Problem 78 - A Complete Resource Book in Chemistry for JEE Main

Question

Solubility of a \(\mathrm{M}_{2} \mathrm{~S}\) salt is \(3.5 \times 10^{-6}\) then find out solubility product. (a) \(1.7 \times 10^{-6}\) (b) \(1.7 \times 10^{-16}\) (c) \(1.7 \times 10^{-18}\) (d) \(1.7 \times 10^{-12}\)

Step-by-Step Solution

Answer

The solubility product \( K_{sp} \) is \( 1.7 \times 10^{-16} \), so the answer is (b).

Step 1: Understanding the Problem

We are given the solubility of a salt \( \text{M}_2\text{S} \), which dissolves in water to form its ions. The solubility of \( \text{M}_2\text{S} \) is given as \( 3.5 \times 10^{-6} \) M. We need to find the solubility product constant \( K_{sp} \) for this salt.

Step 2: Writing the Dissociation Equation

The dissociation of \( \text{M}_2\text{S} \) in water can be represented by the chemical equation: \( \text{M}_2\text{S} \rightarrow 2\text{M}^+ + \text{S}^{2-} \). This tells us that one mole of \( \text{M}_2\text{S} \) produces two moles of \( \text{M}^+ \) ions and one mole of \( \text{S}^{2-} \) ions.

Step 3: Relating Solubility to Ion Concentrations

Given the solubility \( S = 3.5 \times 10^{-6} \) M, the concentration of \( \text{S}^{2-} \) ions in solution will also be \( S \), which is \( 3.5 \times 10^{-6} \) M. The concentration of \( \text{M}^+ \) ions will be twice this amount, \( 2S = 7.0 \times 10^{-6} \) M.

Step 4: Calculating the Solubility Product \( K_{sp} \)

The solubility product \( K_{sp} \) of \( \text{M}_2\text{S} \) can be calculated using the expression \( K_{sp} = [\text{M}^+]^2 [\text{S}^{2-}] \). Substitute the values: \( K_{sp} = (7.0 \times 10^{-6})^2 (3.5 \times 10^{-6}) \).

Step 5: Simplifying the Expression

Calculate the expression: \( K_{sp} = 49.0 \times 10^{-12} \times 3.5 \times 10^{-6} \). This equals \( 171.5 \times 10^{-18} \). Simplifying further, we get, \( K_{sp} = 1.715 \times 10^{-16} \).

Step 6: Choosing the Correct Answer

Considering significant figures, \( K_{sp} = 1.7 \times 10^{-16} \). This matches option (b).

Key Concepts

Dissociation Equation of Salts

To understand the solubility product of a salt like \ \(\text{M}_2\text{S}\), it's crucial to first grasp the dissociation equation. This equation illustrates how the salt breaks down into individual ions when it dissolves in water.
When \ \(\text{M}_2\text{S}\) dissolves, it splits into cations and anions. The chemical dissociation equation is given by: \ \(\text{M}_2\text{S}\rightarrow 2\text{M}^+ + \text{S}^{2-}\). This means:

• One mole of \ \(\text{M}_2\text{S}\) produces two moles of cations \ \(\text{M}^+\).
• It also produces one mole of anions \ \(\text{S}^{2-}\).

This separation is essential as it shows the relationship between the solid salt and the ions in solution.
Recognizing how the dissociation creates ions helps in analyzing their concentrations and the eventual solubility product calculation.

Ion Concentrations in Solution

Once the dissociation equation tells us the formation of ions, we need to determine how these are concentrated in the solution. It's directly connected to the solubility of the original compound.
Given the solubility of \ \(\text{M}_2\text{S}\) as \ \(3.5 \times 10^{-6}\) M, the concentration of ions can be determined as follows:

• The concentration of \ \(\text{S}^{2-}\) ions equals the solubility, thus it is \ \(3.5 \times 10^{-6}\) M.
• For the \ \(\text{M}^+\) ions, since two cations form for every mole of salt, the concentration is twice the solubility: \ \(2 \times 3.5 \times 10^{-6} = 7.0 \times 10^{-6}\) M.

Understanding these concentrations is vital. They represent how many ions are freely moving in the solution, ultimately affecting other calculations related to chemical equilibrium and solubility product.

Solubility Constant Calculation

The calculation of the solubility product constant, \ \(K_{sp}\), is the next step. This constant helps predict how much of the salt can dissolve in a solution.
We calculate \ \(K_{sp}\) by considering the equilibrium reached as the solid dissociates into its ions:
Use the formula: \ \(K_{sp} = [\text{M}^+]^2 [\text{S}^{2-}]\). Substituting the known ion concentrations:

• Put \ \([\text{M}^+] = 7.0 \times 10^{-6}\)
• And \ \([\text{S}^{2-}] = 3.5 \times 10^{-6}\)

The calculation becomes:
\ \(K_{sp} = (7.0 \times 10^{-6})^2 \times 3.5 \times 10^{-6}\)
This simplifies to \ \(1.71 \times 10^{-16}\), confirming the salt's specific solubility capacity. The solubility constant is a powerful tool in chemistry, as it helps understand and predict solubility behaviors in various solutions.