Problem 11 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

If solubility product of \(\mathrm{Zr}_{3}\left(\mathrm{PO}_{4}\right)_{4}\) is denoted by \(\mathrm{K}_{s p}\) and its molar solubility is denoted by \(\mathrm{S}\), then which of the following relation between \(\mathrm{S}\) and \(\mathrm{K}\) is correct? (a) \(\mathrm{S}=\left(\frac{\mathrm{K}_{s p}}{144}\right)^{1 / 6}\) (b) \(\mathrm{S}=\left(\frac{\mathrm{K}_{s p}}{6912}\right)^{1 / 7}\) (c) \(\mathrm{S}=\left(\frac{\mathrm{K}_{s p}}{929}\right)^{1 / 9}\) (d) \(\mathrm{S}=\left(\frac{\mathrm{K}_{s p}}{216}\right)^{1 / 7}\)

Step-by-Step Solution

Answer

The correct relation is (b) \( S = \left( \frac{\mathrm{K}_{sp}}{6912} \right)^{1/7} \)."

Step 1: Write the Dissociation Equation

The dissociation of \( \mathrm{Zr}_3(\mathrm{PO}_4)_4 \) in water can be represented as: \[ \mathrm{Zr}_3( \mathrm{PO}_4 )_4 \rightleftharpoons 3 \mathrm{Zr}^{4+} + 4 \mathrm{PO}_4^{3-} \] This reaction shows that one mole of \( \mathrm{Zr}_3(\mathrm{PO}_4)_4 \) produces 3 moles of \( \mathrm{Zr}^{4+} \) and 4 moles of \( \mathrm{PO}_4^{3-} \).

Step 2: Express Molar Solubility (S) in Terms of Ionic Concentrations

If the molar solubility of \( \mathrm{Zr}_3(\mathrm{PO}_4)_4 \) is \( S \), then the concentration of \( \mathrm{Zr}^{4+} \) ions is \( 3S \) and the concentration of \( \mathrm{PO}_4^{3-} \) ions is \( 4S \).

Step 3: Write the Expression for Solubility Product Constant \(\mathrm{K}_{sp}\)

The solubility product, \( \mathrm{K}_{sp} \), can be expressed as: \[ \mathrm{K}_{sp} = [\mathrm{Zr}^{4+}]^3 \cdot [\mathrm{PO}_4^{3-}]^4 \] Substituting the expressions from Step 2, we have: \[ \mathrm{K}_{sp} = (3S)^3 \cdot (4S)^4 \] This simplifies to \( \mathrm{K}_{sp} = 27S^3 \cdot 256S^4 \).

Step 4: Simplify the Expression for \(\mathrm{K}_{sp}\)

We now simplify the expression: \[ \mathrm{K}_{sp} = 27 \times 256 \times S^{3+4} = 6912 S^7 \]

Step 5: Solve for Molar Solubility (S) in Terms of \(\mathrm{K}_{sp}\)

Rearrange the expression from Step 4 to solve for \( S \): \[ S^7 = \frac{\mathrm{K}_{sp}}{6912} \] Thus, \( S = \left( \frac{\mathrm{K}_{sp}}{6912} \right)^{1/7} \).

Key Concepts

Molar Solubility

Molar solubility refers to the number of moles of a solute that can dissolve in a liter of solution before the solution becomes saturated. It essentially quantifies the amount of substance that can be dissolved. In the context of the exercise, the molar solubility of \( \mathrm{Zr}_3(\mathrm{PO}_4)_4 \) is denoted by \( S \). The equation given relates solubility to the solubility product, allowing us to solve for \( S \) using the solubility product constant, \( \mathrm{K}_{sp} \). Molar solubility is crucial for understanding how substances dissolve and precipitate, and it provides insights into the equilibrium state of a solution.

Dissociation Equation

The dissociation equation is a chemical expression that shows how a compound separates into its ions in a solution. For \( \mathrm{Zr}_3(\mathrm{PO}_4)_4 \), the dissociation equation is: \[ \mathrm{Zr}_3( \mathrm{PO}_4 )_4 \rightleftharpoons 3 \mathrm{Zr}^{4+} + 4 \mathrm{PO}_4^{3-} \]. This indicates that one formula unit of the compound breaks down into three zirconium ions (\( \mathrm{Zr}^{4+} \)) and four phosphate ions (\( \mathrm{PO}_4^{3-} \)). The dissociation equation is critical as it helps us understand the ratio of ions formed, which directly impacts the calculation of ionic concentrations and the solubility product constant.

Ionic Concentration

Ionic concentration refers to the amount of individual ions present in a solution. It is essential for determining how much of each ion is available once the compound has dissociated. From the dissociation of \( \mathrm{Zr}_3(\mathrm{PO}_4)_4 \), the concentration of \( \mathrm{Zr}^{4+} \) is \( 3S \) and for \( \mathrm{PO}_4^{3-} \) is \( 4S \), where \( S \) is the molar solubility. This is because for every mole of \( \mathrm{Zr}_3(\mathrm{PO}_4)_4 \) that dissolves, three moles of \( \mathrm{Zr}^{4+} \) and four moles of \( \mathrm{PO}_4^{3-} \) are produced. Properly understanding the ionic concentration helps in measuring the solubility product accurately.

Solubility Product Constant

The solubility product constant, \( \mathrm{K}_{sp} \), is a special equilibrium constant related to the solubility of a sparingly soluble compound. It quantitatively represents the product of the ionic concentrations, each raised to the power of their coefficients in the dissociation equation. For \( \mathrm{Zr}_3(\mathrm{PO}_4)_4 \), \( \mathrm{K}_{sp} \) is defined as: \[ \mathrm{K}_{sp} = [\mathrm{Zr}^{4+}]^3 \cdot [\mathrm{PO}_4^{3-}]^4 \]. Substituting the ionic concentrations \((3S)^3 \) and \((4S)^4 \) into the equation, we arrived at \( \mathrm{K}_{sp} = 6912 S^7 \). This expression is used to solve for the molar solubility, making \( \mathrm{K}_{sp} \) an essential parameter for predictive calculations about how much of a compound can dissolve under given conditions.