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

Question

Solubility product \(\left(\mathrm{K}_{s}\right)\) of salts \(\mathrm{mx}, \mathrm{mx}_{2}\) and \(\mathrm{mx}_{3}\) 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) \(m x_{2}>m_{3} x>m x\) (d) \(\mathrm{mx}>\mathrm{m}_{3} \mathrm{x}>\mathrm{m} \mathrm{x}_{2}\)

Step-by-Step Solution

Answer

The solubilities decrease in the order: MX > MX_3 > MX_2 (Option d).

Step 1: Understanding solubility product expressions

The solubility product, \(K_s\), is the equilibrium constant for the dissolution of a sparingly soluble salt. For salt of formula \(MX\), \(K_s = s^2\) where \(s\) is the solubility. For \(MX_2\), \(K_s = 4s^3\), and for \(MX_3\), \(K_s = 27s^4\). Here \(s\) is the molar solubility of the salt.

Step 2: Calculating solubility for MX

Given \(K_s (MX) = 4 \times 10^{-8}\), and using the formula \(K_s = s^2\), we solve for \(s\) to find the solubility. \(s = \sqrt{4 \times 10^{-8}} = 2 \times 10^{-4} \text{ M}\).

Step 3: Calculating solubility for MX_2

Given \(K_s (MX_2) = 3.2 \times 10^{-14}\), and using the formula \(K_s = 4s^3\), solve for \(s\) by rearranging to get \(s = \left( \frac{3.2 \times 10^{-14}}{4} \right)^{1/3} = \left(8 \times 10^{-15}\right)^{1/3} \approx 2 \times 10^{-5} \text{ M}\).

Step 4: Calculating solubility for MX_3

Given \(K_s (MX_3) = 2.7 \times 10^{-15}\), using the formula \(K_s = 27s^4\), solve for \(s\): \(s = \left( \frac{2.7 \times 10^{-15}}{27} \right)^{1/4} = \left(1 \times 10^{-16}\right)^{1/4} = 10^{-4} \text{ M}\).

Step 5: Comparing solubilities

The solubilities are \(MX = 2 \times 10^{-4} \text{ M}\), \(MX_2 = 2 \times 10^{-5} \text{ M}\), and \(MX_3 = 10^{-4} \text{ M}\). Comparing these values, the order is \(MX > MX_3 > MX_2\).

Key Concepts

Sparingly Soluble Salts

Sparingly soluble salts are compounds that do not dissolve readily in water. Instead, they establish an equilibrium between the solid salt and its dissolved ions. This means that only a small amount of these salts will typically dissolve in water, creating a saturated solution.
When these salts dissolve, they dissociate into their respective ions in a reversible reaction:

• For a salt with formula \( MX \), it dissociates as \( MX \rightleftharpoons M^+ + X^- \).
• For \( MX_2 \), the dissociation is \( MX_2 \rightleftharpoons M^{2+} + 2X^- \).
• For \( MX_3 \), \( MX_3 \rightleftharpoons M^{3+} + 3X^- \).

The amount that dissolves is often so low it's referred to as 'sparingly soluble'. This behaviour is important when calculating the solubility product, which is our next focus.

Molar Solubility

Molar solubility is the amount of a solute that can dissolve in a liter of water to form a saturated solution. It is typically expressed in moles per liter (M). For sparingly soluble salts, the molar solubility is crucial to determining how much of the salt will actually dissolve.
The calculations for molar solubility differ depending on the salt's formula:

• For \( MX \), the solubility product \( K_s = s^2 \), where \( s \) is the molar solubility.
• For \( MX_2 \), the expression becomes \( K_s = 4s^3 \).
• For \( MX_3 \), it is \( K_s = 27s^4 \).

By plugging in the \( K_s \) values and solving these equations, you can find the molar solubility of each salt.
For instance, given \( K_s (MX) = 4 \times 10^{-8} \), solving \( s^2 = 4 \times 10^{-8} \) gives us \( s \approx 2 \times 10^{-4} \text{ M} \). Similar calculations are made for \( MX_2 \) and \( MX_3 \), resulting in their respective solubilities.

Equilibrium Constant

The equilibrium constant is a crucial concept in chemistry that quantifies the extent of a chemical reaction at equilibrium. For sparingly soluble salts, this constant is known as the solubility product \( K_s \).
The solubility product represents the product of the molar concentrations of the ions, raised to the power of their stoichiometric coefficients from the balanced dissolution equation. It's a specific type of equilibrium constant that applies to the dissolution of salts.

• For \( MX \), \( K_s \) is calculated as \([M^+][X^-] = s^2 \).
• For \( MX_2 \), it becomes \([M^{2+}](2[X^-])^2 = 4s^3 \).
• For \( MX_3 \), this evolves into \([M^{3+}](3[X^-])^3 = 27s^4 \).

Understanding the solubility product helps predict how much salt can dissolve before reaching saturation, allowing chemists to assess solubility trends under different conditions.