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Chapter 1: Problem 3

If equimolar solutions of \(\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}\) and \(\mathrm{NaCl}\) are mixed, which ion will not be present in significant amounts in the resulting solution after equilibrium is established? (A) \(\mathrm{Pb}^{2+}\) (B) \(\mathrm{NO}_{3}^{-}\) (C) \(\mathrm{Na}^{+}\) (D) \(\mathrm{Cl}^{-}\)

Short Answer

Expert verified
The ion not present in significant amounts in the solution after equilibrium is established is (A) \(Pb^{2+}\).

Step by step solution

01

Identify the chemical equation

Firstly, write down the chemical equation of the reaction. When Pb(NO3)2 and NaCl react together, they undergo a double replacement reaction forming PbCl2 and NaNO3. Hence, the chemical equation becomes: Pb(NO3)2 + 2NaCl 鈫� PbCl2 + 2NaNO3.
02

Understand the solubility

According to solubility rules, most nitrate salts are soluble, and most chloride salts are soluble, except for lead(II) chloride (PbCl2). Therefore, in this chemical reaction, Pb(NO3)2 and NaCl are soluble and they dissociate completely. However, PbCl2 is not soluble, and it precipitates out of the solution.
03

Analyze the ions at equilibrium

After reaching the equilibrium, the ions present in the solution will be: Cl-, Na+, NO3-. These ions are from soluble compounds, NaNO3 and remaining NaCl and Pb(NO3)2. Pb2+, from PbCl2, will not be present in significant amounts because PbCl2 is insoluble, and it remains as a precipitate, not an ion in the solution.

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Key Concepts

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

Solubility Rules
Solubility rules help us predict whether a compound will dissolve in water or form a precipitate. These rules are crucial in determining the outcome of reactions involving ionic compounds. Here are some general guidelines:
  • Nitrate (\(\text{NO}_3^-\)) salts and most sodium (\(\text{Na}^+\)) salts are soluble.
  • Chloride (\(\text{Cl}^-\)) salts are mainly soluble, except for those containing lead (\(\text{Pb}^{2+}\)) and silver (\(\text{Ag}^{+}\)), which are usually insoluble.
In the reaction given, \(\text{Pb(NO}_3)_2\) and \(\text{NaCl}\) dissolve because of their high solubility, but when they form \(\text{PbCl}_2\), this compound precipitates as it is insoluble in water. So, solubility rules are essential to identify which compounds remain dissolved and which precipitate.
Precipitation Reactions
In chemistry, precipitation reactions occur when two soluble salts react in solution to form one or more insoluble products, called precipitates. This process is a type of double replacement reaction.
When \(\text{Pb(NO}_3)_2\) and \(\text{NaCl}\) are mixed, the reaction leads to the formation of \(\text{PbCl}_2\), which is insoluble and precipitates out of the solution.
  • PbCl2 is the precipitate because it does not dissolve in water.
  • NaNO3 remains soluble and stays in the solution.
These reactions are crucial for identifying and removing unwanted ions from a solution or for obtaining a specific compound.
Ionic Compounds
Ionic compounds are formed from the electrostatic attraction between cations and anions. These compounds tend to dissolve in water, dissociating into their constituent ions, but their solubility can vary greatly.
  • A simple rule of thumb is that compounds of \(\text{Na}^+\), \(\text{K}^+\), and \(\text{NH}_4^+\) are generally soluble.
  • However, compounds like \(\text{PbCl}_2\) are exceptions鈥攖hey do not dissolve in water.
Understanding the nature of ionic compounds helps in predicting whether a precipitation reaction will occur and what ions will be present in the solution.
Chemical Equations
Chemical equations represent the reactants and products in a chemical reaction using element symbols and chemical formulas. They are balanced to obey the law of conservation of mass, which states that matter cannot be created or destroyed.
For instance, the reaction: \(\text{Pb(NO}_3)_2 + 2\text{NaCl} \to \text{PbCl}_2 + 2\text{NaNO}_3\) illustrates how reactants \(\text{Pb(NO}_3)_2\) and \(\text{NaCl}\) convert to products \(\text{PbCl}_2\) and \(\text{NaNO}_3\).
  • The coefficients (e.g., 2 in front of \(\text{NaCl}\)) ensure the equation is balanced, showing equal numbers of each type of atom on both sides.
Balancing chemical equations is a key skill in chemistry, allowing you to predict the amounts of substances consumed and produced in a reaction.

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Most popular questions from this chapter

The ionization energies for an element are listed in the table below. \(\begin{array}{lllll}{\text { First }} & {\text { Second }} & {\text { Third }} & {\text { Fourth }} & {\text { Fifth }} \\ {8 \mathrm{eV}} & {15 \mathrm{eV}} & {80 \mathrm{eV}} & {109 \mathrm{eV}} & {141 \mathrm{eV}}\end{array}\) Based on the ionization energy table, the element is most likely to be (A) Sodium (B) Magnesium (C) Aluminum (D) Silicon

A 22.0 gram sample of an unknown gas occupies 11.2 liters at standard temperature and pressure. Which of the following could be the identity of the gas? (A) \(\mathrm{CO}_{2}\) (B) \(\mathrm{SO}_{3}\) (C) \(\mathrm{O}_{2}\) (D) He

An electron from which peak would have the greatest velocity after ejection? (A) The peak at 104 \(\mathrm{MJ} / \mathrm{mol}\) (B) The peak at 6.84 \(\mathrm{MJ} / \mathrm{mol}\) (C) The peak at 4.98 \(\mathrm{MJ} / \mathrm{mol}\) (D) The peak at 1.76 \(\mathrm{MJ} / \mathrm{mol}\)

Use the following information to answer questions 25-28. A voltaic cell is created using the following half-cells: \(\begin{array}{ll}{\mathrm{Cr}^{3+}+3 e \rightarrow \mathrm{Cr}(s)} & {E^{\circ}=-0.41 \mathrm{V}} \\ {\mathrm{Pb}^{2+}+2 e \rightarrow \mathrm{Pb}(s)} & {E^{\circ}=-0.12 \mathrm{V}}\end{array}\) The concentrations of the solutions in each half-cell are 1.0 M. Which net ionic equation below represents a possible reaction that takes place when a strip of magnesium metal is oxidized by a solution of chromium (III) nitrate? (A) \(\operatorname{Mg}(s)+\operatorname{Cr}\left(\mathrm{NO}_{3}\right)_{3}(a q) \rightarrow \mathrm{Mg}^{2+}(a q)+\mathrm{Cr}^{3+}(a q)+3 \mathrm{NO}_{3}^{-}(a q)\) (B) \(3 \mathrm{Mg}(s)+2 \mathrm{Cr}^{3+} \rightarrow 3 \mathrm{Mg}^{2+}+2 \mathrm{Cr}(s)\) (C) \(\mathrm{Mg}(s)+\mathrm{Cr}^{3+} \rightarrow \mathrm{Mg}^{2+}+\mathrm{Cr}(s)\) (D) \(3 \mathrm{Mg}(s)+2 \mathrm{Cr}\left(\mathrm{NO}_{3}\right)_{3}(a q) \rightarrow 3 \mathrm{Mg}^{2+}(a q)+2 \mathrm{Cr}(s)+\mathrm{NO}_{3}^{-}(a q)\)

The first ionization energy for a neutral atom of chlorine is 1.25 \(\mathrm{MJ} / \mathrm{mol}\) and the first ionization energy for a neutral atom of argon is 1.52 \(\mathrm{MJ} / \mathrm{mol}\) How would the first ionization energy value for a neutral atom of potassium compare to those values? (A) It would be greater than both because potassium carries a greater nuclear charge then either chlorine or argon. (B) It would be greater than both because the size of a potassium atom is smaller than an atom of either chlorine or argon. (C) It would be less than both because there are more electrons in potassium, meaning they repel each other more effectively and less energy is needed to remove one. (D) It would be less than both because a valence electron of potassium is farther from the nucleus than one of either chlorine or argon.
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