Question

When each of the following pairs of aqueous solutions is mixed, does a precipitation reaction occur? If so, write balanced molecular, total ionic, and net ionic equations: (a) Sodium sulfide + nickel(II) sulfate (b) Lead(II) nitrate + potassium bromide

Short answer

Yes, precipitates form in both reactions. NiS forms in (a) and PbBr₂ forms in (b).

Step-by-step solution

Identify the reactants

The reactants for part (a) are sodium sulfide (Na₂S) and nickel(II) sulfate (NiSO₄). The reactants for part (b) are lead(II) nitrate (Pb(NO₃)₂) and potassium bromide (KBr).

Determine possible products

For part (a), mixing Na₂S and NiSO₄ could yield sodium sulfate (Na₂SO₄) and nickel(II) sulfide (NiS). For part (b), mixing Pb(NO₃)₂ and KBr could yield lead(II) bromide (PbBr₂) and potassium nitrate (KNO₃).

Check solubility of products

Use the solubility rules to determine if any of the products are insoluble and will precipitate. NiS is insoluble, so a precipitate will form in part (a). PbBr₂ is also insoluble, so a precipitate will form in part (b).

Write the balanced molecular equations

For part (a): Na₂S (aq) + NiSO₄ (aq) → Na₂SO₄ (aq) + NiS (s)For part (b): Pb(NO₃)₂ (aq) + 2 KBr (aq) → 2 KNO₃ (aq) + PbBr₂ (s)

Write the total ionic equations

For part (a): 2 Na⁺ (aq) + S²⁻ (aq) + Ni²⁺ (aq) + SO₄²⁻ (aq) → 2 Na⁺ (aq) + SO₄²⁻ (aq) + NiS (s)For part (b): Pb²⁺ (aq) + 2 NO₃⁻ (aq) + 2 K⁺ (aq) + 2 Br⁻ (aq) → 2 K⁺ (aq) + 2 NO₃⁻ (aq) + PbBr₂ (s)

Write the net ionic equations

For part (a): Ni²⁺ (aq) + S²⁻ (aq) → NiS (s)For part (b): Pb²⁺ (aq) + 2 Br⁻ (aq) → PbBr₂ (s)

Key concepts

Solubility Rules

Solubility rules help us determine whether a substance will dissolve in water. These rules are essential when predicting the formation of a precipitate in a chemical reaction.

Here are some general solubility rules you should remember:

• Most nitrates (NO₃⁻), acetates (CH₃COO⁻), and perchlorates (ClO₄⁻) are soluble.
• Most alkali metal salts (like Na⁺, K⁺) and ammonium (NH₄⁺) are soluble.
• Chlorides (Cl⁻), bromides (Br⁻), and iodides (I⁻) are generally soluble, except when paired with silver (Ag⁺), mercury (Hg₂²⁺), or lead (Pb²⁺).
• Sulfates (SO₄²⁻) are usually soluble, but barium (Ba²⁺), calcium (Ca²⁺), and lead (Pb²⁺) sulfates are exceptions.
• Most carbonates (CO₃²⁻), phosphates (PO₄³⁻), and sulfides (S²⁻) are insoluble, except for those of alkali metals and ammonium.

Using these rules, you can predict whether a precipitate will form when two solutions are mixed.
In our example, NiS and PbBr₂ are precipitates because they don't dissolve in water.

Molecular Equation

A molecular equation shows the reactants and products of a reaction in their complete and neutral forms. It’s useful for providing an overall picture of the reaction.

For Example:

• For Sodium sulfide and Nickel(II) sulfate: Na₂S (aq) + NiSO₄ (aq) → Na₂SO₄ (aq) + NiS (s)
• For Lead(II) nitrate and Potassium bromide: Pb(NO₃)₂ (aq) + 2 KBr (aq) → 2 KNO₃ (aq) + PbBr₂ (s)

Notice that all components are shown as complete compounds. The states of matter (aqueous 'aq', solid 's') are indicated, which helps in identifying the precipitate.

Total Ionic Equation

A total ionic equation shows all the soluble ionic substances dissociated into their ions. This provides more detail about what happens during the reaction.

For Example:

• For Sodium sulfide and Nickel(II) sulfate: 2 Na⁺ (aq) + S²⁻ (aq) + Ni²⁺ (aq) + SO₄²⁻ (aq) → 2 Na⁺ (aq) + SO₄²⁻ (aq) + NiS (s)
• For Lead(II) nitrate and Potassium bromide: Pb²⁺ (aq) + 2 NO₃⁻ (aq) + 2 K⁺ (aq) + 2 Br⁻ (aq) → 2 K⁺ (aq) + 2 NO₃⁻ (aq) + PbBr₂ (s)

Here, we see the ions in the solution before they react to form the precipitate. Substances that don't participate in the reaction (i.e., spectator ions) look unchanged.

Net Ionic Equation

A net ionic equation removes the spectator ions, showing only the ions that participate in forming the precipitate. This gives a clearer understanding of the actual chemical change.

For Example:

• For Sodium sulfide and Nickel(II) sulfate: Ni²⁺ (aq) + S²⁻ (aq) → NiS (s)
• For Lead(II) nitrate and Potassium bromide: Pb²⁺ (aq) + 2 Br⁻ (aq) → PbBr₂ (s)

By removing the spectator ions (like Na⁺ and SO₄²⁻ for the first reaction), we focus on the change that occurs to form the solid precipitate. Understanding these reactions is crucial in predicting and explaining chemical behavior in solutions!