Problem 14 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

The minimum volume of water required to dissolve \(0.1 \mathrm{~g}\) lead (II) chloride to get a saturated solution \(\left(K_{\mathrm{Sp}}\right.\) of \(\mathrm{PbCl}_{2}=3.2 \times 10^{-8}\); atomic mass of \(\mathrm{Pb}=207 \mathrm{u}\) ) is : (a) \(1.798 \mathrm{~L}\) (b) \(0.36 \mathrm{~L}\) (c) \(17.95 \mathrm{~L}\) (d) \(0.18 \mathrm{~L}\)

Step-by-Step Solution

Answer

The minimum volume of water required is 0.18 L (option d).

Step 1: Determine Molar Mass of PbCl₂

The molecular weight of PbCl₂ is calculated as follows: a. Atomic mass of Lead (Pb) = 207 u b. Atomic mass of Chlorine (Cl) = 35.5 u Thus, the molar mass of PbCl₂ = 207 + 2(35.5) = 278 g/mol.

Step 2: Find Moles of Lead (II) Chloride

Given 0.1 g of PbCl₂, we find the number of moles using the formula:\[ \text{Moles of PbCl}_2 = \frac{\text{mass}}{\text{molar mass}} = \frac{0.1}{278} \approx 0.00036 \text{ moles} \]

Step 3: Calculate Solubility from Solubility Product

For PbCl₂, the dissociation in water is:\[ \text{PbCl}_2 (s) \rightleftharpoons \text{Pb}^{2+} (aq) + 2\text{Cl}^- (aq) \]The solubility product \(K_{sp}\) is expressed as:\[ K_{sp} = [\text{Pb}^{2+}][\text{Cl}^-]^2 = s \cdot (2s)^2 = 4s^3 \]Given \(K_{sp} = 3.2 \times 10^{-8}\), solve for \(s\), the molarity of a saturated solution:\[ 4s^3 = 3.2 \times 10^{-8} \]

Step 4: Solve for Solubility (s)

We solve the equation \(4s^3 = 3.2 \times 10^{-8}\) by dividing through by 4:\[ s^3 = \frac{3.2 \times 10^{-8}}{4} = 0.8 \times 10^{-8} \]Take the cube root to find \(s\):\[ s = \sqrt[3]{0.8 \times 10^{-8}} \approx 2 \times 10^{-3} \text{ moles/L} \]

Step 5: Determine the Minimum Volume of Water

Using the solubility \(s = 2 \times 10^{-3} \, \text{moles/L}\), calculate the volume needed to dissolve the moles of PbCl₂:\[ \text{Volume} = \frac{\text{Moles of PbCl}_2}{s} = \frac{0.00036}{2 \times 10^{-3}} = 0.18 \text{ L} \]

Step 6: Conclusion: Minimum Volume Required

The minimum volume of water required to dissolve 0.1 g of PbCl₂ to form a saturated solution is therefore \(0.18 \, \text{L}\). Hence, option (d) is correct.

Key Concepts

Molar Mass Calculation

To successfully tackle problems involving solubility and solution calculations, understanding the concept of molar mass is essential. The molar mass of a compound is the sum of the atomic masses of all the atoms in its molecular formula.

In the case of Lead (II) Chloride, denoted as \( \mathrm{PbCl}_2 \), we need to sum the atomic masses of one Lead (Pb) atom and two Chlorine (Cl) atoms:

• The atomic mass of Lead is 207 u.
• The atomic mass of Chlorine is 35.5 u, and since there are two Chlorine atoms, we multiply by 2.

So, the molar mass of \( \mathrm{PbCl}_2 \) is calculated as:\[207 \,\text{u} + 2 \times 35.5 \,\text{u} = 278 \,\text{g/mol}\]This molar mass allows us to convert from mass to moles, which is a key step in determining how much of a substance can be dissolved in a given volume of solvent.

Dissolution Process

The dissolution process is the interaction between a solute and solvent resulting in a solution. For \( \mathrm{PbCl}_2 \), the process involves dissociation into ions when dissolved in water.

The balanced chemical equation for the dissolution of Lead (II) Chloride in water is:\[\mathrm{PbCl}_2 (s) \rightleftharpoons \mathrm{Pb}^{2+} (aq) + 2\mathrm{Cl}^- (aq)\]Initially, \( \mathrm{PbCl}_2 \) is a solid. Upon dissolution, it separates into one lead ion \( \mathrm{Pb}^{2+} \) and two chloride ions \( \mathrm{Cl}^- \) per formula unit.

This release of ions into the solution is critical to achieving saturation. The overall concentration of ions in a saturated solution is determined by the solubility of the compound. Understanding this helps in calculating the exact amount of solvent needed to achieve a saturated solution.

Equilibrium Expressions

When a salt like \( \mathrm{PbCl}_2 \) dissolves in water, an equilibrium is established between the dissolved ions and the undissolved solid. The solubility product constant, \( K_{sp} \), helps quantify this equilibrium.

The expression for the solubility product of \( \mathrm{PbCl}_2 \) is given by:\[K_{sp} = [\mathrm{Pb}^{2+}][\mathrm{Cl}^-]^2\]For \( \mathrm{PbCl}_2 \), we represent its solubility as \( s \). Thus:

• The concentration of \( \mathrm{Pb}^{2+} \) is \( s \).
• The concentration of \( \mathrm{Cl}^- \) is \( 2s \) since there are two chloride ions per formula unit.

Plugging these into the equilibrium expression, we have:\[K_{sp} = s \times (2s)^2 = 4s^3\]To find \( s \), rearrange and solve for \( s \) when \( K_{sp} = 3.2 \times 10^{-8} \). Understanding these expressions is pivotal in solving for the solubility and predicting the amount of a solvent required for dissolution.