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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

Questions 32-36 refer to the following. Two half-cells are set up as follows: Half-Cell A: Strip of \(\mathrm{Cu}(s)\) in \(\mathrm{CuNO}_{3}(a q)\) Half-Cell B: Strip of \(\mathrm{Zn}(s)\) in \(\mathrm{Zn}\left(\mathrm{NO}_{3}\right)_{2}\) (aq) When the cells are connected according to the diagram below, the following reaction occurs: GRAPH CAN'T COPY $$2 \mathrm{Cu}^{+}(a q)+\mathrm{Zn}(s) \rightarrow 2 \mathrm{Cu}(s)+\mathrm{Zn}^{2+}(a q) E^{\circ}=+1.28 \mathrm{V}$$ As the reaction progresses, what will happen to the overall voltage of the cell? (A) It will increase as \(\left[\mathrm{Zn}^{2+}\right]\) increases. (B) It will increase as \(\left[\mathrm{Cu}^{+}\right]\) increases. (C) It will decrease as \(\left[\mathrm{Zn}^{2+}\right]\) increases. (D) The voltage will remain constant.

A student titrates 20.0 \(\mathrm{mL}\) of 1.0 \(M \mathrm{NaOH}\) with 2.0 \(\mathrm{M}\), \(\mathrm{HCO}_{2} \mathrm{H}\left(K_{\mathrm{a}}=1.8 \times 10^{-4}\right) .\) Formic acid is a monoprotic acid. If the formic acid were replaced with a strong acid such as HCl at the same concentration (2.0 M), how would that change the volume needed to reach the equivalence point? (A) The change would reduce the amount as the acid now fully dissociates. (B) The change would reduce the amount because the base will be more strongly attracted to the acid. (C) The change would increase the amount because the reaction will now go to completion instead of equilibrium. (D) Changing the strength of the acid will not change the volume needed to reach equivalence.

Which gas has the strongest IMFs? (A) He (B) Ne (C) NO (D) All gases have identical IMFs.

A gaseous mixture at \(25^{\circ} \mathrm{C}\) contained 1 mole of \(\mathrm{CH}_{4}\) and 2 moles of \(\mathrm{O}_{2}\) and the pressure was measured at 2 \(\mathrm{atm}\) . The gases then underwent the reaction shown below. \(\mathrm{CH}_{4}(g)+2 \mathrm{O}_{2}(g) \rightarrow \mathrm{CO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g)\) What was the pressure in the container after the reaction had gone to completion and the temperature was allowed to return to \(25^{\circ} \mathrm{C} ?\) (A) 1 atm (B) 2 \(\mathrm{atm}\) (C) 3 \(\mathrm{atm}\) (D) 4 \(\mathrm{atm}\)

\(\mathrm{H}_{2} \mathrm{O}(l) \rightarrow \mathrm{H}_{2} \mathrm{O}(s)\) Which of the following is true for the above reaction? (A) The value for \(\Delta S\) is positive. (B) The value for \(\Delta G\) is zero. (C) The value for \(\Delta H\) is positive. (D) The reaction is favored at 1.0 \(\mathrm{atm}\) and 298 \(\mathrm{K}\) .
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