Question

Lead(II) chloride is most soluble in (a) \(0.100 \mathrm{M} \mathrm{NaCl}\); (b) \(0.100 \mathrm{Na}_{2} \mathrm{S}_{2} \mathrm{O}_{3} ;(\mathrm{c}) 0.100 \mathrm{M} \mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2} ;(\mathrm{d}) 0.100 \mathrm{M}\) \(\mathrm{NaNO}_{3} ;(\mathrm{e}) 0.100 \mathrm{MnSO}_{4}\).

Short answer

Lead(II) chloride is most soluble in either (b) \(0.100 \, \mathrm{M} \, \mathrm{Na}_{2}\mathrm{S}_{2}\mathrm{O}_{3}\), (d) \(0.100\, \mathrm{M} \, \mathrm{NaNO}_{3}\), or (e) \(0.100 \, \mathrm{M} \, \mathrm{MnSO}_{4}\)

Step-by-step solution

Identify the chemical formula of Lead(II) chloride

Lead(II) chloride is \( PbCl_2 \). So the ions present in this solution are \( Pb^{2+} \) and \( Cl^- \).

Identify common ions in the given options

Now, look at the ions present in each of the options provided. (a) \( NaCl \) has the common ion \( Cl^- \). (b) \( Na_2S_2O_3 \) does not have any common ions. (c) \( Pb(NO_3)_2 \) has the common ion \( Pb^{2+} \). (d) \( NaNO_3 \) does not have any common ions, and (e) \( MnSO_4 \) does not contain any common ions.

Apply the common ion effect

Applying the common ion effect, additional \( Cl^- \) ions from \( NaCl \) or \( Pb^{2+} \) ions from \( Pb(NO_3)_2 \) in the solution will shift the equilibrium of the dissolution of \( PbCl_2 \) to the left, according to Le Chatelier's principle. This effectively reduces the solubility of \( PbCl_2 \). Since the other options do not contain a common ion with \( PbCl_2 \), they will not affect the solubility.

Determine which salt is most soluble

Therefore, \( PbCl_2 \) is most soluble in a solution that does not contain any common ions. Either (b) \(Na_2S_2O_3\), (d) \(NaNO_3\), or (e) \(MnSO_4\) can be the answer. Without more information or context, any one of the three can be the answer.

Key concepts

Common Ion Effect

Understanding the common ion effect is crucial when delving into solubility and dissolution processes in chemistry. It is a phenomenon that occurs when a compound is dissolved in a solution which already contains one of the ions present in the compound. The presence of a common ion suppresses the dissolution of the compound, affecting its solubility.

For instance, the solubility of Lead(II) chloride, which dissociates into \( Pb^{2+} \) and \( Cl^- \) ions, is decreased when introduced into a solution containing either of these ions from another source, such as \( NaCl \) or \( Pb(NO_3)_2 \).

The principle behind this is firmly related to the dynamics of chemical equilibrium, which can be understood by exploring Le Chatelier's principle next.

Le Chatelier's Principle

To further comprehend the common ion effect, one must grasp Le Chatelier's principle. This principle states that if an external change is applied to a system at equilibrium, the system adjusts in such a way to counteract that change.

When additional common ions are added to an aqueous solution, the equilibrium shifts in order to reduce the disturbance, typically by favoring the formation of the insoluble compound, thus reducing its solubility in the solution.

Applying this to the Lead(II) chloride scenario, the introduction of extra \( Cl^- \) ions from \( NaCl \) or extra \( Pb^{2+} \) ions from \( Pb(NO_3)_2 \) will shift the dissolution equilibrium of \( PbCl_2 \) to form more of the solid salt, hence decreasing its solubility.

Chemical Equilibrium

Chemical equilibrium exists in a reversible reaction when the rates of the forward and reverse reactions are equal, resulting in no net change in the concentrations of reactants and products over time.

The dissolution of salts in water, such as \( PbCl_2 \), is an example of a dynamic equilibrium. This equilibrium can be perturbed by changes in concentration, temperature, or pressure. In the context of our exercise, adding a common ion to the solution shifts the equilibrium position according to Le Chatelier's principle, influencing solubility.

Molarity

Molarity is a way to express concentration, defined as the number of moles of a solute present in one liter of solution. Represented with the unit \( M \), it is a common unit used in chemistry to quantify the strength of a solution.

For instance, a \( 0.100 \ M \) solution of \( NaCl \) indicates there is 0.100 moles of sodium chloride dissolved in one liter of solution. Molarity impacts various properties of solutions, including boiling and freezing points, vapor pressure, and in the context of our discussion, the solubility of substances.

Aqueous Solutions

Aqueous solutions are mixtures where water serves as the solvent. The ability of substances to dissolve in water depends on both the nature of the solute and solvent and the interactions between them.

Water's capacity for solvating ions results from its polarity, which allows it to stabilize ionic compounds in solution. In the case of solubility exercises, understanding how various compounds behave in water is vital for predicting and explaining outcomes, such as in the scenario of solubility with the influence of the common ion effect.