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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) Potassium chloride + iron(III) nitrate (b) Ammonium sulfate + barium chloride

Short Answer

Expert verified
No precipitate forms in (a). In (b), the precipitate is BaSO鈧.

Step by step solution

01

- Identify the reactants

For each pair of solutions, identify the chemical formulas. (a) Potassium chloride: KCl, Iron(III) nitrate: Fe(NO鈧)鈧 (b) Ammonium sulfate: (NH鈧)鈧係O鈧, Barium chloride: BaCl鈧
02

- Determine possible products

When mixing the solutions, use double displacement to determine the products:(a) KCl + Fe(NO鈧)鈧 鈫 KNO鈧 + FeCl鈧 (b) (NH鈧)鈧係O鈧 + BaCl鈧 鈫 BaSO鈧 + 2NH鈧凜l
03

- Solubility rules

Apply the solubility rules to determine if any precipitate forms:(a) Both KNO鈧 and FeCl鈧 are soluble in water, so no precipitation reaction.(b) BaSO鈧 is insoluble in water, so it forms a precipitate while NH鈧凜l remains soluble.
04

- Write balanced molecular equation

Write the balanced molecular equation for (b): (NH鈧)鈧係O鈧(aq) + BaCl鈧(aq) 鈫 BaSO鈧(s) + 2NH鈧凜l(aq)
05

- Write total ionic equation

Write the total ionic equation by splitting aqueous compounds into ions:2NH鈧勨伜(aq) + SO鈧劼测伝(aq) + Ba虏鈦(aq) + 2Cl鈦(aq) 鈫 BaSO鈧(s) + 2NH鈧勨伜(aq) + 2Cl鈦(aq)
06

- Identify and write net ionic equation

Cancel out the spectator ions (ions that appear on both sides) to find the net ionic equation:SO鈧劼测伝(aq) + Ba虏鈦(aq) 鈫 BaSO鈧(s)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Solubility Rules
When working with precipitation reactions, understanding solubility rules is crucial. Solubility rules help predict whether a compound will dissolve in water (aqueous) or form a solid precipitate. Most compounds containing alkali metal ions (like potassium or sodium) and ammonium ion (NH鈧勨伜) are soluble. Nitrates (NO鈧冣伝), acetates (C鈧侶鈧僌鈧傗伝), and most chlorates (ClO鈧冣伝) are also generally soluble. However, some compounds are insoluble. For example, sulfates (SO鈧劼测伝) are generally soluble except for those of barium (Ba虏鈦), lead (Pb虏鈦), and calcium (Ca虏鈦). Chlorides are typically soluble, but exceptions like silver chloride (AgCl) exist. Knowing these rules helps determine if a precipitate forms during a chemical reaction.
Double Displacement Reactions
A double displacement reaction, or metathesis, involves the exchange of ions between two compounds. These reactions generally occur in aqueous solutions. The general form is:

AB + CD 鈫 AD + CB.

Here, the cations A and C swap places. For example, when mixing potassium chloride (KCl) and iron(III) nitrate (Fe(NO鈧)鈧), the reaction forms potassium nitrate (KNO鈧) and iron(III) chloride (FeCl鈧):

KCl + Fe(NO鈧)鈧 鈫 KNO鈧 + FeCl鈧.

A double displacement reaction might yield a precipitate if one of the products is insoluble based on solubility rules. In our example, no precipitate forms since both KNO鈧 and FeCl鈧 are soluble. However, in the reaction between ammonium sulfate ((NH鈧)鈧係O鈧) and barium chloride (BaCl鈧), barium sulfate (BaSO鈧) forms as a precipitate because it is insoluble in water.
Net Ionic Equations
Net ionic equations simplify reactions by showing only the species that change during the reaction. They help focus on the formation of the precipitate. To write a net ionic equation:
  • Write the balanced molecular equation. For ammonium sulfate and barium chloride, it is:

    (NH鈧)鈧係O鈧(aq) + BaCl鈧(aq) 鈫 BaSO鈧(s) + 2NH鈧凜l(aq).

  • Split aqueous compounds into their ions to write the total ionic equation:

    2NH鈧勨伜(aq) + SO鈧劼测伝(aq) + Ba虏鈦(aq) + 2Cl鈦(aq) 鈫 BaSO鈧(s) + 2NH鈧勨伜(aq) + 2Cl鈦(aq).

  • Identify spectator ions (ions that appear on both sides) and cancel them out. Here, NH鈧勨伜 and Cl鈦 are spectator ions.
  • The net ionic equation is:

    SO鈧劼测伝(aq) + Ba虏鈦(aq) 鈫 BaSO鈧(s).

This equation highlights the formation of the barium sulfate precipitate, which is the key change in the reaction.

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Most popular questions from this chapter

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Write two sets of equations (both molecular and total ionic) with different reactants that have the same net ionic equation as the following equation: $$ \mathrm{Ba}\left(\mathrm{NO}_{0}\right)_{2}(a q)+\mathrm{Na}_{2} \mathrm{CO}_{3}(a q) \longrightarrow \mathrm{BaCO}_{3}(s)+2 \mathrm{NaNO}_{3}(a q) $$

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