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Predict whether a precipitation reaction will occur in each of the following situations. Write a net ionic equation for each reaction that occurs. (a) \(\mathrm{NiCl}_{2}(a q)+\left(\mathrm{NH}_{4}\right)_{2} \mathrm{~S}(a q) \longrightarrow ?\) (b) \(\mathrm{Na}_{2} \mathrm{CrO}_{4}(a q)+\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}(a q) \longrightarrow ?\) (c) \(\mathrm{AgClO}_{4}(a q)+\mathrm{CaBr}_{2}(a q) \longrightarrow ?\) (d) \(\mathrm{ZnCl}_{2}(a q)+\mathrm{K}_{2} \mathrm{CO}_{3}(a q) \longrightarrow ?\)

Short Answer

Expert verified
Precipitation reactions occur in all cases, forming solids: NiS, PbCrO4, AgBr, and ZnCO3.

Step by step solution

01

Identify Possible Products for Reaction (a)

Reactants are \( \text{NiCl}_2 (aq) \) and \( (\text{NH}_4)_2 \text{S} (aq) \). The possible products of double displacement are \( \text{NiS} \) and \( \text{NH}_4\text{Cl} \). Check solubility of these products.
02

Determine Precipitate for Reaction (a)

According to solubility rules, \( \text{NH}_4\text{Cl} \) is soluble in water, while \( \text{NiS} \) is insoluble. Therefore, a precipitation reaction occurs forming \( \text{NiS} \).
03

Write Net Ionic Equation for Reaction (a)

For \( \text{NiCl}_2 \ + \ (\text{NH}_4)_2 \text{S} \), the net ionic equation is: \[ \text{Ni}^{2+} (aq) + \text{S}^{2-} (aq) \rightarrow \text{NiS} (s) \]
04

Identify Possible Products for Reaction (b)

Reactants are \( \text{Na}_2\text{CrO}_4 (aq) \) and \( \text{Pb}(\text{NO}_3)_2 (aq) \). Possible products are \( \text{NaNO}_3 \) and \( \text{PbCrO}_4 \). Check solubility of these products.
05

Determine Precipitate for Reaction (b)

\( \text{NaNO}_3 \) is soluble in water. However, \( \text{PbCrO}_4 \) is insoluble in water, thus a precipitation reaction occurs producing \( \text{PbCrO}_4 \).
06

Write Net Ionic Equation for Reaction (b)

For \( \text{Na}_2\text{CrO}_4 \ + \ \text{Pb}(\text{NO}_3)_2 \), the net ionic equation is: \[ \text{Pb}^{2+} (aq) + \text{CrO}_4^{2-} (aq) \rightarrow \text{PbCrO}_4 (s) \]
07

Identify Possible Products for Reaction (c)

Reactants are \( \text{AgClO}_4 (aq) \) and \( \text{CaBr}_2 (aq) \). Possible products are \( \text{Ca}(\text{ClO}_4)_2 \) and \( \text{AgBr} \). Check solubility of these products.
08

Determine Precipitate for Reaction (c)

\( \text{Ca}(\text{ClO}_4)_2 \) is soluble, whereas \( \text{AgBr} \) is insoluble in water. Thus, a precipitation reaction occurs forming \( \text{AgBr} \).
09

Write Net Ionic Equation for Reaction (c)

For \( \text{AgClO}_4 \ + \ \text{CaBr}_2 \), the net ionic equation is: \[ \text{Ag}^{+} (aq) + \text{Br}^{-} (aq) \rightarrow \text{AgBr} (s) \]
10

Identify Possible Products for Reaction (d)

Reactants are \( \text{ZnCl}_2 (aq) \) and \( \text{K}_2\text{CO}_3 (aq) \). Possible products are \( \text{ZnCO}_3 \) and \( \text{KCl} \). Check solubility of these products.
11

Determine Precipitate for Reaction (d)

\( \text{KCl} \) is soluble in water, but \( \text{ZnCO}_3 \) is insoluble. Thus, a precipitation reaction occurs forming \( \text{ZnCO}_3 \).
12

Write Net Ionic Equation for Reaction (d)

For \( \text{ZnCl}_2 \ + \ \text{K}_2\text{CO}_3 \), the net ionic equation is: \[ \text{Zn}^{2+} (aq) + \text{CO}_3^{2-} (aq) \rightarrow \text{ZnCO}_3 (s) \]

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Key Concepts

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

Solubility Rules
Solubility rules are a set of guidelines that help predict whether an ionic compound will dissolve in water or form a precipitate. These rules are important in identifying the outcome of reactions between aqueous solutions. Knowing which compounds are soluble helps determine if a double displacement reaction will result in a precipitate.
  • Most nitrate (NO鈧冣伝), acetate (C鈧侶鈧僌鈧傗伝), and ammonium (NH鈧勨伜) salts are soluble.
  • Salts containing alkali metal ions (like Li鈦, Na鈦, K鈦, etc.) are generally soluble.
  • Chlorides (Cl鈦), bromides (Br鈦), and iodides (I鈦) are usually soluble, except those of silver (Ag鈦), lead (Pb虏鈦), and mercury (Hg虏鈦).
  • Sulfates (SO鈧劼测伝) tend to be soluble, but there are exceptions including those of barium (Ba虏鈦), lead, calcium (Ca虏鈦), and strontium (Sr虏鈦).
  • Carbonates (CO鈧兟测伝), phosphates (PO鈧劼斥伝), sulfides (S虏鈦), and hydroxides (OH鈦) are mostly insoluble, with exceptions often involving alkali metals and ammonium.
These rules provide a quick check on the likely solubility of many common compounds and offer insight into which products of reactions may precipitate.
Net Ionic Equations
Net ionic equations are simplified representations of chemical reactions that show only the ions and molecules directly involved in creating the solid precipitate. By leaving out the spectator ions, which are ions that remain unchanged on both sides of a chemical equation, these equations focus on the formation of the precipitate.
Here's how to write a net ionic equation:
  • Write the balanced molecular equation for the reaction.
  • Dissociate all strong electrolytes (compounds that fully ionize in solution) into their respective ions.
  • Cancel out the spectator ions on both sides of the equation, leaving only the ions that form the precipitate.
  • Write the remaining ions to create the net ionic equation.
Net ionic equations are useful for understanding the core chemical changes occurring in a reaction, especially in solutions where multiple reactions might take place.
Double Displacement Reactions
Double displacement reactions, also known as metathesis reactions, occur when the cations and anions of two different compounds exchange places. This type of reaction often takes place in aqueous solutions and sometimes results in the formation of an insoluble salt, which precipitates out of the solution.
In general, a double displacement reaction can be represented as follows:\[AB + CD \rightarrow AD + CB\]where A and C are cations, and B and D are anions. The key to predicting whether a precipitation reaction will occur lies in the solubility of the products. If one of the products is insoluble according to the solubility rules, a precipitate will form.
For example:In the reaction between Na鈧侰rO鈧 and Pb(NO鈧)鈧, PbCrO鈧 is insoluble, thus leading to a precipitation reaction:
  • The possible products are NaNO鈧 and PbCrO鈧.
  • PbCrO鈧 is insoluble, and this results in the formation of a solid precipitate.
These reactions highlight the importance of understanding solubility and using it to predict the formation of new products in chemical reactions.

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Most popular questions from this chapter

What is the molar concentration of As(III) in a solution if \(22.35 \mathrm{~mL}\) of \(0.100 \mathrm{M} \mathrm{KBr} \mathrm{O}_{3}\) is needed for complete reaction with \(50.00 \mathrm{~mL}\) of the As(III) solution? The balanced equation is: \(3 \mathrm{H}_{3} \mathrm{AsO}_{3}(a q)+\mathrm{BrO}_{3}^{-}(a q) \longrightarrow \mathrm{Br}^{-}(a q)+3 \mathrm{H}_{3} \mathrm{AsO}_{4}(a q)\)

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