Question

Let the solubilities of \(\mathrm{AgCl}\) in \(\mathrm{H}_{2} \mathrm{O}, 0.01 \mathrm{M} \mathrm{CaCl}_{2}\); \(0.01 \mathrm{M} \mathrm{NaCl}\) and \(0.05 \mathrm{M} \mathrm{AgNO}_{3}\) be \(\mathrm{S}_{1}, \mathrm{~S}_{2}, \mathrm{~S}_{3}\) and \(\mathrm{S}_{4}\) respectively. What is the correct relationship between these quantities? (1) \(S_{1}>S_{2}>S_{3}>S_{4}\) (2) \(S_{1}>S_{2}=S_{3}>S_{4}\) (3) \(S_{1}>S_{3}>S_{2}>S_{4}\) (4) \(S_{4}>S_{2}>S_{3}>S_{1}\)

Short answer

The correct relationship is: \( S_1 > S_2 = S_3 > S_4 \).

Step-by-step solution

- Identify the Solubility Product (Ksp) of AgCl

Identify the solubility product constant, Ksp, of AgCl. AgCl dissociates as \(\mathrm{AgCl} \leftrightarrow \mathrm{Ag}^{+} + \mathrm{Cl}^{-}\). The Ksp is given by \(K_{sp} = [\mathrm{Ag}^{+}][\mathrm{Cl}^{-}]\).

- Solubility of AgCl in Pure Water (S1)

For \(\mathrm{AgCl}\) in pure water, consider that there are no common ion effects. Assuming 'S' is the solubility, we have \( [\mathrm{Ag}^{+}] = S_1 \) and \( [\mathrm{Cl}^{-}] = S_1 \).

- Solubility of AgCl in 0.01 M CaCl2 (S2)

Add a common ion effect. \(\mathrm{CaCl}_2\) dissociates into \(\mathrm{Ca}^{2+}\) and \(2\mathrm{Cl}^{-}\). The presence of extra \(\mathrm{Cl}^{-}\) decreases the solubility of \(\mathrm{AgCl}\). So, \(S_2\) will be less than \(S_1\).

- Solubility of AgCl in 0.01 M NaCl (S3)

Again, a common ion effect is present. \(\mathrm{NaCl}\) dissociates into \(\mathrm{Na}^{+}\) and \(\mathrm{Cl}^{-}\), adding more \(\mathrm{Cl}^{-}\) to the solution. Thus, \(S_3\) will be less than \(S_1\). The solubility of \(\mathrm{AgCl}\) will be similarly affected as with \(\mathrm{CaCl}_2\), hence \(S_3\approx S_2\).

- Solubility of AgCl in 0.05 M AgNO3 (S4)

Add a common ion effect of \(\mathrm{Ag}^{+}\). \(\mathrm{AgNO}_3\) dissociates into \(\mathrm{Ag}^{+}\) and \(\mathrm{NO}_3^{-}\). The presence of extra \(\mathrm{Ag}^{+}\) decreases the solubility of \(\mathrm{AgCl}\) significantly. Therefore, \(S_4\) will be much less than \(S_1\).

- Conclusion

From these observations, we can see that \(S_1\) will be greater than \(S_2\approx S_3\) which will be greater than \(S_4\).

Key concepts

Solubility Product Constant (Ksp)

The Solubility Product Constant, commonly abbreviated as Ksp, is a key concept in understanding the solubility of sparingly soluble salts. For a salt like \(\text{AgCl}\), it dissociates in water as:
\[ \text{AgCl} \rightleftharpoons \text{Ag}^{+} + \text{Cl}^{-} \] The Ksp expression for \(\text{AgCl}\) is given by:
\[ K_{sp} = [\text{Ag}^{+}][\text{Cl}^{-}] \] Here, \[ [\text{Ag}^{+}] \] and \[ [\text{Cl}^{-}] \] represent the molar concentrations of the respective ions in a saturated solution at equilibrium. The Ksp value is fixed at a given temperature and indicates how much of the salt can dissolve to form a saturated solution.

Common Ion Effect

The Common Ion Effect refers to the decrease in solubility of an ionic compound when a common ion is added to the solution. For example:

• When \(\text{CaCl}_2\) or \(\text{NaCl}\) is added to a solution containing \(\text{AgCl}\), both dissociate to release \(\text{Cl}^{-}\) ions.
• This increase in \(\text{Cl}^{-}\) ion concentration shifts the equilibrium to the left, reducing the solubility of \(\text{AgCl}\).

Therefore, the solubilities \(\text{S}_2\) and \(\text{S}_3\) are less than \(\text{S}_1\), the solubility of \(\text{AgCl}\) in pure water.

Chemical Equilibrium

Chemical Equilibrium occurs when the rates of the forward and reverse reactions are equal, resulting in constant concentrations of reactants and products.
In the context of solubility, the equilibrium involving \(\text{AgCl}\) is described by the dissociation reaction:
\[ \text{AgCl} \rightleftharpoons \text{Ag}^{+} + \text{Cl}^{-} \] The equilibrium expression derived from this reaction reflects the product of the ion concentrations raised to the power of their coefficients in the balanced chemical equation.

Dissociation Reactions

Dissociation Reactions involve the separation of an ionic compound into its individual ions in solution. For instance:

• \(\text{AgCl}\) dissociates as \[ \text{AgCl} \rightleftharpoons \text{Ag}^{+} + \text{Cl}^{-} \]
• Similarly, \(\text{CaCl}_2\) dissociates into \(\text{Ca}^{2+}\) and \(\text{Cl}^{-}\) ions as \[ \text{CaCl}_2 \rightleftharpoons \text{Ca}^{2+} + 2\text{Cl}^{-} \]

The introduction of additional ions from these dissociations can affect the equilibrium and solubility of other compounds, demonstrating the impact of the common ion effect.

Molar Solubility

Molar Solubility is defined as the number of moles of a solute that can dissolve per liter of solution to form a saturated solution. It directly relates to the Ksp value for a given compound.
If the molar solubility of \(\text{AgCl}\) in pure water is \(\text{S}_1\), then:
\[ K_{sp} = \text{S}_1^2 \] When common ions are added, such as \(\text{Cl}^{-}\) ions from \(\text{CaCl}_2\) or \(\text{NaCl}\), the molar solubility decreases. This effect is due to the shift in equilibrium, aligning with Le Chatelier’s principle, thereby reducing the amount of \(\text{AgCl}\) that can dissolve in the presence of common ions.