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

${\text{Pb(NO}}_{\text{3}}{\text{)}}_{\text{2}}$ on reaction with salt of Chloride gives a precipitate of Lead chloride, that’s insoluble salt by rule 3. The net ionic equation can be given as,

${\text{Pb}}^{\text{2+}}{}_{\left(\text{aq}\right)}{\text{+2Cl}}^{\text{-}}{}_{\left(\text{aq}\right)}\to {\text{PbCl}}_{\text{2(s)}}$

${\text{Hg}}_{\text{2}}{{\text{(NO}}_{\text{3}}\text{)}}_{\text{2}}$ on reaction with salt of Chloride gives a precipitate of Mercuric chloride, that’s insoluble salt by rule 3. The net ionic equation can be given as,

${\text{Hg}}_{\text{2}}{{}^{\text{2+}}}_{\left(\text{aq}\right)}{\text{+2Cl}}^{\text{-}}{}_{\left(\text{aq}\right)}\to {\text{Hg}}_{\text{2}}{\text{Cl}}_{\text{2(s)}}$

Interpretation Introduction

Interpretation: The reagent that helps in the formation of precipitate with their net ionic equation has to be written.

Concept introduction: Salts are formed by the neutralization reaction of acid and base.

The rules for salts to be soluble in water are as follows

1. 1. Most of the nitrate salts are soluble
2. 2. Salts containing the cation as alkali metal such as Sodium, Magnesium, Rubidium etc and Ammonium are said to be water soluble.
3. 3. Salts of Bromide, Chloride and iodide are said to be water soluble except in case of cations such as silver, lead and mercury.
4. 4. Most of the Sulphate salts are soluble except Barium sulphate, Mercury sulphate, Lead sulphate and Calcium sulphate.
5. 5. Most of the Hydroxides are sparingly soluble except Sodium hydroxide and Potassium hydroxide.
6. 6. Most of the Sulphide, Carbonate, Chromates and Phosphates are less soluble except those include in the rule 2.

Answer to Problem 93AE

Calcium: ${\text{Na}}_{\text{2}}{\text{SO}}_{\text{4}}{\text{,Na}}_{\text{2}}{\text{CO}}_{\text{3}}\text{and}{\text{Na}}_{\text{3}}{\text{PO}}_{\text{4}}$ forms precipitate with ${\text{Ca}}^{\text{2+}}$ . The net ionic equations are

$\begin{array}{l}{\text{Ca}}^{\text{2+}}{}_{\left(\text{aq}\right)}{\text{+SO}}_{\text{4}}{{}^{\text{2-}}}_{\left(\text{aq}\right)}\to {\text{CaSO}}_{\text{4(s)}}\\ {\text{Ca}}^{\text{2+}}{}_{\left(\text{aq}\right)}{\text{+CO}}_{\text{3}}{{}^{\text{2-}}}_{\left(\text{aq}\right)}\to {\text{CaCO}}_{\text{3(s)}}\\ {\text{3Ca}}^{\text{2+}}{}_{\left(\text{aq}\right)}{\text{+2PO}}_{\text{4}}{{}^{\text{3-}}}_{\left(\text{aq}\right)}\to {\text{Ca}}_{\text{3}}{{\text{(PO}}_{\text{4}}}_{\text{)}}\end{array}$

Explanation of Solution

To name three reagents that precipitates with ${\text{Ca}}^{\text{2+}}$

The three reagents that precipitate with ${\text{Ca}}^{\text{2+}}$ are ${\text{Na}}_{\text{2}}{\text{SO}}_{\text{4}}{\text{,Na}}_{\text{2}}{\text{CO}}_{\text{3}}\text{and}{\text{Na}}_{\text{3}}{\text{PO}}_{\text{4}}$

${\text{Na}}_{\text{2}}{\text{SO}}_{\text{4}}$ on reaction with salt of Calcium gives Calcium sulphate that is insoluble according to rule 4. The net ionic equation can be given as,

${\text{Ca}}^{\text{2+}}{}_{\left(\text{aq}\right)}{\text{+SO}}_{\text{4}}{{}^{\text{2-}}}_{\left(\text{aq}\right)}\to {\text{CaSO}}_{\text{4(s)}}$

${\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}$ on reaction with salt of Calcium gives Calcium carbonate that is insoluble according to rule 6. The net ionic equation can be given as,

${\text{Ca}}^{\text{2+}}{}_{\left(\text{aq}\right)}{\text{+CO}}_{\text{3}}{{}^{\text{2-}}}_{\left(\text{aq}\right)}\to {\text{CaCO}}_{\text{3(s)}}$

${\text{Na}}_{\text{3}}{\text{PO}}_{\text{4}}$ on reaction with salt of Calcium gives Calcium phosphate that is insoluble according to rule 6. The net ionic equation can be given as,

${\text{3Ca}}^{\text{2+}}{}_{\left(\text{aq}\right)}{\text{+2PO}}_{\text{4}}{{}^{\text{3-}}}_{\left(\text{aq}\right)}\to {\text{Ca}}_{\text{3}}{{\text{(PO}}_{\text{4}}\text{)}}_{\text{2(s)}}$

Interpretation Introduction

Interpretation: The reagent that helps in the formation of precipitate with their net ionic equation has to be written.

Concept introduction: Salts are formed by the neutralization reaction of acid and base.

The rules for salts to be soluble in water are as follows

1. 1. Most of the nitrate salts are soluble
2. 2. Salts containing the cation as alkali metal such as Sodium, Magnesium, Rubidium etc and Ammonium are said to be water soluble.
3. 3. Salts of Bromide, Chloride and iodide are said to be water soluble except in case of cations such as silver, lead and mercury.
4. 4. Most of the Sulphate salts are soluble except Barium sulphate, Mercury sulphate, Lead sulphate and Calcium sulphate.
5. 5. Most of the Hydroxides are sparingly soluble except Sodium hydroxide and Potassium hydroxide.
6. 6. Most of the Sulphide, Carbonate, Chromates and Phosphates are less soluble except those include in the rule 2.

Answer to Problem 93AE

Iron (III): ${\text{NaOH,Na}}_{\text{2}}\text{S}\text{and}{\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}$ forms precipitate with ${\text{Fe}}^{\text{3+}}$ . The net ionic equations are

$\begin{array}{l}{\text{Fe}}^{\text{3+}}{}_{\left(\text{aq}\right)}{\text{+3OH}}^{\text{-}}{}_{\left(\text{aq}\right)}\to {\text{Fe(OH)}}_{\text{3(s)}}\\ {\text{2Fe}}^{\text{3+}}{}_{\left(\text{aq}\right)}{\text{+3S}}^{\text{2-}}{}_{\left(\text{aq}\right)}\to {\text{Fe}}_{\text{2}}{\text{S}}_{\text{3(s)}}\\ {\text{2Fe}}^{\text{3+}}{}_{\left(\text{aq}\right)}{\text{+3CO}}_{\text{3}}{{}^{\text{2-}}}_{\left(\text{aq}\right)}\to {\text{Fe}}_{\text{2}}{{\text{(CO}}_{\text{3}}}_{\text{)}}\end{array}$

Explanation of Solution

To name three reagents that precipitates with Iron (III)

The three reagents that precipitate with Iron (III) are ${\text{NaOH,Na}}_{\text{2}}\text{S}\text{and}{\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}$.

$\text{NaOH}$ on reaction with salt of Iron (III) gives Iron (III) hydroxide that is insoluble according to rule 5. The net ionic equation is

${\text{Fe}}^{\text{3+}}{}_{\left(\text{aq}\right)}{\text{+3OH}}^{\text{-}}{}_{\left(\text{aq}\right)}\to {\text{Fe(OH)}}_{\text{3(s)}}$

${\text{Na}}_{\text{2}}\text{S}$ on reaction with salt of Iron (III) gives Iron (III) sulphide that is insoluble according to rule 6. The net ionic equation is

${\text{2Fe}}^{\text{3+}}{}_{\left(\text{aq}\right)}{\text{+3S}}^{\text{2-}}{}_{\left(\text{aq}\right)}\to {\text{Fe}}_{\text{2}}{\text{S}}_{\text{3(s)}}$

${\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}$ on reaction with salt of Iron (III) gives Iron (III) carbonate that is insoluble according to rule 6. The net ionic equation is

${\text{2Fe}}^{\text{3+}}{}_{\left(\text{aq}\right)}{\text{+3CO}}_{\text{3}}{{}^{\text{2-}}}_{\left(\text{aq}\right)}\to {\text{Fe}}_{\text{2}}{{\text{(CO}}_{\text{3}}\text{)}}_{\text{3(s)}}$

Interpretation Introduction

Interpretation: The reagent that helps in the formation of precipitate with their net ionic equation has to be written.

Concept introduction: Salts are formed by the neutralization reaction of acid and base.

The rules for salts to be soluble in water are as follows

1. 1. Most of the nitrate salts are soluble
2. 2. Salts containing the cation as alkali metal such as Sodium, Magnesium, Rubidium etc and Ammonium are said to be water soluble.
3. 3. Salts of Bromide, Chloride and iodide are said to be water soluble except in case of cations such as silver, lead and mercury.
4. 4. Most of the Sulphate salts are soluble except Barium sulphate, Mercury sulphate, Lead sulphate and Calcium sulphate.
5. 5. Most of the Hydroxides are sparingly soluble except Sodium hydroxide and Potassium hydroxide.
6. 6. Most of the Sulphide, Carbonate, Chromates and Phosphates are less soluble except those include in the rule 2.

Answer to Problem 93AE

Sulphate: ${\text{BaCl}}_{\text{2}}{\text{,Pb(NO}}_{\text{3}}{\text{)}}_{\text{2}}{\text{,Ca(NO}}_{\text{3}}{\text{)}}_{\text{2}}$ forms precipitate with ${\text{SO}}_{\text{4}}{}^{\text{2-}}$ . The net ionic equations are

$\begin{array}{l}{\text{Ba}}^{\text{2+}}{}_{\left(\text{aq}\right)}{\text{+SO}}_{\text{4}}{{}^{\text{2-}}}_{\left(\text{aq}\right)}\to {\text{BaSO}}_{\text{4(s)}}\\ {\text{Pb}}^{\text{2+}}{}_{\left(\text{aq}\right)}{\text{+SO}}_{\text{4}}{{}^{\text{2-}}}_{\left(\text{aq}\right)}\to {\text{PbSO}}_{\text{4(s)}}\\ {\text{Ca}}^{\text{2+}}{}_{\left(\text{aq}\right)}{\text{+SO}}_{\text{4}}{{}^{\text{2-}}}_{\left(\text{aq}\right)}\to {\text{CaSO}}_{\text{4(s)}}\end{array}$

Explanation of Solution

To name three reagents that precipitates with Sulphate

The three reagents that precipitate with Sulphate are ${\text{BaCl}}_{\text{2}}{\text{,Pb(NO}}_{\text{3}}{\text{)}}_{\text{2}}{\text{,Ca(NO}}_{\text{3}}{\text{)}}_{\text{2}}$

${\text{BaCl}}_{\text{2}}$ when treated with a salt of Sulphate gives a precipitate of Barium sulphate that is insoluble according to rule 4. The net ionic equation is

${\text{Ba}}^{\text{2+}}{}_{\left(\text{aq}\right)}{\text{+SO}}_{\text{4}}{{}^{\text{2-}}}_{\left(\text{aq}\right)}\to {\text{BaSO}}_{\text{4(s)}}$

${\text{Pb(NO}}_{\text{3}}{\text{)}}_{\text{2}}$ when treated with a salt of Sulphate gives a precipitate of Lead sulphate that is insoluble according to rule 4. The net ionic equation is

${\text{Pb}}^{\text{2+}}{}_{\left(\text{aq}\right)}{\text{+SO}}_{\text{4}}{{}^{\text{2-}}}_{\left(\text{aq}\right)}\to {\text{PbSO}}_{\text{4(s)}}$

${\text{Ca(NO}}_{\text{3}}{\text{)}}_{\text{2}}$ on reaction with salt of Sulphate gives Calcium sulphate that is insoluble according to rule 4. The net ionic equation can be given as,

${\text{Ca}}^{\text{2+}}{}_{\left(\text{aq}\right)}{\text{+SO}}_{\text{4}}{{}^{\text{2-}}}_{\left(\text{aq}\right)}\to {\text{CaSO}}_{\text{4(s)}}$

Interpretation Introduction

Interpretation: The reagent that helps in the formation of precipitate with their net ionic equation has to be written.

Concept introduction: Salts are formed by the neutralization reaction of acid and base.

The rules for salts to be soluble in water are as follows

1. 1. Most of the nitrate salts are soluble
2. 2. Salts containing the cation as alkali metal such as Sodium, Magnesium, Rubidium etc and Ammonium are said to be water soluble.
3. 3. Salts of Bromide, Chloride and iodide are said to be water soluble except in case of cations such as silver, lead and mercury.
4. 4. Most of the Sulphate salts are soluble except Barium sulphate, Mercury sulphate, Lead sulphate and Calcium sulphate.
5. 5. Most of the Hydroxides are sparingly soluble except Sodium hydroxide and Potassium hydroxide.
6. 6. Most of the Sulphide, Carbonate, Chromates and Phosphates are less soluble except those include in the rule 2.

Answer to Problem 93AE

Mercury (I): ${\text{Na}}_{\text{2}}{\text{SO}}_{\text{4}}\text{,NaCl}\text{and}\text{NaI}$ forms precipitate with ${\text{Hg}}_{\text{2}}{}^{\text{2+}}$ . The net ionic equations are

           $\begin{array}{l}{\text{Hg}}_{\text{2}}{{}^{\text{2+}}}_{\left(\text{aq}\right)}{\text{+SO}}_{\text{4}}{{}^{\text{2-}}}_{\left(\text{aq}\right)}\to {\text{Hg}}_{\text{2}}{\text{SO}}_{\text{4(s)}}\\ {\text{Hg}}_{\text{2}}{{}^{\text{2+}}}_{\left(\text{aq}\right)}{\text{+2Cl}}^{\text{-}}{}_{\left(\text{aq}\right)}\to {\text{Hg}}_{\text{2}}{\text{Cl}}_{\text{2}}{}_{\left(\text{s}\right)}\\ {\text{Hg}}_{\text{2}}{{}^{\text{2+}}}_{\left(\text{aq}\right)}{\text{+2I}}^{\text{-}}{}_{\left(\text{aq}\right)}\to {\text{Hg}}_{\text{2}}{\text{I}}_{\text{2(s)}}\end{array}$

Explanation of Solution

To name three reagents that precipitates with Mercury (I) ion

The three reagents that precipitate with Mercury (I) ion are ${\text{Na}}_{\text{2}}{\text{SO}}_{\text{4}}\text{,NaCl}\text{and}\text{NaI}$

${\text{Na}}_{\text{2}}{\text{SO}}_{\text{4}}$ on treatment with salt of Mercury (I) ion gives Mercury sulphate that is insoluble according to rule 4. The net ionic equation is,

${\text{Hg}}_{\text{2}}{{}^{\text{2+}}}_{\left(\text{aq}\right)}{\text{+SO}}_{\text{4}}{{}^{\text{2-}}}_{\left(\text{aq}\right)}\to {\text{Hg}}_{\text{2}}{\text{SO}}_{\text{4(s)}}$

$\text{NaCl}$ on treatment with salt of Mercury (I) ion gives Mercury chloride that is insoluble according to rule 3. The net ionic equation is,

${\text{Hg}}_{\text{2}}{{}^{\text{2+}}}_{\left(\text{aq}\right)}{\text{+2Cl}}^{\text{-}}{}_{\left(\text{aq}\right)}\to {\text{Hg}}_{\text{2}}{\text{Cl}}_{\text{2}}{}_{\left(\text{s}\right)}$

$\text{NaI}$ on treatment with salt of Mercury (I) ion gives Mercury iodide that is insoluble according to rule 3. The net ionic equation is,

${\text{Hg}}_{\text{2}}{{}^{\text{2+}}}_{\left(\text{aq}\right)}{\text{+2I}}^{\text{-}}{}_{\left(\text{aq}\right)}\to {\text{Hg}}_{\text{2}}{\text{I}}_{\text{2(s)}}$

Interpretation Introduction

Interpretation: The reagent that helps in the formation of precipitate with their net ionic equation has to be written.

Concept introduction: Salts are formed by the neutralization reaction of acid and base.

The rules for salts to be soluble in water are as follows

1. 1. Most of the nitrate salts are soluble
2. 2. Salts containing the cation as alkali metal such as Sodium, Magnesium, Rubidium etc and Ammonium are said to be water soluble.
3. 3. Salts of Bromide, Chloride and iodide are said to be water soluble except in case of cations such as silver, lead and mercury.
4. 4. Most of the Sulphate salts are soluble except Barium sulphate, Mercury sulphate, Lead sulphate and Calcium sulphate.
5. 5. Most of the Hydroxides are sparingly soluble except Sodium hydroxide and Potassium hydroxide.
6. 6. Most of the Sulphide, Carbonate, Chromates and Phosphates are less soluble except those include in the rule 2.

Answer to Problem 93AE

Silver: ${\text{NaBr,Na}}_{\text{2}}{\text{CrO}}_{\text{4}}\text{and}{\text{Na}}_{\text{3}}{\text{PO}}_{\text{4}}$ forms precipitate with ${\text{Ag}}^{\text{+}}$ . The net ionic equations are

$\begin{array}{l}{\text{Ag}}^{\text{+}}{}_{\left(\text{aq}\right)}{\text{+Br}}^{\text{-}}{}_{\left(\text{aq}\right)}\to {\text{AgBr}}_{\left(\text{s}\right)}\\ {\text{2Ag}}^{\text{+}}{}_{\left(\text{aq}\right)}{\text{+CrO}}_{\text{4}}{{}^{\text{2-}}}_{\left(\text{aq}\right)}\to {\text{Ag}}_{\text{2}}{\text{CrO}}_{\text{4(s)}}\\ {\text{3Ag}}^{\text{+}}{}_{\left(\text{aq}\right)}{\text{+PO}}_{\text{4}}{}^{\text{3-}}\to {\text{Ag}}_{\text{3}}{\text{PO}}_{\text{4(s)}}\end{array}$

Explanation of Solution

To name three reagents that precipitates with Silver

The three reagents that precipitate with Silver are ${\text{NaBr,Na}}_{\text{2}}{\text{CrO}}_{\text{4}}\text{and}{\text{Na}}_{\text{3}}{\text{PO}}_{\text{4}}$

$\text{NaBr}$ on treatment with salt of Silver gives Silver bromide that is insoluble according to rule 3. The net ionic equation is,

${\text{Ag}}^{\text{+}}{}_{\left(\text{aq}\right)}{\text{+Br}}^{\text{-}}{}_{\left(\text{aq}\right)}\to {\text{AgBr}}_{\left(\text{s}\right)}$

${\text{Na}}_{\text{2}}{\text{CrO}}_{\text{4}}$ on treatment with salt of Silver gives Silver chromate that is insoluble according to rule 6. The net ionic equation is,

${\text{2Ag}}^{\text{+}}{}_{\left(\text{aq}\right)}{\text{+CrO}}_{\text{4}}{{}^{\text{2-}}}_{\left(\text{aq}\right)}\to {\text{Ag}}_{\text{2}}{\text{CrO}}_{\text{4(s)}}$

${\text{Na}}_{\text{3}}{\text{PO}}_{\text{4}}$ on treatment with salt of Silver gives Silver phosphate that is insoluble according to rule 6. The net ionic equation is,

${\text{3Ag}}^{\text{+}}{}_{\left(\text{aq}\right)}{\text{+PO}}_{\text{4}}{}^{\text{3-}}\to {\text{Ag}}_{\text{3}}{\text{PO}}_{\text{4(s)}}$