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Complete each of the following as a net ionic equation, indicating whether a precipitate forms. If no reaction occurs, so state. (a) \(\mathrm{Na}^{+}+\mathrm{Br}^{-}+\mathrm{Pb}^{2+}+2 \mathrm{NO}_{3}^{-} \longrightarrow\) (b) \(\mathrm{Mg}^{2+}+2 \mathrm{Cl}^{-}+\mathrm{Cu}^{2+}+\mathrm{SO}_{4}^{2-} \longrightarrow\) (c) \(\mathrm{Fe}^{3+}+3 \mathrm{NO}_{3}^{-}+\mathrm{Na}^{+}+\mathrm{OH}^{-} \longrightarrow\)

Short Answer

Expert verified
The net ionic equation for options (a), (b), and (c) are: (a) \( Pb^{2+} + 2 Br^{-} \longrightarrow PbBr2_{(s)} \), (b) No Reaction Occurs, and (c) No Reaction Occurs respectively.

Step by step solution

01

Determine if Formed Compounds are Soluble for (a)

For the reaction \(Na^{+} + Br^{-} + Pb^{2+} + 2 NO_{3}^{-}\), we have to check if the resultant lead bromide (PbBr2) and sodium nitrate (NaNO3) are soluble in water. From solubility rules, we know that most nitrates are soluble and most lead (II) compounds, except Pb(NO3)2 and Pb(C2H3O2)2, are insoluble.
02

Formulate Net Ionic Equation for (a)

Given that PbBr2 is insoluble and NaNO3 is soluble, the net ionic equation will be written as follows: \(Pb^{2+} + 2 Br^{-} \longrightarrow PbBr2_{(s)}\) indicating the formation of a precipitate.
03

Determine if Formed Compounds are Soluble for (b)

For the reaction \( Mg^{2+} + 2 Cl^{-} + Cu^{2+} + SO_{4}^{2-} \), we need to check if the resultant magnesium chlorate (MgCl2) and copper sulfate (CuSO4) are soluble. From solubility rules, we know that most sulfate salts are soluble, except for those with Ba^2+, Pb^2+, Hg2^2+, and Sr^2+. Chlorides are generally soluble, exceptions includes Ag+, Pb2+, and Hg2^2+.
04

Formulate Net Ionic Equation for (b)

Given that both MgCl2 and CuSO4 are soluble, no reaction occurs in this scenario, so we state that no reaction occurs.
05

Determine if Formed Compounds are Soluble for (c)

For the reaction \( Fe^{3+} + 3 NO_{3}^{-} + Na^{+} + OH^{-} \), we have to check if the resultant iron (III) nitrate (Fe(NO3)3) and sodium hydroxide (NaOH) are soluble. From solubility rules, we know that most hydroxides are insoluble, but group 1 element hydroxides (like NaOH) and Ba(OH)2 are soluble. Also, nitrates are generally soluble.
06

Formulate Net Ionic Equation for (c)

Given that NaOH and Fe(NO3)3 are soluble, no reaction occurs in this scenario, so we state that no reaction occurs.

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Key Concepts

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

Solubility Rules
Understanding solubility rules is crucial when dealing with chemical reactions in aqueous solutions. Solubility rules help predict whether a substance will dissolve in water or form an insoluble compound, known as a precipitate.
Generally speaking, certain rules apply:
  • Nitrates (NO3-) are soluble, meaning they dissolve in water.
  • Most chloride (Cl-) and bromide (Br-) salts are soluble, but there are notable exceptions, such as those combined with silver (Ag+), lead (Pb2+), and mercury (Hg22+).
  • Compounds containing alkali metal ions (such as Na+ and K+) and ammonium (NH4+) are generally soluble.
  • Hydroxides (OH-) are mostly insoluble, with the exceptions of those combined with group 1 elements and barium (Ba2+).
  • Sulfate (SO42-) salts tend to be soluble, however, exceptions include combinations with barium (Ba2+), calcium (Ca2+), and lead (Pb2+).
By applying these basic rules, we can determine whether the compounds that form in a reaction will remain dissolved or form a solid precipitate.
Precipitate Formation
A precipitate is a solid that forms in a solution when two aqueous solutions are mixed together and an insoluble product emerges. Precipitation reactions are a common occurrence in chemistry and can be quite dramatic as you see a solid appear seemingly out of nowhere.
To predict if a precipitate will form:
  • Identify the ions present in the given solutions.
  • Use solubility rules to determine if any of the ion combinations will result in an insoluble compound.
  • If an insoluble compound can form, a precipitate will occur.
For example, in the reaction involving lead ions (Pb2+) and bromide ions (Br-), the formation of lead bromide (PbBr2) is predicted due to its insolubility, thus indicating precipitate formation.
This solid is visible as a cloudiness or a layer of sediment in the solution.
Chemical Reactions
Chemical reactions involve the rearrangement of atoms to form new substances. They are represented by chemical equations which depict the reactants turning into products. In the context of aqueous reactions:
  • The reactants and products are often dissolved in water.
  • These solutions can be represented by complete ionic equations, demonstrating all ions present.
  • The net ionic equation, however, simplifies the reaction to only include species directly involved in the formation or dissolution of a precipitate.
For instance, in reaction (a) which involves sodium (Na+), bromide (Br-), lead (Pb2+), and nitrate ions (NO3-), the net ionic equation simplifies to only show the formation of lead bromide precipitate.
Chemical reactions offer insights into how substances interact, forming new materials, which can be useful in various applications from industrial processes to everyday products.

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Most popular questions from this chapter

Phosphorus is essential for plant growth, but an excess of phosphorus can be catastrophic in aqueous ecosystems. Too much phosphorus can cause algae to grow at an explosive rate and this robs the rest of the ecosystem of oxygen. Effluent from sewage treatment plants must be treated before it can be released into lakes or streams because the effluent contains significant amounts of \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\) and \(\mathrm{HPO}_{4}^{2-}\). (Detergents are a major contributor to phosphorus levels in domestic sewage because many detergents contain \(\mathrm{Na}_{2} \mathrm{HPO}_{4}\) ) A simple way to remove \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\) and \(\mathrm{HPO}_{4}^{2-}\) from the effluent is to treat it with lime, \(\mathrm{CaO}\) which produces \(\mathrm{Ca}^{2+}\) and \(\mathrm{OH}^{-}\) ions in water. The \(\mathrm{OH}^{-}\) ions convert \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\) and \(\mathrm{HPO}_{4}^{2-}\) ions into \(\mathrm{PO}_{4}^{3-}\) ions and, finally, \(\mathrm{Ca}^{2+}, \mathrm{OH}^{-}\), and \(\mathrm{PO}_{4}^{3-}\) ions combine to form a precipitate of \(\mathrm{Ca}_{5}\left(\mathrm{PO}_{4}\right)_{3} \mathrm{OH}(\mathrm{s})\) (a) Write balanced chemical equations for the four reactions described above. [Hint: The reactants are \(\mathrm{CaO}\) and \(\mathrm{H}_{2} \mathrm{O} ; \mathrm{H}_{2} \mathrm{PO}_{4}^{-}\) and \(\left.\mathrm{OH}^{-} ; \mathrm{HPO}_{4}^{2-} \text { and } \mathrm{OH}^{-} ; \mathrm{Ca}^{2+}, \mathrm{PO}_{4}^{3-}, \text { and } \mathrm{OH}^{-} .\right]\) (b) How many kilograms of lime are required to remove the phosphorus from a \(1.00 \times 10^{4}\) L holding tank filled with contaminated water, if the water contains \(10.0 \mathrm{mg}\) of phosphorus per liter?

Balance these equations for disproportionation reactions. (a) \(\mathrm{MnO}_{4}^{2-} \longrightarrow \mathrm{MnO}_{2}(\mathrm{s})+\mathrm{MnO}_{4}^{-}\) (basic solution) (b) \(\mathrm{P}_{4}(\mathrm{s}) \longrightarrow \mathrm{H}_{2} \mathrm{PO}_{2}^{-}+\mathrm{PH}_{3}(\mathrm{g})\) (basic solution) (c) \(\mathrm{S}_{8}(\mathrm{s}) \longrightarrow \mathrm{S}^{2-}+\mathrm{S}_{2} \mathrm{O}_{3}^{2-}\) (basic solution) (d) \(\mathrm{As}_{2} \mathrm{S}_{3}+\mathrm{H}_{2} \mathrm{O}_{2} \longrightarrow \mathrm{AsO}_{4}^{3-}+\mathrm{SO}_{4}^{2-}\)

Consider the following redox reaction: $$\begin{array}{r}4 \mathrm{NO}(\mathrm{g})+3 \mathrm{O}_{2}(\mathrm{g})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \longrightarrow 4 \mathrm{NO}_{3}^{-}(\mathrm{aq})+4 \mathrm{H}^{+}(\mathrm{aq})\end{array} $$ (a) Which species is oxidized? (b) Which species is reduced? (c) Which species is the oxidizing agent? (d) Which species is the reducing agent? (e) Which species gains electrons? (f) Which species loses electrons?

Balance these equations for redox reactions in basic solution. (a) \(\mathrm{MnO}_{2}(\mathrm{s})+\mathrm{ClO}_{3}^{-} \longrightarrow \mathrm{MnO}_{4}^{-}+\mathrm{Cl}^{-}\) (b) \(\mathrm{Fe}(\mathrm{OH})_{3}(\mathrm{s})+\mathrm{OCl}^{-} \longrightarrow \mathrm{FeO}_{4}^{2-}+\mathrm{Cl}^{-}\) (c) \(\mathrm{ClO}_{2} \longrightarrow \mathrm{ClO}_{3}^{-}+\mathrm{Cl}\) (d) \(\mathrm{Ag}(\mathrm{s})+\mathrm{CrO}_{4}^{2-} \rightarrow \mathrm{Ag}^{+}+\mathrm{Cr}(\mathrm{OH})_{3}(\mathrm{s})\)

Complete and balance these half-equations. (a) \(\mathrm{SO}_{3}^{2-} \longrightarrow \mathrm{S}_{2} \mathrm{O}_{3}^{2-}\) (acidic solution) (b) \(\mathrm{HNO}_{3} \longrightarrow \mathrm{N}_{2} \mathrm{O}(\mathrm{g})\) (acidic solution) (c) \(\mathrm{Al}(\mathrm{s}) \longrightarrow \mathrm{Al}(\mathrm{OH})_{4}^{-}\) (basic solution) Indicate whether oxidation or reduction is involved.

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