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Chapter 4: Problem 43

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 nitrate \(+\) copper(II) sulfate (b) Ammonium bromide + silver nitrate

Short Answer

Expert verified
No precipitation in (a). Precipitation occurs in (b) with the net ionic equation: Br⁻ (aq) + Ag⁺ (aq) → AgBr (s).

Step by step solution

01

- Determine solubility rules

Check the solubility rules for each of the ions involved in the given pairs of solutions to determine if any insoluble compounds (precipitates) will form when they are mixed.
02

- Analyzing pair (a): Sodium nitrate + copper(II) sulfate

Sodium nitrate (NaNO₃) and copper(II) sulfate (CuSO₄) are both soluble in water. When mixed, the possible products are Na₂SO₄ and Cu(NO₃)₂. According to solubility rules, both of these products are soluble in water, so no precipitation reaction occurs.
03

- Molecular equation for pair (a)

Since there is no precipitation reaction, the molecular equation is simply: NaNO₃ (aq) + CuSO₄ (aq) → No reaction.
04

- Analyzing pair (b): Ammonium bromide + silver nitrate

Ammonium bromide (NH₄Br) and silver nitrate (AgNO₃) are both soluble in water. When mixed, the possible products are NH₄NO₃ and AgBr. According to the solubility rules, AgBr is not soluble in water, so it will form a precipitate. A precipitation reaction occurs.
05

- Molecular equation for pair (b)

Write the balanced molecular equation: NH₄Br (aq) + AgNO₃ (aq) → NH₄NO₃ (aq) + AgBr (s).
06

- Total ionic equation for pair (b)

Write the total ionic equation by separating all aqueous compounds into their ions: NH₄⁺ (aq) + Br⁻ (aq) + Ag⁺ (aq) + NO₃⁻ (aq) → NH₄⁺ (aq) + NO₃⁻ (aq) + AgBr (s).
07

- Net ionic equation for pair (b)

Remove the spectator ions (NH₄⁺ and NO₃⁻) to write the net ionic equation: Br⁻ (aq) + Ag⁺ (aq) → AgBr (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 essential for predicting the solubility of compounds in water.
These rules help us determine whether a compound will dissolve (be soluble) or form a precipitate (be insoluble).
Some basic solubility rules include:
  • Nitrates \((NO_3^-)\) are always soluble.
  • Most salts containing alkali metal ions \((Li^+, Na^+, K^+, etc.)\) and ammonium \((NH_4^+)\) ions are soluble.
  • Most chlorides \((Cl^-)\), bromides \((Br^-)\), and iodides \((I^-)\) are soluble, except those of silver \((Ag^+)\), lead \((Pb^{2+})\), and mercury \((Hg_2^{2+})\).
  • Most sulfates \((SO_4^{2-})\) are soluble, except for barium sulfate \((BaSO_4)\), calcium sulfate \((CaSO_4)\), and lead sulfate \((PbSO_4)\).
  • Most carbonates \((CO_3^{2-})\), phosphates \((PO_4^{3-})\), and sulfides \((S^{2-})\) are generally insoluble, except for those of alkali metals and ammonium.
Knowing these rules helps you quickly figure out if a reaction will produce a precipitate when two solutions are mixed.
It simplifies the process of identifying the products in an aqueous reaction.
molecular equations
A molecular equation is a balanced chemical equation where the compounds are written as molecules, not ions.
It represents the reactants and products as if they were intact, undissociated chemicals.
For example, in the reaction between ammonium bromide and silver nitrate:
  • NH\(_4\)Br (aq) + AgNO\(_3\) (aq) → NH\(_4\)NO\(_3\) (aq) + AgBr (s)
This equation tells us that ammonium bromide and silver nitrate react to form ammonium nitrate and silver bromide.
Here's a breakdown:
  • NH\(_4\)Br and AgNO\(_3\) are the reactants,
  • NH\(_4\)NO\(_3\) and AgBr are the products,
  • (aq) indicates the substance is dissolved in water, and
  • (s) indicates a solid precipitate.
By looking at the molecular equation, we understand the overall process of the reaction without focusing on the specifics of the ions.
total ionic equations
A total ionic equation shows the dissociated ions of soluble compounds in a reaction.
It is useful for understanding the behavior of individual ions during the reaction.
For the reaction between ammonium bromide and silver nitrate, the total ionic equation is:
\( NH_4^+ (aq) + Br^- (aq) + Ag^+ (aq) + NO_3^- (aq) \rightarrow NH_4^+ (aq) + NO_3^- (aq) + AgBr (s) \)
In this equation:
  • NH\(_4^+\), Br\(^-\), Ag\(^+\), and NO\(_3^-\) are the individual ions in the reactants,
  • NH\(_4^+\) and NO\(_3^-\) are also present in the products, indicating they don't participate directly in the formation of the precipitate
  • AgBr (s) is the solid precipitate formed.
This version of the equation shows which ions are present in the solution and which compound precipitates, making it clear that the only product forming a solid is AgBr.
net ionic equations
Net ionic equations highlight the ions directly involved in the formation of a precipitate.
They exclude the spectator ions, which do not change during the reaction.
For the reaction between ammonium bromide and silver nitrate, the net ionic equation is:
\( Br^- (aq) + Ag^+ (aq) \rightarrow AgBr (s) \)
The spectator ions NH\(_4^+\) and NO\(_3^-\) are omitted because they don't participate in the formation of the precipitate.
This net ionic equation simplifies the reaction to its most fundamental components, showing only the ions that form the precipitate:
  • Br\(^-\) ions from ammonium bromide,
  • Ag\(^+\) ions from silver nitrate,
  • AgBr (s) as the precipitate.
By focusing on the ions that change state, net ionic equations help you understand the essence of the chemical reaction.

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Most popular questions from this chapter

The amount of ascorbic acid (vitamin \(\left.\mathrm{C}, \mathrm{C}_{6} \mathrm{H}_{8} \mathrm{O}_{6}\right)\) in tablets is determined by reaction with bromine and then titration of the hydrobromic acid with standard base: $$ \begin{array}{l} \mathrm{C}_{6} \mathrm{H}_{8} \mathrm{O}_{6}(a q)+\mathrm{Br}_{2}(a q) \longrightarrow \mathrm{C}_{6} \mathrm{H}_{6} \mathrm{O}_{6}(a q)+2 \mathrm{HBr}(a q) \\ \mathrm{HBr}(a q)+\mathrm{NaOH}(a q) \longrightarrow \mathrm{NaBr}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \end{array} $$ A certain tablet is advertised as containing \(500 \mathrm{mg}\) of vitamin \(\mathrm{C}\). One tablet was dissolved in water and reacted with \(\mathrm{Br}_{2}\). The solution was then titrated with \(43.20 \mathrm{~mL}\) of \(0.1350 \mathrm{M} \mathrm{NaOH}\). Did the tablet contain the advertised quantity of vitamin C?

Over time, as their free fatty acid (FFA) content increases, edible fats and oils become rancid. To measure rancidity, the fat or oil is dissolved in ethanol, and any FFA present is titrated with KOH dissolved in ethanol. In a series of tests on olive oil, a stock solution of \(0.050 \mathrm{M}\) ethanolic \(\mathrm{KOH}\) was prepared at \(25^{\circ} \mathrm{C},\) stored at \(0^{\circ} \mathrm{C},\) and then placed in a \(100-\mathrm{mL}\) buret to titrate oleic acid [an FFA with formula \(\left.\mathrm{CH}_{3}\left(\mathrm{CH}_{2}\right)_{7} \mathrm{CH}=\mathrm{CH}\left(\mathrm{CH}_{2}\right)_{7} \mathrm{COOH}\right]\) in the oil. Each of four \(10.00-\mathrm{g}\) samples of oil took several minutes to titrate: the first required \(19.60 \mathrm{~mL}\), the second \(19.80 \mathrm{~mL},\) and the third and fourth \(20.00 \mathrm{~mL}\) of the ethanolic \(\mathrm{KOH}\). (a) What is the apparent acidity of each sample, in terms of mass \(\%\) of oleic acid? (Note: As the ethanolic KOH warms in the buret, its volume increases by a factor of \(0.00104 /{ }^{\circ} \mathrm{C}\).) (b) Is the variation in acidity a random or systematic error? Explain. (c) What is the actual acidity? How would you demonstrate this?

Identify the oxidizing and reducing agents in the following reactions: $$ \begin{array}{l} \text { (a) } 5 \mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}(a q)+2 \mathrm{MnO}_{4}^{-}(a q)+6 \mathrm{H}^{*}(a q) \longrightarrow \\ 2 \mathrm{Mn}^{2+}(a q)+10 \mathrm{CO}_{2}(g)+8 \mathrm{H}_{2} \mathrm{O}(l) \end{array} $$ (b) \(3 \mathrm{Cu}(s)+8 \mathrm{H}^{+}(a q)+2 \mathrm{NO}_{3}^{-}(a q) \longrightarrow\) $$ 3 \mathrm{Cu}^{2+}(a q)+2 \mathrm{NO}(g)+4 \mathrm{H}_{2} \mathrm{O}(l) $$

How many grams of iron(III) sulfide form when \(62.0 \mathrm{~mL}\) of \(0.135 \mathrm{M}\) iron(III) chloride reacts with \(45.0 \mathrm{~mL}\) of \(0.285 \mathrm{M}\) calcium sulfide?

Give the oxidation number of arsenic in each of the following: (a) \(\mathrm{AsH}_{3}\) (b) \(\mathrm{H}_{2} \mathrm{~A} \mathrm{~s} \mathrm{O}_{4}^{-}\) (c) \(\mathrm{AsCl}_{3}\)
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