Chapter 7, Problem 16QAP

(a)

Interpretation Introduction

Interpretation : According to the general solubility rule, the participate formed in the solution of sodium carbonate and manganese(II) chloride needs to be determined.

Concept Introduction : There are following solubility rules:

1. The salts formed from sodium, potassium and ammonium ions are soluble in nature.
2. All sulfate salts are soluble except ${\text{CaSO}}_4$ , ${\text{PbSO}}_4$ and ${\text{BaSO}}_4$
3. All nitrate salts are soluble. The formula of nitrate polyatomic ion is ${\text{NO}}_3^{-}$ .
4. All the salts formed from chloride, bromide and iodide are soluble except those with ${\text{Pb}}^{2+}$ and ${\text{Ag}}^{+}$ .
5. All sulfate salts are insoluble except ${\text{Na}}_{\text{2}}{\text{SO}}_{\text{3}}$ , ${\text{K}}_{\text{2}}{\text{SO}}_{\text{3}}$ and ${\left({\text{NH}}_4\right)}_{\text{2}}{\text{SO}}_{\text{3}}$ .
6. All sulfides are insoluble except ${\text{Na}}_{\text{2}}\text{S}$ , ${\text{K}}_{\text{2}}\text{S}$ and ${\left({\text{NH}}_4\right)}_{\text{2}}\text{S}$ .
7. All oxides and hydroxides are insoluble except ${\text{Na}}_{\text{2}}\text{O}$ /NaOH, ${\text{K}}_{\text{2}}\text{O}$ /KOH and $\text{BaO}/\text{Ba}{\left(\text{OH}\right)}_{\text{2}}$ .
8. All carbonates are insoluble except ${\text{Na}}_{\text{2}}{\text{CO}}_3$ , ${\text{K}}_{\text{2}}{\text{CO}}_{\text{3}}$ and ${\left({\text{NH}}_4\right)}_{\text{2}}{\text{CO}}_{\text{3}}$ .

Answer to Problem 16QAP

  ${\text{MnCO}}_{\text{3}}$

Explanation of Solution

In order to identify the component which will precipitate it is important to identify the ions in solution first.

  $\begin{array}{l}{\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}\to 2{\text{Na}}^{+}(aq)+{\text{CO}}_3{}^{-2}(aq)\\ {\text{MnCl}}_{\text{2}}\to {\text{Mn}}^{+2}(aq)+2{\text{Cl}}^{-}(aq)\end{array}$

From here, we need to identify the pair of cation-ion that might precipitate given the conditions.

According to rule 8: All carbonates are insoluble except ${\text{Na}}_{\text{2}}{\text{CO}}_3$ , ${\text{K}}_{\text{2}}{\text{CO}}_{\text{3}}$ and ${\left({\text{NH}}_4\right)}_{\text{2}}{\text{CO}}_{\text{3}}$ .

Thus, ${\text{Na}}_{\text{2}}{\text{CO}}_3$ is soluble and the following is likely to precipitate:

  ${\text{Mn}}^{+2}(aq)+{\mathrm{CO}}_3{}^{-2}(aq)\to {\text{MnCO}}_3(s)$

(b)

Interpretation Introduction

Interpretation : According to the general solubility rule, the participate formed in the given solution of potassium sulfate and calcium acetate needs to be determined.

Concept Introduction : There are following solubility rules:

1. The salts formed from sodium, potassium and ammonium ions are soluble in nature.
2. All sulfate salts are soluble except ${\text{CaSO}}_4$ , ${\text{PbSO}}_4$ and ${\text{BaSO}}_4$
3. All nitrate salts are soluble. The formula of nitrate polyatomic ion is ${\text{NO}}_3^{-}$ .
4. All the salts formed from chloride, bromide and iodide are soluble except those with ${\text{Pb}}^{2+}$ and ${\text{Ag}}^{+}$ .
5. All sulfate salts are insoluble except ${\text{Na}}_{\text{2}}{\text{SO}}_{\text{3}}$ , ${\text{K}}_{\text{2}}{\text{SO}}_{\text{3}}$ and ${\left({\text{NH}}_4\right)}_{\text{2}}{\text{SO}}_{\text{3}}$ .
6. All sulfides are insoluble except ${\text{Na}}_{\text{2}}\text{S}$ , ${\text{K}}_{\text{2}}\text{S}$ and ${\left({\text{NH}}_4\right)}_{\text{2}}\text{S}$ .
7. All oxides and hydroxides are insoluble except ${\text{Na}}_{\text{2}}\text{O}$ /NaOH, ${\text{K}}_{\text{2}}\text{O}$ /KOH and $\text{BaO}/\text{Ba}{\left(\text{OH}\right)}_{\text{2}}$ .
8. All carbonates are insoluble except ${\text{Na}}_{\text{2}}{\text{CO}}_3$ , ${\text{K}}_{\text{2}}{\text{CO}}_{\text{3}}$ and ${\left({\text{NH}}_4\right)}_{\text{2}}{\text{CO}}_{\text{3}}$ .

Answer to Problem 16QAP

  ${\text{CaSO}}_4$

Explanation of Solution

In order to identify the component which will precipitate it is important to identify the ions in solution first.

  $\begin{array}{l}{\text{K}}_{\text{2}}{\text{SO}}_{\text{4}}\to 2{\text{K}}^{+}(aq)+{\text{SO}}_4{}^{-2}(aq)\\ {\text{Ca(C}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}{\text{)}}_{\text{2}}\to {\text{Ca}}^{+2}(aq)+2({\text{C}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_2{}^{-})(aq)\end{array}$

From here, we need to identify the pair of cation-ion that might precipitate given the conditions.

According to rule 2, all sulfate salts are soluble except ${\text{CaSO}}_4$ , ${\text{PbSO}}_4$ and ${\text{BaSO}}_4$ . Thus, ${\text{CaSO}}_4$ is formed as a precipitate as follows:

  ${\text{Ca}}^{+2}(aq)+{\text{SO}}_4{}^{-2}(aq)\to {\text{CaSO}}_{\text{4}}\left(s\right)$

(c)

Interpretation Introduction

Interpretation : According to the general solubility rule, the participate formed in the given solution of hydrochloric acid and mercurous acetate needs to be determined.

Concept Introduction : There are following solubility rules:

1. The salts formed from sodium, potassium and ammonium ions are soluble in nature.
2. All sulfate salts are soluble except ${\text{CaSO}}_4$ , ${\text{PbSO}}_4$ and ${\text{BaSO}}_4$
3. All nitrate salts are soluble. The formula of nitrate polyatomic ion is ${\text{NO}}_3^{-}$ .
4. All the salts formed from chloride, bromide and iodide are soluble except those with ${\text{Pb}}^{2+}$ and ${\text{Ag}}^{+}$ .
5. All sulfate salts are insoluble except ${\text{Na}}_{\text{2}}{\text{SO}}_{\text{3}}$ , ${\text{K}}_{\text{2}}{\text{SO}}_{\text{3}}$ and ${\left({\text{NH}}_4\right)}_{\text{2}}{\text{SO}}_{\text{3}}$ .
6. All sulfides are insoluble except ${\text{Na}}_{\text{2}}\text{S}$ , ${\text{K}}_{\text{2}}\text{S}$ and ${\left({\text{NH}}_4\right)}_{\text{2}}\text{S}$ .
7. All oxides and hydroxides are insoluble except ${\text{Na}}_{\text{2}}\text{O}$ /NaOH, ${\text{K}}_{\text{2}}\text{O}$ /KOH and $\text{BaO}/\text{Ba}{\left(\text{OH}\right)}_{\text{2}}$ .
8. All carbonates are insoluble except ${\text{Na}}_{\text{2}}{\text{CO}}_3$ , ${\text{K}}_{\text{2}}{\text{CO}}_{\text{3}}$ and ${\left({\text{NH}}_4\right)}_{\text{2}}{\text{CO}}_{\text{3}}$ .

Answer to Problem 16QAP

  ${\text{Hg}}_{\text{2}}{\text{Cl}}_{\text{2}}$

Explanation of Solution

In order to identify the component which will precipitate it is important to identify the ions in solution first.

  $\begin{array}{l}\text{HCl}\to {\text{H}}^{+}(aq)+{\text{Cl}}^{-}(aq)\\ {\text{Hg}}_{\text{2}}{{\text{(C}}_{\text{2}}}_{{\text{H}}_{\text{3}}}\to {\text{Hg}}_2{}^{+2}(aq)+2({\text{C}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_2{}^{-})(aq)\end{array}$

From here, we need to identify the pair of cation-ion that might precipitate given the conditions.

According to rule 4, all the salts formed from chloride, bromide and iodide are soluble except those with ${\text{Pb}}^{2+}$ and ${\text{Ag}}^{+}$ .

The following is likely to precipitate:

  ${\text{Hg}}_{\text{2}}{}^{+2}(aq)+{\text{Cl}}^{-}(aq)\to {\text{Hg}}_{\text{2}}{\text{Cl}}_2(s)$

(d)

Interpretation Introduction

Interpretation : According to the general solubility rule, the participate formed in the given solution of sodium nitrate and lithium sulfate needs to be determined.

Concept Introduction : There are following solubility rules:

1. The salts formed from sodium, potassium and ammonium ions are soluble in nature.
2. All sulfate salts are soluble except ${\text{CaSO}}_4$ , ${\text{PbSO}}_4$ and ${\text{BaSO}}_4$
3. All nitrate salts are soluble. The formula of nitrate polyatomic ion is ${\text{NO}}_3^{-}$ .
4. All the salts formed from chloride, bromide and iodide are soluble except those with ${\text{Pb}}^{2+}$ and ${\text{Ag}}^{+}$ .
5. All sulfate salts are insoluble except ${\text{Na}}_{\text{2}}{\text{SO}}_{\text{3}}$ , ${\text{K}}_{\text{2}}{\text{SO}}_{\text{3}}$ and ${\left({\text{NH}}_4\right)}_{\text{2}}{\text{SO}}_{\text{3}}$ .
6. All sulfides are insoluble except ${\text{Na}}_{\text{2}}\text{S}$ , ${\text{K}}_{\text{2}}\text{S}$ and ${\left({\text{NH}}_4\right)}_{\text{2}}\text{S}$ .
7. All oxides and hydroxides are insoluble except ${\text{Na}}_{\text{2}}\text{O}$ /NaOH, ${\text{K}}_{\text{2}}\text{O}$ /KOH and $\text{BaO}/\text{Ba}{\left(\text{OH}\right)}_{\text{2}}$ .
8. All carbonates are insoluble except ${\text{Na}}_{\text{2}}{\text{CO}}_3$ , ${\text{K}}_{\text{2}}{\text{CO}}_{\text{3}}$ and ${\left({\text{NH}}_4\right)}_{\text{2}}{\text{CO}}_{\text{3}}$ .

Answer to Problem 16QAP

No precipitate

Explanation of Solution

In order to identify the component which will precipitate it is important to identify the ions in solution first.

  $\begin{array}{l}{\text{NaNO}}_3\to {\text{Na}}^{+}(aq)+{\text{NO}}_3{}^{-}(aq)\\ {\text{Li}}_{\text{2}}{\text{SO}}_{\text{4}}\to 2{\text{Li}}^{+}(aq)+{\text{SO}}_4{}^{-2}(aq)\end{array}$

From here, we need to identify the pair of cation-ion that might precipitate given the conditions.

According to rule 3, all nitrate salts are soluble. The formula of nitrate polyatomic ion is ${\text{NO}}_3^{-}$ and according to rule 2, all sulfate salts are soluble except ${\text{CaSO}}_4$ , ${\text{PbSO}}_4$ and ${\text{BaSO}}_4$ thus, no precipitate will formed.

(e)

Interpretation Introduction

Interpretation : According to the general solubility rule, the participate formed in the given solution of potassium hydroxide and nickel (II) chloride needs to be determined.

Concept Introduction : There are following solubility rules:

1. The salts formed from sodium, potassium and ammonium ions are soluble in nature.
2. All sulfate salts are soluble except ${\text{CaSO}}_4$ , ${\text{PbSO}}_4$ and ${\text{BaSO}}_4$
3. All nitrate salts are soluble. The formula of nitrate polyatomic ion is ${\text{NO}}_3^{-}$ .
4. All the salts formed from chloride, bromide and iodide are soluble except those with ${\text{Pb}}^{2+}$ and ${\text{Ag}}^{+}$ .
5. All sulfate salts are insoluble except ${\text{Na}}_{\text{2}}{\text{SO}}_{\text{3}}$ , ${\text{K}}_{\text{2}}{\text{SO}}_{\text{3}}$ and ${\left({\text{NH}}_4\right)}_{\text{2}}{\text{SO}}_{\text{3}}$ .
6. All sulfides are insoluble except ${\text{Na}}_{\text{2}}\text{S}$ , ${\text{K}}_{\text{2}}\text{S}$ and ${\left({\text{NH}}_4\right)}_{\text{2}}\text{S}$ .
7. All oxides and hydroxides are insoluble except ${\text{Na}}_{\text{2}}\text{O}$ /NaOH, ${\text{K}}_{\text{2}}\text{O}$ /KOH and $\text{BaO}/\text{Ba}{\left(\text{OH}\right)}_{\text{2}}$ .
8. All carbonates are insoluble except ${\text{Na}}_{\text{2}}{\text{CO}}_3$ , ${\text{K}}_{\text{2}}{\text{CO}}_{\text{3}}$ and ${\left({\text{NH}}_4\right)}_{\text{2}}{\text{CO}}_{\text{3}}$ .

Answer to Problem 16QAP

  $\text{Ni(OH}{)}_2$

Explanation of Solution

In order to identify the component which will precipitate it is important to identify the ions in solution first.

  $\begin{array}{l}\text{NaOH}\to {\text{Na}}^{+}(aq)+{\text{OH}}^{-}(aq)\\ {\text{NiCl}}_{\text{2}}\to {\text{Ni}}^{+2}(aq)+2{\text{Cl}}^{-}(aq)\end{array}$

From here, we need to identify the pair of cation-ion that might precipitate given the conditions.

According to the rule, transition metal hydroxide is not soluble.

The following is likely to precipitate:

  ${\text{Ni}}^{+2}(aq)+{\text{OH}}^{-}(aq)\to {\text{Ni(OH)}}_{\text{2}}(s)$

(f)

Interpretation Introduction

Interpretation : According to the general solubility rule, the participate formed in the given solution of sulfuric acid and barium chloride needs to be determined.

Concept Introduction : There are following solubility rules:

1. The salts formed from sodium, potassium and ammonium ions are soluble in nature.
2. All sulfate salts are soluble except ${\text{CaSO}}_4$ , ${\text{PbSO}}_4$ and ${\text{BaSO}}_4$
3. All nitrate salts are soluble. The formula of nitrate polyatomic ion is ${\text{NO}}_3^{-}$ .
4. All the salts formed from chloride, bromide and iodide are soluble except those with ${\text{Pb}}^{2+}$ and ${\text{Ag}}^{+}$ .
5. All sulfate salts are insoluble except ${\text{Na}}_{\text{2}}{\text{SO}}_{\text{3}}$ , ${\text{K}}_{\text{2}}{\text{SO}}_{\text{3}}$ and ${\left({\text{NH}}_4\right)}_{\text{2}}{\text{SO}}_{\text{3}}$ .
6. All sulfides are insoluble except ${\text{Na}}_{\text{2}}\text{S}$ , ${\text{K}}_{\text{2}}\text{S}$ and ${\left({\text{NH}}_4\right)}_{\text{2}}\text{S}$ .
7. All oxides and hydroxides are insoluble except ${\text{Na}}_{\text{2}}\text{O}$ /NaOH, ${\text{K}}_{\text{2}}\text{O}$ /KOH and $\text{BaO}/\text{Ba}{\left(\text{OH}\right)}_{\text{2}}$ .
8. All carbonates are insoluble except ${\text{Na}}_{\text{2}}{\text{CO}}_3$ , ${\text{K}}_{\text{2}}{\text{CO}}_{\text{3}}$ and ${\left({\text{NH}}_4\right)}_{\text{2}}{\text{CO}}_{\text{3}}$ .

Answer to Problem 16QAP

  ${\text{Ba(SO}}_{\text{4}}{\text{)}}_{\text{2}}$

Explanation of Solution

In order to identify the component which will precipitate it is important to identify the ions in solution first.

  $\begin{array}{l}{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}\to 2{\text{H}}^{+}(aq)+{\text{SO}}_4{}^{2-}(aq)\\ {\text{BaCl}}_{\text{2}}\to {\text{Ba}}^{+2}(aq)+2{\mathrm{Cl}}^{-}(aq)\end{array}$

From here, we need to identify the pair of cation-ion that might precipitate given the conditions.

According to the rule,

The following is likely to precipitate:

  ${\text{Ba}}^{+2}(aq)+{\text{SO}}_{\text{4}}{}^{-2}(aq)\to {\text{BaSO}}_4(s)$