Problem 208 - A Complete Resource Book in Chemistry for JEE Main

Question

Three sparingly soluble salts \(\mathrm{M}_{2} \mathrm{~B}, \mathrm{MB}\) and \(\mathrm{MB}_{3}\) have the same solubility product. Their solubilities will be in the order (a) \(\mathrm{MB}_{3}>\mathrm{M}_{2} \mathrm{~B}>\mathrm{MB}\) (b) \(\mathrm{MB}>\mathrm{M}_{2} \mathrm{~B}>\mathrm{MB}_{3}\) (c) \(\mathrm{MB}_{3}>\mathrm{MB}>\mathrm{M}_{2} \mathrm{~B}\) (d) \(\mathrm{MB}>\mathrm{MB}_{3}>\mathrm{M}_{2} \mathrm{~B}\)

Step-by-Step Solution

Answer

(b) \(\mathrm{MB} > \mathrm{M}_{2} \mathrm{~B} > \mathrm{MB}_{3}\).

Step 1: Understand the Solubility Product

The solubility product (Ksp) of a salt is a constant at a given temperature that indicates the product of the concentrations of the ions in a saturated solution. It reflects how much a salt can dissolve in water.

Step 2: Write the Ionization Equations

For each salt, write its ionization equation.- For \(\mathrm{M}_{2} \mathrm{~B}\), the equation is \(\mathrm{M}_{2} \mathrm{~B} \rightleftharpoons 2\mathrm{M}^{+} + \mathrm{B}^{2-}\).- For \(\mathrm{MB}\), the equation is \(\mathrm{MB} \rightleftharpoons \mathrm{M}^{+} + \mathrm{B}^{-}\).- For \(\mathrm{MB}_{3}\), the equation is \(\mathrm{MB}_{3} \rightleftharpoons \mathrm{M}^{3+} + 3\mathrm{B}^{-}\).

Step 3: Express Solubility in Terms of Ion Concentration

Solubility refers to the amount of salt that can dissolve to form a saturated solution. Let the solubility of \(\mathrm{MB}\) be \(s\). Thus, the solubility of salts can be expressed based on these equations:- For \(\mathrm{M}_{2} \mathrm{~B}\), if solubility is \(s\), concentration of ions is \([2s]^2 \cdot [s] = 4s^3\).- For \(\mathrm{MB}\), concentration of ions is \([s] [s] = s^2\).- For \(\mathrm{MB}_{3}\), concentration is \([s^3] [3s]^3 = 27s^4\).

Step 4: Compare Solubilities

Given the same \(\mathrm{K_{sp}}\) for all three salts, we have:- \(s_{MB}^2 = s^{2+}\)- \(4s_{M_{2}B}^3 = \mathrm{K_{sp}}\)- \(27s_{MB_3}^4 = \mathrm{K_{sp}}\)From the equations:- \(s_{M_{2}B} = \left(\frac{\mathrm{K_{sp}}}{4}\right)^{1/3}\)- \(s_{MB} = \sqrt{\mathrm{K_{sp}}}\)- \(s_{MB_3} = \left(\frac{\mathrm{K_{sp}}}{27}\right)^{1/4}\)Since \(s_{MB} > s_{M_{2}B} > s_{MB_3}\), the solubilities follow this order.

Key Concepts

Ion Concentration

Ion concentration is a fundamental concept in understanding how salts dissolve in water. It represents the number of ions present in a solution when a salt undergoes dissociation. Each type of ion has a specific charge, often represented as positive or negative depending on its electron gain or loss.
Understanding ion concentration is key to determining the solubility of different salts. For the salts in our example \(\text{M}_2 \text{B}, \text{MB}, \text{MB}_3\), the concentration of ions in a saturated solution helps calculate how much of each salt can dissolve before reaching equilibrium.

• For \(\text{M}_2 \text{B}\), this involves two \(\text{M}^+\) ions and one \(\text{B}^{2-}\) ion.
• In the case of \(\text{MB}\), it yields one \(\text{M}^+\) ion and one \(\text{B}^-\) ion in its ion concentration expression.
• While \(\text{MB}_3\) produces one \(\text{M}^{3+}\) ion and three \(\text{B}^-\) ions each contributing uniquely to the total concentration.

By calculating these concentrations, students can better understand the behavior of each solution and predict which salts form more ions; thus having a higher apparent solubility.

Solubility Order

The order of solubility denotes which salts dissolve more in comparison to others. Understanding this helps chemists rank different substances by their ability to ionize in water. The solubility order can depend heavily on the complexity of ion interaction and balance described by the solubility product, \(K_{sp}\).
In the exercise given, we observe how even if the solubility product is the same for different salts, the way each compound dissociates affects the solubility order. Here is how they compare:

• \(\text{MB}\), with the simplest dissociation into two parts, proves to have the greatest solubility.
• \(\text{M}_2 \text{B}\) falls in the middle category owing to its slightly more complex dissociation.
• Finally, \(\text{MB}_3\), with its ions forming in a more complex relationship, results in it being less soluble than the others.

Understanding solubility order helps predict how these compounds will behave in mixtures, and is crucial information when selecting salts for industrial or chemical processes.

Sparingly Soluble Salts

Sparingly soluble salts, also known as slightly soluble salts, are those that do not dissolve well in water. Unlike highly soluble salts, these salts reach a saturation point quickly and can be accurately described by their solubility product \(K_{sp}\).
While their low solubility may seem limiting, these salts are very important in fields such as environmental science, medicine, and geology, as they often participate in reactions that form precipitates.
Characteristics of sparingly soluble salts include:

• They form equilibrium quickly between dissolved ions and undissolved solid.
• A small increase in ion concentration can lead to precipitation.
• Their solubility is sensitive to environmental changes like temperature and pressure.

Through understanding their behavior, one can control reactions by manipulating external conditions to create or prevent precipitates, aiding in processes such as water treatment and metal extraction.

Ionization Equations

Ionization equations represent how a salt dissociates into its constituent ions when dissolved in water. Writing these equations accurately is crucial in predicting and calculating the solubility and ionic concentration of salts in a solution.
For the salts provided in our example, these equations offer insight into each substance's structure and dissociation pattern:

• **\(\text{M}_2 \text{B}\):** \(\text{M}_2 \text{B} \rightleftharpoons 2\text{M}^{+} + \text{B}^{2-}\) indicates double the concentration of \(\text{M}^{+}\) ions.
• **\(\text{MB}\):** \(\text{MB} \rightleftharpoons \text{M}^{+} + \text{B}^{-}\) shows a direct simple dissociation that is reflected in higher solubility.
• **\(\text{MB}_3\):** \(\text{MB}_3 \rightleftharpoons \text{M}^{3+} + 3\text{B}^{-}\) illustrates a complex separation that results in a lower solubility due to the greater formation of individual ions.

Learning to write and interpret ionization equations helps students understand how different ions contribute to a solution and assists in calculating equilibrium constants vital for scientific and industrial applications.