Problem 58 - A Complete Resource Book in Chemistry for JEE Main

Question

Solubility product \(\left(\mathrm{K}_{s}\right)\) of salts \(\mathrm{mx}, \mathrm{m} \mathrm{x}_{2}\) and \(\mathrm{m} \mathrm{x}_{3}\) at T temperature are \(4 \times 10^{-8}, 3.2 \times 10^{-14}\) and \(2.7 \times 10^{-15}\) respectively. The solubilities of these salts are in the decreasing order? (a) \(\mathrm{mx}>\mathrm{mx}_{2}>\mathrm{m}_{3} \mathrm{x}\) (b) \(m_{3} x>m x_{2}>m x\) (c) \(\mathrm{mx}_{2}>\mathrm{m}_{3} \mathrm{x}>\mathrm{mx}\) (d) \(m x>m_{3} x>m x_{2}\)

Step-by-Step Solution

Answer

The solubilities decrease in the order: \(mx > mx_3 > mx_2\).

Step 1: Understanding Solubility Product

The solubility product (\(K_s\)) for a salt \(mx\) is related to its solubility \(s\) as \(K_s = s^2\) for binary salts, \(K_s = 4s^3\) for ternary salts like \(mx_2\), and \(K_s = 27s^4\) for quaternary salts like \(mx_3\). This relationship changes based on the stoichiometry of the salt.

Step 2: Calculate Solubility for Binary Salt \(mx\)

For the salt \(mx\), the solubility \(s\) is given by \(s^2 = 4 \times 10^{-8}\). Therefore, \(s = \sqrt{4 \times 10^{-8}} = 2 \times 10^{-4}\).

Step 3: Calculate Solubility for Ternary Salt \(mx_2\)

For the salt \(mx_2\), the solubility \(s\) satisfies \(4s^3 = 3.2 \times 10^{-14}\). Solving gives \(s^3 = 0.8 \times 10^{-14}\), hence, \(s = \sqrt[3]{0.8 \times 10^{-14}} \approx 2 \times 10^{-5}\).

Step 4: Calculate Solubility for Quaternary Salt \(mx_3\)

For the salt \(mx_3\), the solubility \(s\) satisfies \(27s^4 = 2.7 \times 10^{-15}\). Solving gives \(s^4 = 10^{-16}\), hence, \(s = \sqrt[4]{10^{-16}} = 10^{-4}\).

Step 5: Compare Solubilities

The calculated solubilities are: \(mx = 2 \times 10^{-4}\), \(mx_2 = 2 \times 10^{-5}\), and \(mx_3 = 10^{-4}\). The order of decreasing solubility is \(mx > mx_3 > mx_2\).

Key Concepts

Solubility Calculation

Solubility calculations help us understand how much of a substance can dissolve in a solvent at a given temperature. This involves using the solubility product constant \(K_s\). In essence, \(K_s\) is an equilibrium constant that applies to the dissolving process of salts and indicates product concentration levels. For each salt, the relationship between \(K_s\) and its solubility depends on the salt's stoichiometry.

For binary salts like \(mx\), this relationship is expressed as \(K_s = s^2\). For ternary salts such as \(mx_2\), it becomes \(K_s = 4s^3\). For quaternary salts like \(mx_3\), the formula used is \(K_s = 27s^4\). These equations help in calculating the solubility \(s\) from the given \(K_s\) value.

Knowing how to manipulate and calculate with these formulas is crucial in chemistry. The solubility values computed from these calculations give insights into how much the compounds dissolve under specific conditions, ultimately guiding us in reactions involving these substances.

Chemical Equilibria

Chemical equilibria refer to the state in which the concentrations of reactants and products do not change over time. This occurs because the forward and reverse reactions proceed at equal rates. In the context of solubility, equilibria help define the point where a solution can no longer dissolve additional solute. At this point, the solution reaches its solubility product\(K_s\).

During equilibrium, the dynamic state allows ongoing reactions at equal rates resulting in stable concentrations:

• Forward reaction: Dissolution of salt forming ions in solution.
• Reverse reaction: Formation of solid salt from ions in solution.

Understanding equilibria is essential when calculating solubility, as it explains why saturated solutions sustain a balance of solid and dissolved particles.

By grasping these concepts, students can predict and calculate outcomes in various chemical processes, a skill critical when dealing with intricate chemical systems.

Precipitation Reactions

Precipitation reactions occur when cations and anions in an aqueous solution combine to form an insoluble solid called a precipitate. These reactions are highly dependent on solubility rules, and the solubility product \(K_s\) is what determines if precipitation will occur.

If the product of ionic concentrations in the solution exceeds the \(K_s\), the solution becomes supersaturated, resulting in precipitation. However, if ionic concentration products are less than \(K_s\), the ions remain dissolved, and no precipitate forms.

• Supersaturated solutions lead to precipitation forming a solid.
• Unsaturated or saturated solutions allow ions to remain dissolved.

By understanding these principles, students can assess scenarios where a change in concentration leads to precipitation.

Overall, mastering these concepts helps one predict the formation of precipitates, essential for experiments and applications in laboratory settings.