Question

Write the balanced chemical equation, write the overall ionic equation, identify the spectator ions and possible precipitates, and write the net ionic equation for each of the following reactions.

Mercury (II) Chloride (aq) + Potassium sulfide (aq) = ???

Answer 1

Mercury (II) Chloride reacts with Potassium sulfide to form Mercury (II) sulfide as a precipitate. The net ionic equation is Hg2+(aq) + S2−(aq) → HgS(s), with potassium and chloride ions acting as spectator ions.

Step-by-step explanation:

When Mercury (II) Chloride (aq) reacts with Potassium sulfide (aq), a precipitation reaction occurs. The balanced chemical equation for the reaction is:

HgCl2(aq) + K2S(aq) → HgS(s) + 2KCl(aq)

The complete ionic equation would break all soluble ionic compounds into their constituent ions:

Hg2+(aq) + 2Cl−(aq) + 2K+(aq) + S2−(aq) → HgS(s) + 2K+(aq) + 2Cl−(aq)

The spectator ions here are the potassium and chloride ions, as they do not participate in forming the precipitate. The possible precipitate formed is Mercury (II) sulfide (HgS), which is insoluble in water. Thus, the net ionic equation, which shows only the species that undergo a chemical change, is: Hg2+(aq) + S2−(aq) → HgS(s)

Answer 2

Balanced chemical equation:

`HgCl_2_(_a_q_)~+~K_2S_(_a_q_)->~2KCl_(_a_q_)~+~HgS_(_s_)`

Overall ionic equation:

`Hg^+^2_(_a_q_)~+~Cl^-^1~_(_a_q_)~+~K^+^1_(_a_q_)~+~S^-^2_(_a_q_)->~K^+^1_(_a_q_)~+~Cl^-^1_(_a_q_)~+~HgS_(_s_)`

Spectator ions

`Cl^-^1~_(_a_q_)`

`K^+^1_(_a_q_)`

Possible precipitates

`HgS_(_s_)`

Net ionic equation

`Hg^+^2_(_a_q_)~+~S^-^2_(_a_q_)->~HgS_(_s_)`

Step-by-step explanation:

For the chemical equation we will have:

`HgCl_2_(_a_q_)~+~K_2S_(_a_q_)->~2KCl_(_a_q_)~+~HgS_(_s_)`

If we follow the solubility rules, we can find the state of the products KCl and HgS. For the case of KCl we will have an aqueosus state since all the salts of
`K^+` are soluble. For HgS we will have a solid state because all the sulfide salts are non-soluble (except the ones made of
`K^+~Na^+~NH_4^+`).

For the overall ionic equation we will have:

`Hg^+^2_(_a_q_)~+~Cl^-^1~_(_a_q_)~+~K^+^1_(_a_q_)~+~S^-^2_(_a_q_)->~K^+^1_(_a_q_)~+~Cl^-^1_(_a_q_)~+~HgS_(_s_)`

We have to keep in mind that the compounds that can be broken into his ions are the ones that have the aqueous state (aq).

For the spectator ions we have to search the compounds that are in both sides:

`Cl^-^1~_(_a_q_)`

`K^+^1_(_a_q_)`

For the net ionic equation we have to remove the spectator ions:

`Hg^+^2_(_a_q_)~+~S^-^2_(_a_q_)->~HgS_(_s_)`