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Solubility rules are essential guidelines that help predict whether a compound will dissolve in water. These rules are based on empirical data and experience, allowing chemists to foresee the outcomes of mixing solutions.
When you mix two solutions, the ions may combine to form new compounds. Solubility rules help determine if these new compounds will stay dissolved or form a precipitate, which is a solid material that falls out of the solution.
Here are some general solubility rules you can rely on:

• Nitrates (NO鈧冣伝) are always soluble.
• Sulfates (SO鈧劼测伝) are generally soluble, with exceptions for barium, lead, and calcium.
• Sulphides (S虏鈦�) are generally insoluble except when paired with alkali metals or ammonium.
• Most chloride (Cl鈦�), bromide (Br鈦�), and iodide (I鈦�) salts are soluble, except those of silver, lead, and mercury.

By using these rules, you can quickly decide if a precipitation reaction is likely when mixing two solutions. Just like in our exercise! Sodium sulfide and iron(III) nitrate react, and by applying the rules, we predict iron(III) sulfide precipitates because it's generally insoluble.

Net ionic equations provide a simplified version of a chemical reaction by showing only the ions that participate in forming a product. Here's how you can write one:
First, you list all ions present in the reaction. Then, you identify and eliminate the spectator ions鈥攖hese are ions that do not change during the reaction.
In the example of sodium sulfide and iron(III) nitrate, we start by writing the complete ionic equation:

• For sodium sulfide: 2 Na鈦� and S虏鈦�
• For iron(III) nitrate: Fe鲁鈦� and 3 NO鈧冣伝
• The products will include Na鈦�, NO鈧冣伝, and the solid Fe鈧係鈧�

The spectator ions here are sodium (Na鈦�) and nitrate (NO鈧冣伝) because they don鈥檛 form the solid product.
Thus, the net ionic equation, which shows only the ions forming the precipitate, is:\[\text{Fe}^{3+} + 3 \, \text{S}^{2-} \rightarrow \text{Fe}_{2}\text{S}_{3}\]

The molecular equation represents the complete chemical equation using the molecular formulas of the reactants and products. It shows the overall reaction happening between the substances.
For our exercise, when sodium sulfide reacts with iron(III) nitrate, you write everything using their chemical formulas, without splitting them into ions.\[ 2 \, \text{Na}_{2}\text{S} + \text{Fe}(\text{NO}_{3})_{3} \rightarrow 4 \, \text{NaNO}_{3} + \text{Fe}_{2}\text{S}_{3} \]
This equation provides an overview of the reaction, making it easy to see what starts and what ends the process. It is crucial as it sets the groundwork for deriving both complete and net ionic equations.
In this reaction, we establish that sodium sulfide and iron (III) nitrate transform into sodium nitrate and the insoluble iron(III) sulfide, indicating a precipitation occurs.

Ions are charged particles that form when atoms gain or lose electrons. They play a key role in chemical reactions, especially in aqueous solutions where they're often responsible for conductivity and reactivity.
In an ionic compound, these particles either carry a positive charge, making them cations, or a negative charge, turning them into anions.

• Examples of cations include Na鈦� and Fe鲁鈦�.
• Common anions are S虏鈦� and NO鈧冣伝.

When these ions dissolve in water, they separate from their compound and move freely, allowing them to engage in reactions.
In precipitation reactions, specific cations and anions in the solution come together to form an insoluble compound. Like in our problem, where the Fe鲁鈦� (from iron(III) nitrate) and S虏鈦� (from sodium sulfide) ions combine to form the insoluble iron(III) sulfide precipitate.