Question

Three solutions are mixed together to form a single solution; in the final solution, there are 0.2 mol pb(ch3coo)2, 0.1 mol na2s, and 0.1 mol cacl2 present. part a what solid(s) will precipitate? express your answer as a chemical formula. if there is more than one answer, separate them by a comma.

Answer 1

We are given the compounds:

0.2 mol pb(ch3coo)2, 0.1 mol na2s, and 0.1 mol cacl2

Pb will react with both S and Cl to form a solid compound, therefore:

Pb+2 + S-2 ---> PbS(s) 
Pb+2 + 2Cl- ---> PbCl2(s) 

Lead(II) sulfide and Lead(II) chloride will precipitate.

Answer 2

The solids that are precipitated in the given solution are  $\boxed{{\text{PbS, PbC}}{{\text{l}}_2}}$

Further Explanation:

Precipitation reaction:

It is the type of reaction in which an insoluble salt is formed by the combination of two solutions containing soluble salts. That insoluble salt is known as precipitate and therefore such reactions are named precipitation reactions. An example of precipitation reaction is,

${\text{AgN}}{{\text{O}}_3}\left( {aq} \right)+{\text{KBr}}\left({aq}\right)\to {\text{AgBr}}\left( s\right)+{\text{KN}}{{\text{O}}_3}\left({aq}\right)$

Here, AgBr is a precipitate.

The solubility rules to determine the solubility of the compound are as follows:

1. The common compounds of group 1A are soluble.

2. All the common compounds of ammonium ion and all acetates, chlorides, nitrates, bromides, iodides, and perchlorates are soluble in nature. Only the chlorides, bromides, and iodides of ${\text{A}}{{\text{g}}^ + }$ , ${\text{P}}{{\text{b}}^{2 + }}$ , ${\text{C}}{{\text{u}}^ + }$ and ${\text{Hg}}_2^{2 + }$ are not soluble.

3. All the acetates and chlorates are soluble in nature.

4. All sulfides, except those formed by group 1A, 2A, and ammonium ion are insoluble.

5. The chlorides, bromides, and iodides of all the metals are soluble in water, except for silver, lead, and mercury (II). Mercury (II) iodide is water insoluble. Lead halides are soluble in hot water.

6. Almost all the sulfides of transition metals are highly insoluble. These include CdS, FeS, ZnS, and ${\text{A}}{{\text{g}}_2}{\text{S}}$.The sulfides of arsenic, antimony, bismuth, and lead are also insoluble.

We have ${\text{Pb}}{\left({{\text{C}}{{\text{H}}_3}{\text{COO}}}\right)_2}$ ,${\text{N}}{{\text{a}}_2}{\text{S}}$ and ${\text{CaC}}{{\text{l}}_2}$ in the final solution. So ${\text{Pb}}{\left({{\text{C}}{{\text{H}}_3}{\text{COO}}}\right)_2}$  reacts with both ${\text{N}}{{\text{a}}_2}{\text{S}}$ as well as ${\text{CaC}}{{\text{l}}_2}$ .

In the solution, $0.1\;{\text{mol}}$ of ${\text{Pb}}{\left({{\text{C}}{{\text{H}}_3}{\text{COO}}}\right)_2}$ will react with $0.1\;{\text{mol}}$ of ${\text{N}}{{\text{a}}_2}{\text{S}}$ to form $0.1\;{\text{mol}}$ of ${\text{PbS}}$ and $0.2\;{\text{mol}}$ of ${\text{C}}{{\text{H}}_3}{\text{COONa}}$ .

1. The reaction between ${\text{Pb}}{\left({{\text{C}}{{\text{H}}_3}{\text{COO}}}\right)_2}$ and ${\text{N}}{{\text{a}}_2}{\text{S}}$  is as follows:

${\text{Pb}}{\left({{\text{C}}{{\text{H}}_3}{\text{COO}}}\right)_2}\left({aq} \right)+{\text{N}}{{\text{a}}_2}{\text{S}}\left( {aq}\right)\to{\text{PbS}}\left( s\right)+2{\text{C}}{{\text{H}}_3}{\text{COONa}}\left( {aq}\right)$

According to the solubility rules, PbS is an insoluble salt. The acetates of group 1A are soluble and therefore ${\text{C}}{{\text{H}}_3}{\text{COONa}}$ is soluble in water. Hence PbS will form precipitate in the above reaction.

Also in the solution, $0.1\;{\text{mol}}$ of ${\text{Pb}}{\left({{\text{C}}{{\text{H}}_3}{\text{COO}}}\right)_2}$ will react with $0.1\;{\text{mol}}$ of ${\text{CaC}}{{\text{l}}_2}$ to form $0.1\;{\text{mol}}$ of ${\text{PbC}}{{\text{l}}_2}$ and $0.1\;{\text{mol}}$ of ${\text{Ca}}{\left({{\text{C}}{{\text{H}}_3}{\text{COO}}}\right)_2}$ .

2. The reaction between  ${\text{Pb}}{\left({{\text{C}}{{\text{H}}_3}{\text{COO}}}\right)_2}$ and ${\text{CaC}}{{\text{l}}_2}$ is as follows:

${\text{Pb}}{\left( {{\text{C}}{{\text{H}}_3}{\text{COO}}} \right)_2}\left({aq}\right)+{\text{CaC}}{{\text{l}}_2}\left( {aq} \right)\to{\text{PbC}}{{\text{l}}_2}\left( s \right)+ {\text{Ca}}{\left({{\text{C}}{{\text{H}}_3}{\text{COO}}}\right)_2}\left( {aq}\right)$

According to the solubility rules, ${\text{PbC}}{{\text{l}}_2}$ is an insoluble salt. The acetates of group 2A are soluble and therefore ${\text{Ca}}{\left( {{\text{C}}{{\text{H}}_3}{\text{COO}}} \right)_2}$ is soluble in water. Hence ${\mathbf{PbC}}{{\mathbf{l}}_{\mathbf{2}}}$ will form precipitate in the above reaction.

Learn more:

1. Balanced chemical equation brainly.com/question/1405182

2. Oxidation and reduction reaction brainly.com/question/2973661

Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: Chemical reaction and equation

Keywords: precipitate, PbS, PbCl2, Na2S, CaCl2, PbCl2, Ca(CH3COO)2, Pb(CH3COO)2, CH3COONa, soluble, insoluble, solubility rules, solubility, precipitation reaction.